Trigonometric Identities

Lesson · Intermediate

Trigonometry

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

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

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

\sin 2\alpha = 2\sin\alpha\cos\alpha

\cos 2\alpha = \cos^2\alpha-\sin^2\alpha = 2\cos^2\alpha-1 = 1-2\sin^2\alpha

\cos 3\alpha = 4\cos^3\alpha-3\cos\alpha

\sin 3\alpha = 3\sin\alpha-4\sin^3\alpha

\tan x = \frac{1-\cos 2x}{\sin 2x}

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

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

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

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

\sin x+\cos x = \sqrt2\sin(x+45^\circ)

\sin x-\cos x = \sqrt2\sin(x-45^\circ)

\tan x+\cot x=\frac{2}{\sin 2x}

On the unit circle, the point corresponding to α is

(\cos\alpha,\sin\alpha)

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

By Euler's formula,

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

Since

e^{i(\alpha+\beta)} = e^{i\alpha}e^{i\beta}

we obtain

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

Equating real and imaginary parts gives

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

and

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

Setting

\beta=\alpha

we obtain

\sin2\alpha = 2\sin\alpha\cos\alpha

and

\cos2\alpha = \cos^2\alpha-\sin^2\alpha

Using

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

we also get

\cos2\alpha = 2\cos^2\alpha-1 = 1-2\sin^2\alpha

Next,

\begin{aligned} \cos3\alpha &= \cos(\alpha+2\alpha)\\ &= \cos\alpha\cos2\alpha-\sin\alpha\sin2\alpha\\ &= \cos\alpha(2\cos^2\alpha-1) - 2\sin^2\alpha\cos\alpha\\ &= 2\cos^3\alpha-\cos\alpha - 2(1-\cos^2\alpha)\cos\alpha\\ &= 4\cos^3\alpha-3\cos\alpha \end{aligned}

For sine,

\begin{aligned} \sin3\alpha &= \sin(\alpha+2\alpha)\\ &= \sin\alpha\cos2\alpha+\cos\alpha\sin2\alpha\\ &= \sin\alpha(1-2\sin^2\alpha) + \cos\alpha(2\sin\alpha\cos\alpha)\\ &= \sin\alpha-2\sin^3\alpha + 2\sin\alpha\cos^2\alpha\\ &= \sin\alpha-2\sin^3\alpha + 2\sin\alpha(1-\sin^2\alpha)\\ &= \sin\alpha-2\sin^3\alpha + 2\sin\alpha-2\sin^3\alpha\\ &= 3\sin\alpha-4\sin^3\alpha \end{aligned}

For tangent,

\begin{aligned} \frac{1-\cos2x}{\sin2x} &= \frac{1-(1-2\sin^2x)} {2\sin x\cos x}\\ &= \frac{2\sin^2x} {2\sin x\cos x}\\ &= \frac{\sin x}{\cos x}\\ &= \tan x \end{aligned}

For tangent,

\begin{aligned} \tan(\alpha+\beta) &= \frac{\sin(\alpha+\beta)} {\cos(\alpha+\beta)}\\ &= \frac{ \sin\alpha\cos\beta+\cos\alpha\sin\beta }{ \cos\alpha\cos\beta-\sin\alpha\sin\beta } \end{aligned}

Dividing numerator and denominator by

\cos\alpha\cos\beta

we obtain

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

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

Since

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

we get

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

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

Finally,

\begin{aligned} \sqrt2\sin(x+45^\circ) &= \sqrt2(\sin x\cos45^\circ+\cos x\sin45^\circ)\\ &= \sin x+\cos x \end{aligned}

and

\begin{aligned} \sqrt2\sin(x-45^\circ) &= \sqrt2(\sin x\cos45^\circ-\cos x\sin45^\circ)\\ &= \sin x-\cos x \end{aligned}

\tan x+\cot x = \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}

= \frac{\sin^2x+\cos^2x}{\cos x\sin x}

= \frac{1}{\cos x\sin x}

= \frac{2}{2\cos x\sin x} = \frac{2}{\sin 2x}

Thus all the stated identities follow