Trigonometric Equations - Double Angle Formula

The trigonometric double angle formulas give a relationship between the basic trigonometric functions applied to twice an angle in terms of trigonometric functions of the angle itself.

Double angle formulas:

We have

$$\begin{aligned} \sin 2 x &= 2\sin x \cos x \\ \\ \cos 2x &= \cos^2 x - \sin^2 x \\ &= 2\cos^2 x - 1 \\ &= 1 - 2\sin^2 x \\ &= \frac{1 - \tan^2 x}{1 + \tan^2 x} \\ \\ \tan 2x &= \frac{2\tan x}{1 - \tan^2 x}. \ _\square \end{aligned}$$

We can prove the double angle identities using the sum formulas for sine and cosine:

$$\begin{aligned} \sin 2x &= \sin(x+x)\\ &= \sin x \cos x + \cos x \sin x\\ &= 2 \sin x \cos x\\ \\ \cos 2x &= \cos(x+x)\\ &= \cos x\cos x - \sin x \sin x\\ &= \cos^2 x - \sin^2 x. \ _\square \end{aligned}$$

From these formulas, we also have the following identities:

$$\begin{aligned} \sin^2 x&=\frac{1}{2}(1-\cos 2x)\\ \cos^2 x&=\frac{1}{2}(1+\cos 2x)\\ \sin x \cos x&=\frac{1}{2}(\sin 2x)\\ \tan^2 x&=\frac{1-\cos 2x}{1+\cos 2x}. \end{aligned}$$

We have

$$\begin{aligned}\sinh 2x&=2\sinh x \cosh x\\ \\ \cosh 2x &=\cosh^2 x+\sinh^2 x\\ &=2\cosh^2 x-1\\ &=2\sinh^2 x+1\\ \\ \tanh 2x &=\frac{2\tanh x}{1+\tanh^2 x}.\end{aligned}$$

Prove the identity $$$$

$$\begin{aligned}(\sin x + \cos x)^2=1+\sin 2x.\end{aligned}$$

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We have

$$\begin{aligned} (\sin x+\cos x)^2 &=\sin^2 x+\cos^2 x+2\sin x \cos x \\ &=1+2\sin x \cos x \\ &=1+\sin 2x. \ _\square \end{aligned}$$

Prove the identity $$$$

$$\sin 2x=\tan x (1+\cos 2x).$$

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We have

$$\begin{aligned} \frac{\sin x}{\cos x}(1+\cos 2x) &=\frac{\sin x}{\cos x}\left(1+2\cos^2 x-1\right)\\ &=\frac{\sin x}{\cos x}(2\cos^2 x)\\ &=2\sin x\cos x \\ &=\sin 2x. \ _\square \end{aligned}$$

Prove the identity $$$$

$$\cos 2x=\frac{\cot x-\tan x}{\cot x+\tan x}.$$

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We have

$$\begin{aligned} \frac{\cot x-\tan x}{\cot x+\tan x} &= \frac{\frac{\cos x}{\sin x} -\frac{\sin x}{\cos x}}{\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}}\\ &=\frac{\cos^2 x-\sin^2 x}{\cos^2 x+\sin^2 x}\\ &=\cos 2x. \ _\square \end{aligned}$$

Here goes some important terminology:

• versed sine $x:$ $\text{vers}(x) = 1 - \cos x = 2\sin^2 \frac{x}{2}$
• versed cosine $x:$ $\text{vercos}(x)= 1 + \cos x = 2\cos^2\frac{x}{2}$
• coversed sine $x:$ $\text{covers}(x)= 1 - \sin x = \left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)^2$
• coversed cosine $x:$ $\text{covercos}(x)= 1 + \sin x = \left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^2$
• $\text{chord}(x)= \text{crd}(x) = 2\sin\frac{x}{2}.$