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Toolkit 38

Trigonometric Identities

sin⁡2α+cos⁡2α=1\sin^2\alpha+\cos^2\alpha=1
sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta
cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta
sin⁡2α=2sin⁡αcos⁡α\sin 2\alpha=2\sin\alpha\cos\alpha
cos⁡2α=cos⁡2α−sin⁡2α=2cos⁡2α−1=1−2sin⁡2α\cos 2\alpha=\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha
cos⁡3α=4cos⁡3α−3cos⁡α\cos 3\alpha=4\cos^3\alpha-3\cos\alpha
tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}
cot⁡(α+β)=cot⁡αcot⁡β−1cot⁡α+cot⁡β\cot(\alpha+\beta)=\frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta}
sin⁡x+cos⁡x=2sin⁡(x+45∘)\sin x+\cos x=\sqrt{2}\sin(x+45^\circ)
sin⁡x−cos⁡x=2sin⁡(x−45∘)\sin x-\cos x=\sqrt{2}\sin(x-45^\circ)

Proof

On the unit circle, the point corresponding to α\alpha is

(cos⁡α,sin⁡α).(\cos\alpha, \sin\alpha).

By the Pythagorean theorem,

sin⁡2α+cos⁡2α=1.\sin^2\alpha + \cos^2\alpha = 1.

By Euler's formula,

eiθ=cos⁡θ+isin⁡θ.e^{i\theta} = \cos\theta + i\sin\theta.

Since ei(α+β)=eiαeiβe^{i(\alpha+\beta)} = e^{i\alpha}e^{i\beta}, we obtain

cos⁡(α+β)+isin⁡(α+β)=(cos⁡α+isin⁡α)(cos⁡β+isin⁡β)=cos⁡αcos⁡β−sin⁡αsin⁡β+i(sin⁡αcos⁡β+cos⁡αsin⁡β).\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⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta

and

sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β.\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta.

Setting β=α\beta = \alpha, we obtain

sin⁡2α=2sin⁡αcos⁡α\sin 2\alpha = 2\sin\alpha\cos\alpha

and

cos⁡2α=cos⁡2α−sin⁡2α.\cos 2\alpha = \cos^2\alpha - \sin^2\alpha.

Using sin⁡2α+cos⁡2α=1\sin^2\alpha + \cos^2\alpha = 1, we also get

cos⁡2α=2cos⁡2α−1=1−2sin⁡2α.\cos 2\alpha = 2\cos^2\alpha - 1 = 1 - 2\sin^2\alpha.

Next,

cos⁡3α=cos⁡(α+2α)=cos⁡αcos⁡2α−sin⁡αsin⁡2α=cos⁡α(2cos⁡2α−1)−2sin⁡2αcos⁡α=2cos⁡3α−cos⁡α−2(1−cos⁡2α)cos⁡α=4cos⁡3α−3cos⁡α.\begin{aligned} \cos 3\alpha &= \cos(\alpha+2\alpha) \\ &= \cos\alpha\cos 2\alpha - \sin\alpha\sin 2\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 tangent,

tan⁡(α+β)=sin⁡(α+β)cos⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡βcos⁡αcos⁡β−sin⁡αsin⁡β.\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⁡αcos⁡β\cos\alpha\cos\beta, we obtain

tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β.\tan(\alpha+\beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}.

Since cot⁡(α+β)=1/tan⁡(α+β)\cot(\alpha+\beta) = 1/\tan(\alpha+\beta), we get

cot⁡(α+β)=1−tan⁡αtan⁡βtan⁡α+tan⁡β=cot⁡αcot⁡β−1cot⁡α+cot⁡β.\cot(\alpha+\beta) = \frac{1-\tan\alpha\tan\beta}{\tan\alpha+\tan\beta} = \frac{\cot\alpha\cot\beta - 1}{\cot\alpha + \cot\beta}.

Finally,

2sin⁡(x+45∘)=2(sin⁡xcos⁡45∘+cos⁡xsin⁡45∘)=sin⁡x+cos⁡x,\begin{aligned} \sqrt{2}\sin(x+45^\circ) &= \sqrt{2}\left(\sin x\cos 45^\circ + \cos x\sin 45^\circ\right) \\ &= \sin x + \cos x, \end{aligned}

and

2sin⁡(x−45∘)=2(sin⁡xcos⁡45∘−cos⁡xsin⁡45∘)=sin⁡x−cos⁡x.\begin{aligned} \sqrt{2}\sin(x-45^\circ) &= \sqrt{2}\left(\sin x\cos 45^\circ - \cos x\sin 45^\circ\right) \\ &= \sin x - \cos x. \end{aligned}

Thus all the stated identities follow.

□\square
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