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Trigonometric identity

In mathematics, trigonometric identities are equalities involving trigonometric functions that are true for all values of the occurring variables. These identities are useful whenever expressions involving trigonometric functions need to be simplified. An important application is the integration of non-trigonometric functions: a common trick involves first using the substitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity.

Notation: With trigonometric functions, we define functions sin2, cos2, etc., such that sin2(x) = (sin(x))2.

From the Definitions

$\tan (x) = \frac {\sin (x)} {\cos(x)} \qquad \operatorname{cot}(x) = \frac{1} {\tan(x)}$

$\operatorname{sec}(x) = \frac{1} {\cos(x)} \qquad \operatorname{csc}(x) = \frac{1} {\sin(x)}$

Periodicity, Symmetry and Shifts

These are easiest proven from the unit circle:

$\sin(x) = \sin(x + 2\pi) \qquad \cos(x) = \cos(x + 2\pi) \qquad \tan(x) = \tan(x + \pi)$

$\sin(-x) = -\sin(x) \qquad \cos(-x) = \cos(x)$

$\tan(-x) = -\tan(x) \qquad \cot(-x) = -\cot(x)$

$\sin(x) = \cos\left(\frac{\pi}{2} - x\right)  \qquad \cos(x) = \sin\left(\frac{\pi}{2}-x\right) \qquad \tan(x) = \cot\left(\frac{\pi}{2} - x\right) $

For some purposes it is important to know that any linear combination of sine waves of the same period but different phase shifts is also a sine wave with the same period, but a different phase shift. In other words, we have

$a\sin x+b\cos x=\sqrt{a^2+b^2}\cdot\sin(x-\varphi)$
where
$\varphi=-{\rm arctan}(b/a).$

From the Pythagorean Theorem

$\sin^2(x) + \cos^2(x) = 1 \qquad \tan^2(x) + 1 = \sec^2(x) \qquad \cot^2(x) + 1 = \csc^2(x)$

The quickest way to prove these is Euler's formula. The tangent formula follows from the other two.

$\sin(x + y) = \sin(x) \cos(y) + \cos(x) \sin(y)$

$\cos(x + y) = \cos(x) \cos(y) - \sin(x) \sin(y)$

$\tan(x + y) = \frac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)}$

${\rm cis}(x+y)={\rm cis}(x)\,{\rm cis}(y)$
where
${\rm cis}(x)=\cos(x)+i\sin(x).$

Double-Angle Formulas

These can be shown by substituting $x = y$ in the addition theorems, and using the Pythagorean formula for the latter two. Or use de Moivre's formula with $n = 2$.

$\sin(2x) = 2 \sin (x) \cos(x)$

$\cos(2x) = \cos^2(x) - \sin^2(x) = 2 \cos^2(x) - 1 = 1 - 2 \sin^2(x)$

$\tan(2x) = \frac{2 \tan (x)} {1 - \tan^2(x)}$

Multiple-Angle Formulas

If Tn is the nth Chebyshev polynomial then

$\cos(nx)=T_n(\cos(x)).$
$\cos(nx)+i\sin(nx)=(\cos(x)+i\sin(x))^n$

The Dirichlet kernel Dn(x) is the function occuring on both sides of the next identity:

$1+2\cos(x)+2\cos(2x)+2\cos(3x)+\cdots+2\cos(nx)=\frac{\sin\left(\left(n+\frac{1}{2}\right)x\right)}{\sin(x/2)}$
The convolution of any square-integrable function of period 2π with the Dirichlet kernel coincides with the function's nth-degree Fourier approximation.

Power-Reduction Formulas

Solve the third and fourth double angle formula for cos2(x) and sin2(x).

$\cos^2(x) = {1 + \cos(2x) \over 2}$

$\sin^2(x) = {1 - \cos(2x) \over 2}$

Half-Angle Formulas

Substitute x/2 for x in the power reduction formulas, then solve for cos(x/2) and sin(x/2).

$\left|\cos\left(\frac{x}{2}\right)\right| = \sqrt{\left(\frac{1 + \cos(x)}{2}\right)}$

$\left|\sin\left(\frac{x}{2}\right)\right| = \sqrt{\left(\frac{1 - \cos(x)}{2}\right)}$

Multiply tan(x/2) by 2cos(x/2) / ( 2cos(x/2)) and substitute sin(x/2) / cos(x/2) for tan(x/2). The numerator is then sin(x) via the double angle formula, and the denominator is 2cos2(x/2) - 1 + 1 which is cos(x) + 1 by the double angle formulae. The second formula comes from the first formula multiplied by sin(x) / sin(x) and simplified using the pythagorean trig identity.

$\tan\left(\frac{x}{2}\right) = \frac{\sin(x)}{\cos(x) + 1} = \frac{1 - \cos(x)}{\sin(x)}$

Products to Sums

These can be proven by expanding their right-hand-sides using the addition theorems.

$\cos(x) \cos(y) = {\cos(x + y) + \cos(x - y) \over 2}$

$\sin(x) \sin(y) = {\cos(x - y) - \cos(x + y) \over 2}$

$\sin(x) \cos(y) = {\sin(x + y) + \sin(x - y) \over 2}$

Sums to Products

Replace x by (x + y) / 2 and y by (xy) / 2 in the Product-to-Sum formulas.

$\sin(x) + \sin(y) = 2 \sin\left( \frac{x + y}{2} \right) \cos\left( \frac{x - y}{2} \right)$
$\cos(x) + \cos(y) = 2 \cos\left( \frac{x + y}{2} \right) \cos\left( \frac{x - y}{2} \right)$

Inverse Trigonometric Functions

If x > 0 then

$\arctan(x)+\arctan(1/x)=\frac{\pi}{2}.$

If x < 0 then the right side of the equality is -π/2.

$\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)$

Several identities similar to the following can be derived from the Pythagorean theorem:

$\cos(\arcsin(x))=\sqrt{1-x^2}$

Identities with no variables

Richard Feynman is reputed to have learned as a boy, and always remembered, the following curious identity:

$\cos 20^\circ\cdot\cos 40^\circ\cdot\cos 80^\circ=1/8.$
However, this identity is a special case of one that does contain a variable:
$\prod_{j=0}^{k-1}\cos(2^j x)=\frac{\sin(2^k x)}{2^k\sin(x)}.$
The following are perhaps not as readily generalized to identities with variables in them:
$\cos 36^\circ+\cos 108^\circ=1/2.$
$\cos 24^\circ+\cos 48^\circ+\cos 96^\circ+\cos 168^\circ=1/2.$
Degree-measure ceases to be more felicitous than radian measure when we consider this identity with 21 in the denominators:
$\cos\frac{2\pi}{21}+\cos\frac{2(2\pi)}{21}+\cos\frac{4(2\pi)}{21}+\cos\frac{5(2\pi)}{21}+\cos\frac{8(2\pi)}{21}+\cos\frac{10(2\pi)}{21}=1/2.$
The factors 1, 2, 4, 5, 8, 10 may start to make the pattern clear: They are the integers less than 21/2 that have no prime factors in common with 21. The last several examples are corollaries of a basic fact about the irreducible cyclotomic polynomials; the cosines are the real parts of the zeroes of those polynomials; the sum of the zeroes is the Möbius function evaluated at (in the very last case above) 21; only half of the zeroes are present above.

Calculus

In calculus it is essential that angles that are arguments to trigonometric functions be measured in radians; if they are measured in degrees or any other units, then the relations stated below fail. If the trigonometric functions are defined in terms of geometry, then their derivatives can be found by first verifying that

$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1,$
and
$\lim_{x\rightarrow 0}\frac{1-\cos(x)}{x}=0,$
and then using the limit definition of the derivative and the addition theorems; if they are defined by their Taylor series, then the derivatives can be found by differentiating the power series term by term.

${d \over dx}\sin(x) = \cos(x)$

The rest of the trig functions can be differentiated using the above identities and the rules of differentiation, for instance

${d \over dx}\cos(x) = -\sin(x)$

${d \over dx}\tan(x) = \sec^2(x)$

${d \over dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}}$

${d \over dx}\arctan(x)=\frac{1}{1+x^2}$

The integral identities can be found in Wikipedia's table of integrals.

All Wikipedia text is available under the terms of the GNU Free Documentation License

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