3.4: More Trigonometric Equations

When the solution to a trigonometric equation is one of the quadrantal angles (left(0^{circ}, 90^{circ}, 180^{circ}, 270^{circ} ext { and } mathrm{so} ext { on } ight),) then determining all the solutions between (0^{circ}) and (360^{circ}) can work a little differently. If we go back to the unit circle, we can see this more clearly:

In the diagram above we can see the sine and cosine for (0^{circ}, 90^{circ}, 180^{circ},) and (270^{circ}) since ( an heta=frac{sin heta}{cos heta},) then we can see that ( an 0^{circ}=0, an 90^{circ}) is undefined, ( an 180^{circ}=)
0 and ( an 270^{circ}) is also undefined.

The real issue with the quadrantal angles is finding (sin ^{-1}(0), cos ^{-1}(0)) or ( an ^{-1}(0)) The calculator returns values of:
sin ^{-1}(0)=0^{circ}
cos ^{-1}(0)=90^{circ}
an ^{-1}(0)=0^{circ}
]In each case, there is another possibility than differs from the given angle by (180^{circ}) so:

sin ^{-1}(0)=0^{circ}, 180^{circ}
cos ^{-1}(0)=90^{circ}, 270^{circ}
an ^{-1}(0)=0^{circ}, 180^{circ}
]Let's look at how this is used in solving an equation:

Example 1
Solve the given equation for (0^{circ} leq x<360^{circ})
[ an ^{2} x- an x=0
]We could use the quadratic formula to solve this, but we can also solve by factoring:
an ^{2} x- an x=0
an x( an x-1)=0
an x=0 ext { or } an x=1
]Using a calculator to find ( an ^{-1}(0)) and ( an ^{-1}(1)) returns values of ( an ^{-1}(0)=0^{circ}) and ( an ^{-1}(1)=45^{circ} .) Once we know the reference angle for ( an ^{-1}(1),) then we know that since the tangent is also positive in Quadrant (Pi I), the solutions here are (45^{circ}) and (225^{circ} .) The calculator returns an answer of (0^{circ}) for ( an ^{-1}(0),) but we just saw that ( an 180^{circ}=0) as well.
The answers for this equation are (x=45^{circ}, 225^{circ}, 0^{circ}, 180^{circ})

Another approach to solving trigonometric equations involves using Pythagorean Identities to make a substitution that so that the equation can be simply solved by the quadratic formula. Here's an example:

Example 2
Solve the given equation for (0^{circ} leq x<360^{circ})
[sin ^{2} heta-6 cos heta=4
]Notice that, unlike the problems we saw in the previous section, this equation involves both the sine and the cosine. To remedy this, we can replace the (sin ^{2} heta) term with the expression (1-cos ^{2} heta)
sin ^{2} heta-6 cos heta=4
1-cos ^{2} heta-6 cos heta=4
0=cos ^{2} heta+6 cos heta+3
]using the quadratic formula:
[cos heta approx-5.449,-0.5505

since (cos ^{-1}(-5.449)) is not a real-valued angle, we can focus on the other answer:
(cos ^{-1}(-0.5505) approx 123.4^{circ} .) since the cosine function is also negative in the third quadrant, we need to find the reference angle that will help us identify the third quadrant angle that is a solution for this equation:
]So the reference angle is (56.6^{circ})
]The solutions are ( heta approx 123.4^{circ}, 236.6^{circ})

Example 3
Solve the given equation for (0^{circ} leq x<360^{circ})
[2 cos ^{2} heta-sin heta=sin ^{2} heta+1

First, we'll substitute (1-sin ^{2} heta) for the (cos ^{2} heta)
2 cos ^{2} heta-sin heta=sin ^{2} heta+1
2left(1-sin ^{2} heta ight)-sin heta=sin ^{2} heta+1
2-2 sin ^{2} heta-sin heta=sin ^{2} heta+1
0=3 sin ^{2} heta+sin heta-1
]Solving this with the quadratic formula gives us solutions of (sin heta approx-0.7676,0.43426)
[sin ^{-1}(-0.7676) approx-50.1^{circ}
][sin ^{-1}(0.43426) approx 25.7^{circ}

We'll work with the positive solution first. since the sine is also positive in Quadrant II, the other angle will be (180^{circ}-25.7^{circ}=154.3^{circ})

For the negative solution, we know that the sine is negative in Quadrants III and (mathrm{IV},) so with a reference angle of (50.1^{circ},) in the third quadrant (180^{circ}+50.1^{circ}=230.1^{circ}) and in the fourth quadrant (360^{circ}-50.1^{circ}=309.9^{circ})
The solution set is ( heta approx 25.7^{circ}, 154.3^{circ}, 230.1^{circ}, 309.9^{circ})

Exercises 3.4
Solve the given equations for (0^{circ} leq x<360^{circ})
1. (quad 9 sin ^{2} heta-6 sin heta=1)
2. (quad 4 cos ^{2} heta+4 cos heta=1)
3. (quad sec ^{2} alpha-2 sec alpha-3=0)
4. (quad csc ^{2} eta+4 csc eta-10=0)
5. (quad csc ^{2} x+4 csc x-7=0)
6. (quad 3 cot ^{2} x-3 cot x-1=0)
7. (quad 2 sin ^{2} x=1-cos x)
8. (quad cos ^{2} alpha+4=2 sin alpha-3)
9. (quad cos ^{2} eta-3 sin eta+2 sin ^{2} eta=0)
10. (quad sin ^{2} heta=2 cos heta+3 cos ^{2} heta)
11. (quad sec ^{2} x=2 an x+4)
12. (quad 3 an ^{2} x=sec x+2)
13. (quad cos alpha+1=2 cos 2 alpha)
14. (quad cos 2 x-3 sin x-2=0)
15. (quad csc ^{2} heta=cot heta+5)
16. (quad csc heta+5=2 cot ^{2} heta+2)

Continuous Everywhere but Differentiable Nowhere

I have to say that we’re doing some pretty neat stuff for trig this year in precalculus. I’m working with two other teachers and totally writing everything we’re doing from scratch. I had about 3 days to teach solve some basic trigonometric equations. They are basic. Like . But we’ve put a lot of thought into what we’re teaching, how we’re teaching it, and why we’re teaching it — and more complicated trig equations just didn’t make the cut. [1]

Besides not-a-lot-of-time, the other bugaboo I was contending was how to deal with inverse trig. Long story short, I’ve found a way to teach inverse trig which makes me fairly happy in my advanced precalculus class. But because of our time constraints, I decided that we could get my standard precalculus kids solving trig equations without understanding the theory behind the restricted domain of inverse trig functions. :) Why? They learned years ago in geometry that if they have a triangle like the one below

they could get an angle, like angle , by writing: . And then using the inverse of sine, they could get . They know about the inverse trig functions already. So I wanted to exploit that fact. And if organically a question about what the calculator was doing when spitting out an answer, and why it only gave one answer, I promised myself I would address it. (This year, no question like that arose.)

A quick last note, before I shared how I approached these few days in class, I decided to totally eliminate the use of the term “reference angle.” Kids would discover the relationships among the solutions of trig equations on their own. No need for new terminology here. Just logic.

Day 1: Three important “do nows”

This led to a great discussion. Every group decided the “top left equation” was going to be the easiest. And every group decided that the log and tangent equations were going to be the hardest. When I pressed them on why, they said it’s because they forgot logarithms from last year, and that tangent was just kinda tricky. They could “undo” a square root or a square, but they didn’t really know how to “undo” a logarithm or tangent function.

Next I threw up this slide. I just wanted to remind kids that sometimes there are more than one solution to equations — even simple equations they know. I also wanted them to see that they knew something about the tangent equation. They knew it had infinitely many solutions — even though they might (right now) know what those solutions are!

Finally, I wanted to do a serious review of special angles and their relationship with the unit circle. So I had kids spend 5 minutes solving these basic trig equations.

Obviously I put the unit circle on there as a prompt to get them thinking. And YES, that last trig equation, with the 3/7ths, was done on purpose. I asked kids after they got stuck on it if there were some of these they would not want to appear on a pop quiz. They all recognized that the 3/7th one was bad because it wasn’t one of the coordinates associated with the special angles.

This laid the groundwork for the packet.

Kids had good conversations and were able to solve equations like and using the unit circle/protractor, a detailed graph of the sine and cosine waves, and using their calculators to get fairly precise answers.

Their nightly work was simply to finish the sine and cosine questions in Part 1 (questions #1-4).

Day 2: Expanding Understanding

I started with an awesome “do now.”

I thought this was going to be a quick 4-5 minute discussion. But kids took 3-4 minutes just to really talk in their groups. And I had them share their thinking. It led to kids talking about “efficiency” and “conceptual understanding” themselves! They all pretty much though the unit circle was the best way to solve it — even with the annoyance of the protractor — because they liked the conceptual understanding it provided. They thought the calculator did the work quickly, and was more accurate, but it annoyingly only gave one of the solutions (so you had to use logic and the unit circle to figure out the second solution), and you could easily forget the meaning of what you were doing. I was so proud of what they were saying. Super awesome metacognition! All in all, this was probably 7-8 minutes.

Then I let them loose on the tangent questions in the packet (Part I #5 and 6). They initially had to solve using a protractor. Every single group remembered tangent represented slope. Most groups reasoned that if , they would get and as their solutions. And since this slope was slightly greater than 1, the angles would be slightly different, just a few degrees higher. It was lovely. (And exactly what I hoped would happen, which is why I chose to use 1.1 in the equation.) But one group literally drew a line with a slope of 1.1 and measured the angles associated with that. I wasn’t surprised that a group did that, but I expected a few more to do so. (I had this group share their thinking with the rest of the class, at the end of the period.)

Then kids spent the rest of the class working on select questions in Part II (8, 9, 11) and Part III (13, 14, 15).

For nightly work, kids finished any of those problems (#8, 9, 11, 13, 14, 15) that they didn’t finish up in class.

Day 3: Polishing Things Off

I started with a question that I wanted to reinforce after the previous class:

We did a bit of review of some unrelated Algebra II ideas to help set them up for our next unit on polynomials. And then…

… to work! I had kids discuss problem 13 in their groups first (since I could see that being a place where a kid, at home, might get trapped… and I wanted them to use each other to get unstuck). And then they compared their answers to the other nightly work questions — and used a solution sheet I gave them to see if they were correct. Then I set them loose on using Desmos to do Part IV. The rest of the period was spent working on finishing up the problems that weren’t assigned in the packet (the ones they skipped).

Pretty much all groups were working together amazingly, and when I went around to check in on different groups, everyone was getting all the questions correct. The biggest problem was actually finding a good window in Desmos! If that’s the biggest problem, I’m golden.

What I loved:

Okay, so I’m going to toot my own horn here. Although the packet may “look” simple, I have to say the only way to see why it’s so awesome is to actually do it. The choice of having kids solve and then immediately solve was on purpose, to generate good conversations with kids about reference angles without using that term. The choice of was done specifically to exploit their understanding of . And the fact that they’re constantly looking at the same question through three different lenses (unit circle, wave, calculator) is deliciously sweet. And then — at the very end — they get to see the solution a fourth way, by using Desmos to graph these equations to find a solution? SO COOL. Because the very last thing we had done in this class was learning transformations of sine and cosine graphs! [2]

This packet, and associated “do nows” and conversations, did what I was hoping for. It highlighted multiple representations. It had kids thinking deeply about the meaning of sine, cosine, and tangent. It had kids develop a way to understand multiple solutions to trig equations by simply using logic and what they know. It had kids recognize that the more they understand trigonometry, the more ways they have to solve a trig problem. And no kid got derailed because they didn’t understand inverses deeply.

[1] I could argue a case for these type of equations, as well as a case against them. But considering our goals and what we’ve already done with trig, I think we’re making the right decision. Why? Because our goal isn’t solving algebraic equations writ large, and I could see solving something like being useful for that. But for getting a deeper understanding of the trigonometric functions? I see less value. (Not no value, mind you, but less…)

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To define the sine function of an acute angle α, start with a right triangle that contains an angle of measure α in the accompanying figure, angle α in triangle ABC is the angle of interest. The three sides of the triangle are named as follows:

  • The opposite side is the side opposite to the angle of interest, in this case side a.
  • The hypotenuse is the side opposite the right angle, in this case side h. The hypotenuse is always the longest side of a right-angled triangle.
  • The adjacent side is the remaining side, in this case side b. It forms a side of (and is adjacent to) both the angle of interest (angle A) and the right angle.

Once such a triangle is chosen, the sine of the angle is equal to the length of the opposite side, divided by the length of the hypotenuse: [5]

The other trigonometric functions of the angle can be defined similarly for example, the cosine of the angle is the ratio between the adjacent side and the hypotenuse, while the tangent gives the ratio between the opposite and adjacent sides. [5]

In trigonometry, a unit circle is the circle of radius one centered at the origin (0, 0) in the Cartesian coordinate system.

Let a line through the origin intersect the unit circle, making an angle of θ with the positive half of the x-axis. The x- and y-coordinates of this point of intersection are equal to cos(θ) and sin(θ) , respectively. This definition is consistent with the right-angled triangle definition of sine and cosine when 0° < θ < 90°: because the length of the hypotenuse of the unit circle is always 1, sin ⁡ ( θ ) = opposite hypotenuse = opposite 1 = opposite >< extrm >>=< frac < extrm ><1>>=< extrm >> . The length of the opposite side of the triangle is simply the y-coordinate. A similar argument can be made for the cosine function to show that cos ⁡ ( θ ) = adjacent hypotenuse >< extrm >>> when 0° < θ < 90°, even under the new definition using the unit circle. tan(θ) is then defined as sin ⁡ ( θ ) cos ⁡ ( θ ) >> , or, equivalently, as the slope of the line segment.

Using the unit circle definition has the advantage that the angle can be extended to any real argument. This can also be achieved by requiring certain symmetries, and that sine be a periodic function.

These apply for all values of θ .

Reciprocal Edit

The reciprocal of sine is cosecant, i.e., the reciprocal of sin(A) is csc(A) , or cosec(A) . Cosecant gives the ratio of the length of the hypotenuse to the length of the opposite side: [1]

Inverse Edit

The inverse function of sine is arcsine (arcsin or asin) or inverse sine ( sin −1 ). [1] As sine is non-injective, it is not an exact inverse function, but a partial inverse function. For example, sin(0) = 0 , but also sin(π) = 0 , sin(2π) = 0 etc. It follows that the arcsine function is multivalued: arcsin(0) = 0 , but also arcsin(0) = π , arcsin(0) = 2π , etc. When only one value is desired, the function may be restricted to its principal branch. With this restriction, for each x in the domain, the expression arcsin(x) will evaluate only to a single value, called its principal value.

where (for some integer k):

By definition, arcsine satisfies the equation:

Calculus Edit

Other trigonometric functions Edit

It is possible to express any trigonometric function in terms of any other (up to a plus or minus sign, or using the sign function).

The following table documents how sine can be expressed in terms of the other common trigonometric functions:

f θ Using plus/minus (±) Using sign function (sgn)
f θ = ± per Quadrant f θ =
cos sin ⁡ ( θ ) = ± 1 − cos 2 ⁡ ( θ ) ( heta )>>> + + = sgn ⁡ ( cos ⁡ ( θ − π 2 ) ) 1 − cos 2 ⁡ ( θ ) left(cos left( heta -<2>> ight) ight)( heta )>>>
cos ⁡ ( θ ) = ± 1 − sin 2 ⁡ ( θ ) ( heta )>>> + + = sgn ⁡ ( sin ⁡ ( θ + π 2 ) ) 1 − sin 2 ⁡ ( θ ) left(sin left( heta +<2>> ight) ight)( heta )>>>
cot sin ⁡ ( θ ) = ± 1 1 + cot 2 ⁡ ( θ ) ( heta )>>>> + + = sgn ⁡ ( cot ⁡ ( θ 2 ) ) 1 1 + cot 2 ⁡ ( θ ) left(cot left(<2>> ight) ight)( heta )>>>>
cot ⁡ ( θ ) = ± 1 − sin 2 ⁡ ( θ ) sin ⁡ ( θ ) ( heta )>>>> + + = sgn ⁡ ( sin ⁡ ( θ + π 2 ) ) 1 − sin 2 ⁡ ( θ ) sin ⁡ ( θ ) left(sin left( heta +<2>> ight) ight)( heta )>>>>
tan sin ⁡ ( θ ) = ± tan ⁡ ( θ ) 1 + tan 2 ⁡ ( θ ) ( heta )>>>> + + = sgn ⁡ ( tan ⁡ ( 2 θ + π 4 ) ) tan ⁡ ( θ ) 1 + tan 2 ⁡ ( θ ) left( an left(<4>> ight) ight)( heta )>>>>
tan ⁡ ( θ ) = ± sin ⁡ ( θ ) 1 − sin 2 ⁡ ( θ ) ( heta )>>>> + + = sgn ⁡ ( sin ⁡ ( θ + π 2 ) ) sin ⁡ ( θ ) 1 − sin 2 ⁡ ( θ ) left(sin left( heta +<2>> ight) ight)( heta )>>>>
sec sin ⁡ ( θ ) = ± sec 2 ⁡ ( θ ) − 1 sec ⁡ ( θ ) ( heta )-1>>>> + + = sgn ⁡ ( sec ⁡ ( 4 θ − π 2 ) ) sec 2 ⁡ ( θ ) − 1 sec ⁡ ( θ ) left(sec left(<2>> ight) ight)( heta )-1>>>>
sec ⁡ ( θ ) = ± 1 1 − sin 2 ⁡ ( θ ) ( heta )>>>> + + = sgn ⁡ ( sin ⁡ ( θ + π 2 ) ) 1 1 − sin 2 ⁡ ( θ ) left(sin left( heta +<2>> ight) ight)( heta )>>>>

For all equations which use plus/minus (±), the result is positive for angles in the first quadrant.

The basic relationship between the sine and the cosine can also be expressed as the Pythagorean trigonometric identity: [2]

where sin 2 (x) means (sin(x)) 2 .

Sine squared function Edit

The graph shows both the sine function and the sine squared function, with the sine in blue and sine squared in red. Both graphs have the same shape, but with different ranges of values, and different periods. Sine squared has only positive values, but twice the number of periods.

The sine squared function can be expressed as a modified sine wave from the Pythagorean identity and power reduction—by the cosine double-angle formula: [6]

The table below displays many of the key properties of the sine function (sign, monotonicity, convexity), arranged by the quadrant of the argument. For arguments outside those in the table, one may compute the corresponding information by using the periodicity sin ⁡ ( α + 360 ∘ ) = sin ⁡ ( α ) )=sin(alpha )> of the sine function.

Quadrant Degrees Radians Value Sign Monotony Convexity
1st Quadrant 0 ∘ < x < 90 ∘ <x<90^> 0 < x < π 2 <2>>> 0 < sin ⁡ ( x ) < 1 + increasing concave
2nd Quadrant 90 ∘ < x < 180 ∘ <x<180^> π 2 < x < π <2>><x<pi > 0 < sin ⁡ ( x ) < 1 + decreasing concave
3rd Quadrant 180 ∘ < x < 270 ∘ <x<270^> π < x < 3 π 2 <2>>> − 1 < sin ⁡ ( x ) < 0 decreasing convex
4th Quadrant 270 ∘ < x < 360 ∘ <x<360^> 3 π 2 < x < 2 π <2>><x<2pi > − 1 < sin ⁡ ( x ) < 0 increasing convex

The following table gives basic information at the boundary of the quadrants.

Degrees Radians sin ⁡ ( x ) Point type
0 ∘ > 0 0 Root, Inflection
90 ∘ > π 2 <2>>> 1 Maximum
180 ∘ > π 0 Root, Inflection
270 ∘ > 3 π 2 <2>>> − 1 Minimum

Using only geometry and properties of limits, it can be shown that the derivative of sine is cosine, and that the derivative of cosine is the negative of sine.

Using the reflection from the calculated geometric derivation of the sine is with the (4n+k)-th derivative at the point 0:

This gives the following Taylor series expansion at x = 0. One can then use the theory of Taylor series to show that the following identities hold for all real numbers x (where x is the angle in radians): [7]

If x were expressed in degrees then the series would contain factors involving powers of π/180: if x is the number of degrees, the number of radians is y = πx /180, so

The series formulas for the sine and cosine are uniquely determined, up to the choice of unit for angles, by the requirements that

The radian is the unit that leads to the expansion with leading coefficient 1 for the sine and is determined by the additional requirement that

The coefficients for both the sine and cosine series may therefore be derived by substituting their expansions into the Pythagorean and double angle identities, taking the leading coefficient for the sine to be 1, and matching the remaining coefficients.

In general, mathematically important relationships between the sine and cosine functions and the exponential function (see, for example, Euler's formula) are substantially simplified when angles are expressed in radians, rather than in degrees, grads or other units. Therefore, in most branches of mathematics beyond practical geometry, angles are generally assumed to be expressed in radians.

A similar series is Gregory's series for arctan, which is obtained by omitting the factorials in the denominator.

Continued fraction Edit

The sine function can also be represented as a generalized continued fraction:

The continued fraction representation can be derived from Euler's continued fraction formula and expresses the real number values, both rational and irrational, of the sine function.

Zero is the only real fixed point of the sine function in other words the only intersection of the sine function and the identity function is sin ⁡ ( 0 ) = 0 .

The arc length of the sine curve from 0 to x is L x / ( 2 π ) , plus a correction that varies periodically in x with period π . The Fourier series for this correction can be written in closed form using special functions, but it is perhaps more instructive to write the decimal approximations of the Fourier coefficients. The sine curve arc length from 0 to x is

1.21600672 x + 0.10317093 sin ⁡ ( 2 x ) − 0.00220445 sin ⁡ ( 4 x ) + 0.00012584 sin ⁡ ( 6 x ) − 0.00001011 sin ⁡ ( 8 x ) + ⋯

The leading term in the above equation, and the limit of arc length to distance ratio is given by:

The law of sines states that for an arbitrary triangle with sides a, b, and c and angles opposite those sides A, B and C:

This is equivalent to the equality of the first three expressions below:

where R is the triangle's circumradius.

It can be proven by dividing the triangle into two right ones and using the above definition of sine. The law of sines is useful for computing the lengths of the unknown sides in a triangle if two angles and one side are known. This is a common situation occurring in triangulation, a technique to determine unknown distances by measuring two angles and an accessible enclosed distance.

For certain integral numbers x of degrees, the value of sin(x) is particularly simple. A table of some of these values is given below.

x in degrees 90° 180° 270° 360°
x in radians 0 π/2 π 3π/2
x in gons 0 100 g 200 g 300 g 400 g
x in turns 0 1/4 1/2 3/4 1
sin x 0 1 0 −1 0

Other values not listed above:

Sine is used to determine the imaginary part of a complex number given in polar coordinates (r, φ):

z = r ( cos ⁡ ( φ ) + i sin ⁡ ( φ ) )

r and φ represent the magnitude and angle of the complex number respectively. i is the imaginary unit. z is a complex number.

Although dealing with complex numbers, sine's parameter in this usage is still a real number. Sine can also take a complex number as an argument.

Sine with a complex argument Edit

Domain coloring of sin(z) in the complex plane. Brightness indicates absolute magnitude, saturation represents complex argument.

The definition of the sine function for complex arguments z:

where i 2 = −1, and sinh is hyperbolic sine. This is an entire function. Also, for purely real x,

For purely imaginary numbers:

It is also sometimes useful to express the complex sine function in terms of the real and imaginary parts of its argument:

Partial fraction and product expansions of complex sine Edit

Using the partial fraction expansion technique in complex analysis, one can find that the infinite series

Using product expansion technique, one can derive

Alternatively, the infinite product for the sine can be proved using complex Fourier series.

Using complex Fourier series, the function cos ⁡ ( z x ) can be decomposed as

Usage of complex sine Edit

sin(z) is found in the functional equation for the Gamma function,

which in turn is found in the functional equation for the Riemann zeta-function,

As a holomorphic function, sin z is a 2D solution of Laplace's equation:

The complex sine function is also related to the level curves of pendulums. [ how? ] [9] [ better source needed ]

Complex graphs Edit

While the early study of trigonometry can be traced to antiquity, the trigonometric functions as they are in use today were developed in the medieval period. The chord function was discovered by Hipparchus of Nicaea (180–125 BCE) and Ptolemy of Roman Egypt (90–165 CE).

The function of sine and versine (1 − cosine) can be traced to the jyā and koṭi-jyā functions used in Gupta period (320 to 550 CE) Indian astronomy (Aryabhatiya, Surya Siddhanta), via translation from Sanskrit to Arabic and then from Arabic to Latin. [3]

All six trigonometric functions in current use were known in Islamic mathematics by the 9th century, as was the law of sines, used in solving triangles. [10] With the exception of the sine (which was adopted from Indian mathematics), the other five modern trigonometric functions were discovered by Arabic mathematicians, including the cosine, tangent, cotangent, secant and cosecant. [10] Al-Khwārizmī (c. 780–850) produced tables of sines, cosines and tangents. [11] [12] Muhammad ibn Jābir al-Harrānī al-Battānī (853–929) discovered the reciprocal functions of secant and cosecant, and produced the first table of cosecants for each degree from 1° to 90°. [12]

The first published use of the abbreviations 'sin', 'cos', and 'tan' is by the 16th century French mathematician Albert Girard these were further promulgated by Euler (see below). The Opus palatinum de triangulis of Georg Joachim Rheticus, a student of Copernicus, was probably the first in Europe to define trigonometric functions directly in terms of right triangles instead of circles, with tables for all six trigonometric functions this work was finished by Rheticus' student Valentin Otho in 1596.

In a paper published in 1682, Leibniz proved that sin x is not an algebraic function of x. [13] Roger Cotes computed the derivative of sine in his Harmonia Mensurarum (1722). [14] Leonhard Euler's Introductio in analysin infinitorum (1748) was mostly responsible for establishing the analytic treatment of trigonometric functions in Europe, also defining them as infinite series and presenting "Euler's formula", as well as the near-modern abbreviations sin., cos., tang., cot., sec., and cosec. [15]

Etymology Edit

Etymologically, the word sine derives from the Sanskrit word for chord, jiva*(jya being its more popular synonym). This was transliterated in Arabic as jiba جيب, which however is meaningless in that language and abbreviated jb جب . Since Arabic is written without short vowels, "jb" was interpreted as the word jaib جيب, which means "bosom". When the Arabic texts were translated in the 12th century into Latin by Gerard of Cremona, he used the Latin equivalent for "bosom", sinus (which means "bosom" or "bay" or "fold"). [16] [17] Gerard was probably not the first scholar to use this translation Robert of Chester appears to have preceded him and there is evidence of even earlier usage. [18] The English form sine was introduced in the 1590s.

There is no standard algorithm for calculating sine. IEEE 754-2008, the most widely used standard for floating-point computation, does not address calculating trigonometric functions such as sine. [19] Algorithms for calculating sine may be balanced for such constraints as speed, accuracy, portability, or range of input values accepted. This can lead to different results for different algorithms, especially for special circumstances such as very large inputs, e.g. sin(10 22 ) .

A common programming optimization, used especially in 3D graphics, is to pre-calculate a table of sine values, for example one value per degree, then for values in-between pick the closest pre-calculated value, or linearly interpolate between the 2 closest values to approximate it. This allows results to be looked up from a table rather than being calculated in real time. With modern CPU architectures this method may offer no advantage. [ citation needed ]

The CORDIC algorithm is commonly used in scientific calculators.

The sine function, along with other trigonometric functions, is widely available across programming languages and platforms. In computing, it is typically abbreviated to sin .

Some CPU architectures have a built-in instruction for sine, including the Intel x87 FPUs since the 80387.

In programming languages, sin is typically either a built-in function or found within the language's standard math library.

For example, the C standard library defines sine functions within math.h: sin(double) , sinf(float) , and sinl(long double) . The parameter of each is a floating point value, specifying the angle in radians. Each function returns the same data type as it accepts. Many other trigonometric functions are also defined in math.h, such as for cosine, arc sine, and hyperbolic sine (sinh).

Similarly, Python defines math.sin(x) within the built-in math module. Complex sine functions are also available within the cmath module, e.g. cmath.sin(z) . CPython's math functions call the C math library, and use a double-precision floating-point format.

Turns based implementations Edit

Some software libraries provide implementations of sine using the input angle in half-turns, a half-turn being an angle of 180 degrees or π radians. Representing angles in turns or half-turns has accuracy advantages and efficiency advantages in some cases. [20] [21]

Environment Function name Angle units
MATLAB sinpi [22] half-turns
OpenCL sinpi [23] half-turns
R sinpi [24] half-turns
Julia sinpi [25] half-turns
CUDA sinpi [26] half-turns
ARM sinpi [27] half-turns

The accuracy advantage stems from the ability to perfectly represent key angles like full-turn, half-turn, and quarter-turn losslessly in binary floating-point or fixed-point. In contrast, representing 2 π , π , and π 2 <2>>> in binary floating-point or binary scaled fixed-point always involves a loss of accuracy.

Turns also have an accuracy advantage and efficiency advantage for computing modulo to one period. Computing modulo 1 turn or modulo 2 half-turns can be losslessly and efficiently computed in both floating-point and fixed-point. For example, computing modulo 1 or modulo 2 for a binary point scaled fixed-point value requires only a bit shift or bitwise AND operation. In contrast, computing modulo π 2 <2>>> involves inaccuracies in representing π 2 <2>>> .

For applications involving angle sensors, the sensor typically provides angle measurements in a form directly compatible with turns or half-turns. For example, an angle sensor may count from 0 to 4096 over one complete revolution. [28] If half-turns are used as the unit for angle, then the value provided by the sensor directly and losslessly maps to a fixed-point data type with 11 bits to the right of the binary point. In contrast, if radians are used as the unit for storing the angle, then the inaccuracies and cost of multiplying the raw sensor integer by an approximation to π 2048 <2048>>> would be incurred.

There is at least two strategies:

  • For identities like $sin(3t)=3 sin(t)-4sin^3(t)$, i.e expressing $sin(nt)$ as a sum of powers of $sin(t)$, this is actually the Chebyshev polynomials.
  • For expressing $sin^n(t)$ as a sum of sinuses $sin(kt)$ you can use Moivre formula and a bynomial expansion: $sin^3(t)=left(frac<>-e^<-it>><2i> ight)^3=frac<1><-8i>left( e^<3it>-3e^<(2i-i)t>+3e^<(i-2i)t>-e^<-3it> ight)=frac<1><4>left(-frac-e^<-3it>><2i>+3frac<>-e^<-it>><2i> ight)$ so: $sin^3(t)=frac<1><4>(-sin(3t)+3 sin(t))$

You can use the definition of sine and cosine with Euler's formula for a more "constructed" way of handling such identities, I'd not algorithmics. For example, sine can be defined as $sint =frac < e^- e^<-it>><2i>$. As such we can see that:

As seen by the identities you provided.

Without using higher level mathematics I am wondering whether a binary type algorithm might help where the fundamental forms are idenities like

$sin(apm b)=sin a cos b pm cos a sin b$ and $cos(apm b)=cos a cos b mp sin a sin b$

Since $sin (2^n a)=2 sin (2^ a)cos(2^ a)$ and $cos (2^n a)= cos^2 (2^ a)-sin^2(2^ a)$

For example in the case of proving $sin 6a=32 cos^5 a sin a-32 cos^3 sin a +6 cos a sin a ag<1>$

We can start with the left hand side of $(1)$ $ sin 6a=sin 4a cos 2a + cos 4a sin 2a$

Quickly getting to $ sin 6a=sin 2a left(3cos^2 2a - sin^2 2a ight)=sin 2aleft( 4 cos^2 2a - 1 ight)$

[or alternatively $ sin 6a = sin 2aleft( 3- 4 sin^2 2a ight)$]

Then continue with the r.h.s. of $(1)$ $32 cos^5 a sin a-32 cos^3 sin a +6 cos a sin a = sin 2a left( 16 cos^4 a - 16 cos^2 a +3 ight)$

and we are then simply left to prove $ 4 cos^2 2a - 1 = 16 cos^4 a - 16 cos^2 a +3 $

An identity involving $sin 13a$ for example would decompose as

$sin 13a =sin 12a cos a + cos 12a sin a$ then $sin 13a =left(sin 8a cos 4a+ cos 8a sin 4a ight) cos a + left(cos 8a cos 4a- sin 8a sin 4a ight) sin a$ and so on.

3.4: More Trigonometric Equations

In the previous two sections we’ve looked at a couple of Calculus I topics in terms of parametric equations. We now need to look at a couple of Calculus II topics in terms of parametric equations.

In this section we will look at the arc length of the parametric curve given by,

[x = fleft( t ight)hspace<0.5in>y = gleft( t ight)hspace<0.5in>alpha le t le eta ]

We will also be assuming that the curve is traced out exactly once as (t) increases from (alpha) to (eta). We will also need to assume that the curve is traced out from left to right as (t) increases. This is equivalent to saying,

So, let’s start out the derivation by recalling the arc length formula as we first derived it in the arc length section of the Applications of Integrals chapter.

We will use the first (ds) above because we have a nice formula for the derivative in terms of the parametric equations (see the Tangents with Parametric Equations section). To use this we’ll also need to know that,

The arc length formula then becomes,

This is a particularly unpleasant formula. However, if we factor out the denominator from the square root we arrive at,

Now, making use of our assumption that the curve is being traced out from left to right we can drop the absolute value bars on the derivative which will allow us to cancel the two derivatives that are outside the square root and this gives,

Arc Length for Parametric Equations

Notice that we could have used the second formula for (ds) above if we had assumed instead that

If we had gone this route in the derivation we would have gotten the same formula.

Let’s take a look at an example.

We know that this is a circle of radius 3 centered at the origin from our prior discussion about graphing parametric curves. We also know from this discussion that it will be traced out exactly once in this range.

So, we can use the formula we derived above. We’ll first need the following,


> = 3cos left( t ight)hspace<0.5in>hspace<0.25in>frac<><
> = - 3sin left( t ight)]

Since this is a circle we could have just used the fact that the length of the circle is just the circumference of the circle. This is a nice way, in this case, to verify our result.

Let’s take a look at one possible consequence if a curve is traced out more than once and we try to find the length of the curve without taking this into account.

Notice that this is the identical circle that we had in the previous example and so the length is still 6(p). However, for the range given we know it will trace out the curve three times instead once as required for the formula. Despite that restriction let’s use the formula anyway and see what happens.

In this case the derivatives are,

and the length formula gives,

The answer we got form the arc length formula in this example was 3 times the actual length. Recalling that we also determined that this circle would trace out three times in the range given, the answer should make some sense.

If we had wanted to determine the length of the circle for this set of parametric equations we would need to determine a range of (t) for which this circle is traced out exactly once. This is, (0 le t le frac<<2pi >><3>). Using this range of (t) we get the following for the length.

which is the correct answer.

Be careful to not make the assumption that this is always what will happen if the curve is traced out more than once. Just because the curve traces out (n) times does not mean that the arc length formula will give us (n) times the actual length of the curve!

Before moving on to the next section let’s notice that we can put the arc length formula derived in this section into the same form that we had when we first looked at arc length. The only difference is that we will add in a definition for (ds) when we have parametric equations.

egin 16(sin^<2>x)^5+sin^2 x-1&=0 , (2sin^2x-1) (8sin^8x+4sin^6x+2sin^4x+sin^2x+1) &=0 , cos2x,(8sin^8x+4sin^6x+2sin^4x+sin^2x+1) &=0. end

egin 8sin^8x+4sin^6x+2sin^4x+sin^2x+1&>0quad forall xinmathbb , end

egin cos2x&=0 , 2x&= fracpi2+pi k,quad k=0,1,dots , x&= fracpi4+ frac2,quad k=0,1,dots . end

For $x$ restricted to the interval $[0,2pi]$, we have four solutions egin x_1&= fracpi4=45^circ , x_2&= fracpi4+ fracpi2= frac<3pi>4=135^circ . x_3&= fracpi4+pi= frac<5pi>4=225^circ . x_4&= fracpi4+ frac<3pi>2= frac<7pi>4=315^circ . end

Substitution of $x_1,dots,x_4$ into original equation shows that only solutions $x_1,x_3$ are valid:

As @Malcolm noted, two extra solutions were introduced by the squaring the original equation, that's why the final check of the found roots is essential.

By letting $x=arctan u$ the problem boils down to solving $ (u-1)(4u^4+3u^3+3u^2+u+1)=0$ and $ 4u^4+3u^3+3u^2+u+1 = left(2u^2+frac<3><4>u ight)^2+left(u+frac<1><2> ight)^2+frac<23><16>u^2+frac<3><4>geq frac<3> <4>$ is always positive, hence $u=1$ (i.e. $xin<4>,frac<5pi><4>>$) is the only solution.
Completing the square is a powerful technique: for instance, the decomposition $ 4u^4+3u^3+3u^2+u+1 = left(2u^2+frac<3><4>u ight)^2+frac<39><16>left(u+frac<8><39> ight)^2+frac<35> <39>$ proves that the LHS is $geqfrac<35><39>$: this is pretty close to the actual value of the minimum.

Note that $ an x =1 implies x= pi /4 ext < and >5pi/4 $ are solutions.

In the interest of completeness I am posting this solution although I don't think that this is necessarily a high school approach. The approach is based on the idea that equations with trigonometric functions are much more amenable to manipulation if there is only one trigonometric function involved. $ sin^5(x) = frac 4 4 = cos x csc^5 x 4= cot x csc^4 x 4 = cot x (1+cot^2 x)^2 $ Or writing $u = cot x$, $ u^5 + 2u^3 + u - 4 =0 $ If we notice that $1+2+1 = 4$ it becomes easier to guess $u=1$ is a solution. In terms of our original variable this is the solution $cot x =1$ or $x = 45°$ 0r $225°$.

To see if there are any more solutions we could try to factor the other factor in the polynomial, $u^4+u^3+3u^2+3u+4$. We would find that it has no real roots, but it is probably easier to consider the graphs of $y=sin^5 x$ and of $y=frac<4>$. We can easily convince ourselves that there are only two intersections in $[0,360]$, so we have found all the solutions.

This is too long for a comment.

The exact way for solving the problem has been given by gt6989b in his/her answer.

Under some conditions, you could approximate the solution.

Let me change notation and define $y=frac<2 (r-s)>implies w=frac-frac<2 (r-s)>y$ The equation becomes $frac-frac<2 (r-s)>y=2r sin(y)$ For leq yleq pi$, a very crude approximation of mine is $sin(y)approx frac<120 >(pi -y) y$ which would let you with a quadratic equation in $y$. $240 r x y^2+ left(-240 pi r x-2 pi ^4 r+2 pi ^4 s ight)y+pi ^5 r=0$

Much better would be, for the same range, the magnificent $sin(y) simeq frac<16 (pi -y) ,y><5 pi ^2-4 (pi -y), y>qquad (0leq yleqpi)$ proposed, more than $1400$ years ago, by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (see here). This will give you a cubic equation to solve.$8 (s-r)y^3+4 pi (r (8 x+3)-2 s)y^2+2 pi ^2 (5 s-r (16 x+7))y+5 pi ^3 r=0$