/math/ - Trigonometric functions are used to relate the angles of a right triangle to its sides. ,,\qquad \sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\quad\cos(\text{angle}) =\frac{\text{adjacent}}{\text{hypotenuse}}\quad\tan(\text{angle}) = \ - Mathematics

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Anonymous June 16, 2022 (Thu) 16:55:45 No. 189

Trigonometric functions are used to relate the angles of a right triangle to its sides.

\qquad \sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\quad\cos(\text{angle}) =\frac{\text{adjacent}}{\text{hypotenuse}}\quad\tan(\text{angle}) = \frac{\sin(\text{angle})}{\cos(\text{angle})} = \frac{\text{opposite}}{\text{adjacent}}\quad

With respect to any of the two angles

\alpha

or

\beta

(the third angle of a right triangle is, of course,

90^\text{o}

thereofre it's irrelevant) three sides can be identified with the following names: hypotenuse, opposite and adjacent. The hypotenuse is easy to identify and it's the longest i.e.

c

. The adjacent side is the one that's "touching" the angle; for angle

\alpha

that would be

b

, while for angle

\beta

that would be

a

. The opposite side is the one that's NOT "touching" the angle; for angle

\alpha

that would be

a

, while for angle

\beta

that would be

b

.

Therefore, the complete set of trigonometric functions for the triangle in the pic related is:

\qquad \sin(\alpha) = \frac{a}{c}\quad \cos(\alpha) = \frac{b}{c}\quad \tan(\alpha) = \frac{a}{b}

\qquad \sin(\beta) = \frac{b}{c}\quad \cos(\beta) = \frac{a}{c}\quad \tan(\beta) = \frac{b}{a}

Thus, just knowing the angle

\alpha

allows conversion of any side to any other side:

	$a$	$b$	$c$
$a=$	$a$	$b\tan(\alpha)$	$c\sin(\alpha)$
$b=$	$\frac{a}{\tan(\alpha)}$	$b$	$c\cos(\alpha)$
$c=$	$\frac{a}{\sin(\alpha)}$	$\frac{b}{\cos(\alpha)}$	$c$

And similarly, just knowing the angle

\beta

also allows conversion of any side to any other side:

	$a$	$b$	$c$
$a=$	$a$	$\frac{b}{\tan(\beta)}$	$c\cos(\beta)$
$b=$	$a\tan(\beta)$	$b$	$c\sin(\beta)$
$c=$	$\frac{a}{\cos(\beta)}$	$\frac{b}{\sin(\beta)}$	$c$

Intuition behind this is straightforward: For angles of a triangle, sine and cosine are bounded functions between zero and one.

(0 \leq \sin(x)\leq 1,\; 0\leq \cos(x) \leq1)

. Multiplying something with these functions should produce a smaller or equal value, while dividing something with these functions should produce a larger or equal value. Since hypotenuse is the longest, multiplying it with sine or cosine will "reduce" it to a leg. Similarly, dividing a leg by sine or cosine will "grow" it into a hypotenuse. A leg can be converted to another leg by "growing" it into a hypotenuse first, then "reducing" it to another leg which effectively is the same as multiplying the leg with a tangent value of the angle that the leg opposes.

>>

Anonymous June 16, 2022 (Thu) 16:57:15 No. 190

Dividing both sides of Pythagorean theorem

a^2 + b^2 = c^2

with

c^2

a fundamental trigonometric identity is obtained

\qquad \sin^2(\alpha) + \cos^2(\alpha) = 1

Solving equation invarious ways produces conversion formulas

\qquad |\sin(\alpha)| = \sqrt{1 - \cos^2(\alpha)} = \frac{\tan(x)}{\sqrt{\tan^2(x) + 1}}

\qquad |\cos(\alpha)| = \sqrt{1 - \sin^2(\alpha)} = \frac{1}{\sqrt{\tan^2(x) + 1}}

Trigonometric function values

$\alpha=$	$0^\text{o}$	$30^\text{o}$	$45^\text{o}$	$60^\text{o}$	$90^\text{o}$
$\sin\alpha =$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$
$\cos\alpha =$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	$1$
$\tan\alpha =$	$0$	$\frac{\sqrt{3}}{3}$	$1$	$\sqrt{3}$	$+\infty$

Easy way to remember the values

\frac{\sqrt{0}}{2}

\frac{\sqrt{1}}{2}

\frac{\sqrt{2}}{2}

\frac{\sqrt{3}}{2}

\frac{\sqrt{4}}{2}

Of course, this table is only a mnemonic and

\sin(15^\text{o})

is NOT

\frac{\sqrt{0.5}}{2}

, although the value can be found through the half angle formula

\qquad \sin(15^\text{o}) = \frac{1 - \cos(30^\text{o})}{2} = \frac{2 - \sqrt{3}}{ 4}

>>

Anonymous June 16, 2022 (Thu) 16:58:05 No. 191

Using the Euler identity

e^{i\theta} = \cos\theta + i\sin\theta

:

\begin{aligned}\qquad \cos(\alpha + \beta) + i\sin(\alpha + \beta) &= e^{i(\alpha + \beta)} = e^{i\alpha}e^{i\beta} \\&= (\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta)\\&=\cos\alpha\cos\beta - \sin\alpha\sin\beta + i(\sin\alpha\cos\beta + \cos\alpha\sin\beta) \end{aligned}

Matching real and imaginary parts left and right, we obtain the angle sum formulas

\cos(\alpha + \beta)

and

\sin(\alpha + \beta)

.
Difference formulas are obtained by substituting

\beta

for

-\beta

and remembering that:

\qquad\sin(-\theta) = -\cos(\theta)

\qquad\cos(-\theta) = +\cos(\theta)

Angle sum and difference formulas:

\qquad \sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)

\qquad \sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)

\qquad \cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)

\qquad \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)

By dividing (1) and (3), or (2) and (4):

\qquad\tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}

\qquad \tan(\alpha - \beta) = \frac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}

And from these, other formulas can be derived.

>>

Anonymous June 16, 2022 (Thu) 16:58:37 No. 192

Half-angle formulas

\qquad \sin^2\left(\frac{\alpha}{2}\right) = \frac{1 - \cos(2\alpha)}{2}

\qquad \cos^2\left(\frac{\alpha}{2}\right) = \frac{1 + \cos(2\alpha)}{2}

Notice that both formulas above use the

\cos(2\alpha)

Half-angle tangent formula is better given by:

\qquad \tan\left(\frac{\alpha + \beta}{2}\right) = \frac{\sin(\alpha) + \sin(\beta)}{\cos(\alpha) + \cos(\beta)}

By setting

\beta = 0

:

\qquad \tan\left(\frac{\alpha}{2}\right) = \frac{\sin(\alpha)}{\cos(\alpha) + 1}

By substituting

\beta

for

-\beta

\qquad \tan\left(\frac{\alpha - \beta}{2}\right) = \frac{\sin(\alpha) - \sin(\beta)}{\cos(\alpha) + \cos(\beta)}

>>

Anonymous June 16, 2022 (Thu) 16:59:09 No. 193

Double angle formulas:

By setting

\beta = \alpha

we get double angle formulas. There is one double-angle formula for sine but three useful double angle formulas for cosine. The three are easily from each other using the fundamental identity

\sin^2(\alpha) + \cos^2(\alpha) = 1

\qquad \sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)

\qquad\begin{aligned} \cos(2\alpha) &= \cos^2(\alpha) - \sin^2(\alpha)\\&= 1 - 2\sin^2(\alpha)\\&= 2\cos^2(\alpha) - 1\end{aligned}

In additon, double angle formulas can be expressed in terms of a tangent:

\qquad \sin(2\alpha) = \frac{2\tan(\alpha)}{1 + \tan^2(\alpha)}

\qquad \cos(2\alpha) = \frac{2\tan(\alpha)}{1 + \tan^2(\alpha)}

\qquad \tan(2\alpha) = \frac{2\tan\alpha}{1 - \tan^2\alpha}

>>

Anonymous June 16, 2022 (Thu) 16:59:44 No. 194

Triple angle formulas:

\qquad \sin(3\alpha) = 3\sin(\alpha) - 4\sin^3(\alpha)

\qquad \cos(3\alpha) = 4\cos(\alpha) - 3\cos(\alpha)

>What about quadruple and quintuple angle formulas?
Coefficients for n-angle formulas are the coefficients of Chebyshev polynomials.
Specifically Chebyshev polynomials of the first kind for sine, and Chebyshev polynomials of the second kind for cosine.

Triple angle formula for tangnet:

\qquad \tan(3\alpha) = \frac{3\tan(\alpha) - \tan^3{\alpha}}{1 - 3\tan^2{\alpha}}

>>

Anonymous June 16, 2022 (Thu) 17:00:18 No. 195

Product to sum formulas

By summing formulas

\sin(\alpha+\beta)

and

\sin(\alpha-\beta)

:

\qquad 2\sin(\alpha)\cos(\beta) = \sin(\alpha + \beta) + \sin(\alpha - \beta)

\qquad 2\cos(\alpha)\sin(\beta) = \sin(\alpha + \beta) - \sin(\alpha - \beta)

By summing formulas

\cos(\alpha+\beta)

and

\cos(\alpha-\beta)

:

\qquad 2\sin(\alpha)\sin(\beta) = \cos(\alpha - \beta) - \cos(\alpha + \beta)

\qquad 2\cos(\alpha)\cos(\beta) = \cos(\alpha - \beta) + \cos(\alpha + \beta)

Tangent formula can be obtained by dividing the previous two formulas

\qquad\tan(\alpha)\tan{\beta} = \frac{\cos(\alpha - \beta) - \cos(\alpha + \beta)}{\cos(\alpha - \beta) + \cos(\alpha + \beta)}

>>

Anonymous June 16, 2022 (Thu) 17:00:56 No. 196

Sum to product formulas

These are obtained by setting

x = \alpha + \beta,\enspace y = \alpha - \beta

, solving

\alpha = \frac{x + y}{2},\enspace\beta=\frac{x - y}{2}

then substituting this in the "product-to-sum" formulas above.
After renaming

x, y

back to

\alpha,\beta

:

\qquad \sin(\alpha) + \sin(\beta) = 2\sin\left(\frac{\alpha + \beta}{2}\right)\cos\left(\frac{\alpha - \beta}{2}\right)

\qquad \sin(\alpha) - \sin(\beta) = 2\sin\left(\frac{\alpha - \beta}{2}\right)\cos\left(\frac{\alpha + \beta}{2}\right)

\qquad \cos(\alpha) + \cos(\beta) = 2\cos\left(\frac{\alpha + \beta}{2}\right)\cos\left(\frac{\alpha - \beta}{2}\right)

\qquad \cos(\alpha) - \cos(\beta) = -2\sin\left(\frac{\alpha + \beta}{2}\right)\sin\left(\frac{\alpha - \beta}{2}\right)

Tangent formula can be obtained in the following way:

\qquad \tan(\alpha) + \tan(\beta) = \frac{\sin(\alpha)}{\cos(\alpha)} + \frac{\sin(\beta)}{\cos(\beta)} = \frac{\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)} = \frac{\sin(\alpha + \beta)}{\cos(\alpha)\cos(\beta)}

>>

Anonymous June 16, 2022 (Thu) 17:01:28 No. 197

Taylor series of trigonometric functions

Sine and cosine have the following Maclaurin series (which are Taylor series around

x_0 = 0)

:

\qquad \sin(x) = \sum_{n=0}^\infty\frac{(-1)^n}{(2n + 1)!}x^{2n + 1}

\qquad \cos(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}

This is the same as writing:

\qquad \sin(x) = x -\frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} -\frac{x^{11}}{11!} +\dots

\qquad \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \frac{x^{10}}{10}+\dots

By summing the Maclaurin series of

\cos(x)

and

i\sin(x)

it's possible to prove Euler's formula

e^{ix} = \cos(x) + i\sin(x)

\qquad \cos(x) + i\sin(x) = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \dots = e^{ix}

Maclaurin series can be used to approximate trigonometric functions to an arbitrary degree
For small angles (i.e.

\alpha\approx 0

), good approximations (often used in physics) are:

\qquad \sin(\alpha)\approx x

\qquad \cos(\alpha)\approx 1

Which are just Maclaurin series with only 1 term.

>>

Anonymous June 16, 2022 (Thu) 17:02:01 No. 198 >>211

Computation of trigonometric functions

Maclaurin series lend an easy way to numerically compute.
We need a helper function

ifac(n)

that computes

n!,\; n\in\mathbb{N}

and a helper function

fpow(x, n)

that computes

x^n,\;x\in\mathbb{R}, n\in\mathbb{N}

:

int ifac(int n) {

    return (n > 0) ? (n * ifac(n - 1)) : 1;

}

float fpow(float x, int n) {

    return (n < 0) ? fpow(1/x, -n)

        : (n == 0) ? 1

        : (n == 1) ? x

        : (n > 1)  ? x * fpow(x, n - 1)

        : 1;

}

Now

\sin(x)

can be computed with

fsin(x, prec)

:

float fsin(float x, int termCount) {

    int n, sign = 1;

    float numer  = x,

          denom  = 1,

          result = sign * (numer / denom);



    for (n = 1; n < termCount; n++)  {

        sign *= -1;

        numer = fpow(x, 2*n+1);

        denom = ifac(2*n+1);

        result += sign * (numer / denom);

    }



    return result;

}

While

\cos(x)

can be computed with

fcos(x, prec)

:

float fcos(float x, int termCount) {

    int n, sign = 1;

    float numer = 1,

          denom = 1,

          result = sign * (numer / denom);

    for (n = 1; n < termCount; n++)  {

        sign *= -1;

        numer = fpow(x, 2*n);

        denom = ifac(2*n);

        result += sign * (numer / denom);

    }

    return result;

}

While Maclaurin series for

\sin(x)

and

\cos(x)

converge on entire

\mathbb{R}

, they converge the fastest when

x\in(-\pi,\pi]

so its good to bound

x

to that interval considering trigonometric functions are periodic. In fact, only 5-15 are iterations are usually good enough in double-precision floating point precision.

>>

Anonymous June 23, 2022 (Thu) 18:40:06 No. 211

File: 8ce0ff601f99fdbf275b63c9b3b1944e.jpg ( 172.93 KB , 1040x1136 , 1656009606752.jpg )

>>198
You touched on what nobody actually mentions, how these functions are computed. Excellent exposition.

>>

Anonymous September 04, 2022 (Sun) 09:48:21 No. 289

File: 1648530726809.png ( 25.95 KB , 644x800 , 1662284901464.png )

gemmy though, thanks for the full info

>>

Anonymous September 04, 2022 (Sun) 21:57:36 No. 292

great thread