# 3.4.3: Simplifying Trigonometric Expressions with Double-Angle Identities

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Simplify sine, cosine, and tangent of angles multiplied or divided by 2.

As Agent Trigonometry, you are given the following cryptic clue. How could you simplify this clue?

$$\dfrac{\tan 2x}{\dfrac{\tan x}{1+\tan x}}$$

### Simplifying Trigonometric Expressions

We can also use the double-angle and half-angle formulas to simplify trigonometric expressions.

Let's simplify $$\dfrac{\cos 2x}{\sin x\cos x}$$.

Use $$\cos 2a=\cos ^2a−\sin ^2a$$ and then factor.

\begin{aligned} \dfrac{\cos 2x}{\sin x\cos x}&=\dfrac{\cos ^2x−\sin ^2x}{\sin x+\cos x} \\&=\dfrac{(\cos x−\sin x)\cancel{(\cos x+\sin x)}}{\cancel{\sin x+\cos x }}\\&=\cos x−\sin x \end{aligned}

Now, let's find the formula for $$\sin 3x$$.

You will need to use the sum formula and the double-angle formula. \sin 3x=\sin (2x+x)

\begin{aligned} \sin 3x&=\sin (2x+x) \\&=\sin 2x\cos x+\cos 2x\sin x \\&=2\sin x\cos x\cos x+\sin x(2\cos ^2x−1) \\ &=2\sin x\cos ^2x+2\sin x\cos ^2x−\sin x \\&=4\sin x\cos ^2x−\sin x \\&=\sin x(4\cos ^2x−1) \end{aligned}

Finally, let's verify the identity $$\cos x+2\sin ^2 \dfrac{x}{2}=1$$.

Simplify the left-hand side use the half-angle formula.

\begin{aligned} &\cos x+2\sin ^2\dfrac{x}{2} \\ &\cos x+2\left(\sqrt{\dfrac{1−\cos x}{2}}\right)^2 \\ &\cos x+2\cdot \dfrac{1−\cos x}{2} \\ &\cos x+1−\cos x \\ &1 \end{aligned}

##### Example $$\PageIndex{1}$$

Earlier, you were asked to simplify $$\dfrac{\tan 2x}{\dfrac{\tan x}{1+\tan x}}$$.

Solution

Use $$\tan 2a=\dfrac{2\tan a}{1−\tan ^2a}$$ and then factor.

\begin{aligned}\dfrac{\tan 2x}{\dfrac{\tan x}{1+\tan x}}&=\dfrac{2\tan x}{1−\tan ^2x}\cdot \dfrac{1+\tan x}{\tan x} \\&=\dfrac{2\tan x}{(1+\tan x)(1−\tan x)}\cdot \dfrac{1+\tan x}{\tan x}=\dfrac{2}{1−\tan x}\end{aligned}

##### Example $$\PageIndex{2}$$

Simplify $$\dfrac{\sin 2x}{\sin x}$$.

Solution

$$\dfrac{\sin 2x}{\sin x}=\dfrac{2\sin x\cos x}{\sin x}=2\cos x$$

##### Example $$\PageIndex{3}$$

Verify $$\cos x+2\cos ^2 \dfrac{x}{2}=1+2\cos x$$.

Solution

\begin{aligned} \cos x+2\cos ^2\dfrac{x}{2}&=1+2\cos x \\ \cos x+2 \sqrt{\dfrac{1+\cos x}{2}}&= \\ \cos x+1+\cos x&= \\ 1+2\cos x&= \end{aligned}

## Review

Simplify the following expressions.

1. $$\sqrt{2+2\cos x} \left(\cos \dfrac{x}{2}\right)$$
2. $$\dfrac{\cos 2x}{\cos ^2x}$$
3. $$\tan 2x(1+\tan x)$$
4. $$\cos 2x−3\sin ^2x$$
5. $$\dfrac{1+\cos 2x}{\cot x}$$
6. $$(1+\cos x)^2 \tan \dfrac{x}{2}$$

Verify the following identities.

1. $$\cot \dfrac{x}{2}=\dfrac{\sin x}{1−\cos x}$$
2. $$\dfrac{\sin x}{1+\cos x}=\dfrac{1−\cos x}{\sin x}$$
3. $$\dfrac{\sin 2x}{1+\cos 2x}=\tan x$$
4. $$(\sin x+\cos x)^2=1+\sin 2x$$
5. $$\sin x\tan \dfrac{x}{2}+2\cos x=2\cos ^2 \dfrac{x}{2}$$
6. $$\cot x+\tan x=2\csc 2x$$
7. $$\cos 3x=4\cos ^3x−3\cos x$$
8. $$\cos 3x=\cos ^3x−3\sin ^2x\cos x$$
9. $$\sin 2x−\tan x=\tan x\cos 2x$$
10. $$\cos ^4x−\sin ^4x=\cos 2x$$