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Simplifying Trigonometric Expressions with Double-Angle Identities

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

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Simplifying Trig Expressions using Double and Half Angle Formulas

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

\begin{align*}\frac{\tan 2x}{\frac{tan x}{1 + \tan x}}\end{align*}

Simplifying Trigonometric Expressions

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

Simplify using the double angle and half angle formulas

Simplify \begin{align*}\frac{\cos 2x}{\sin x \cos x}\end{align*}.

Use \begin{align*}\cos 2a=\cos^2a-\sin^2a\end{align*} and then factor.

\begin{align*}\frac{\cos 2x}{\sin x \cos x}&=\frac{\cos^2x- \sin^2x}{\sin x+ \cos x} \\ &=\frac{\left(\cos x- \sin x\right) \cancel{\left(\cos x + \sin x\right)}}{\cancel{\sin x+ \cos x}} \\ &=\cos x- \sin x\end{align*}

Find the formula for \begin{align*}\sin 3x\end{align*}.

You will need to use the sum formula and the double-angle formula. \begin{align*}\sin 3x=\sin(2x+x)\end{align*}

\begin{align*}\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^2 x- \sin x \\ &=4 \sin x \cos^2x- \sin x \\ &=\sin x(4 \cos^2x-1)\end{align*}

We will explore other possibilities for the \begin{align*}\sin 3x\end{align*} because of the different formulas for \begin{align*}\cos 2a\end{align*} in the Problem Set.

Verify the identity \begin{align*}\cos x+2 \sin^2 \frac{x}{2}=1\end{align*}.

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

\begin{align*}& \cos x+2 \sin^2 \frac{x}{2} \\ & \cos x+2 \left(\sqrt{\frac{1- \cos x}{2}}\right)^2 \\ & \cos x+2 \cdot \frac{1- \cos x}{2} \\ & \cos x+1- \cos x \\ & 1\end{align*}

Examples

Example 1

Earlier, you were asked how could you simplify \begin{align*}\frac{\tan 2x}{\frac{tan x}{1 + \tan x}}\end{align*}

Use \begin{align*}\tan 2a=\frac {2\tan a}{1 - \tan^2 a}\end{align*} and then factor.

\begin{align*}\frac{\tan 2x}{\frac{tan x}{1 + \tan x}}=\frac{2\tan x}{1-\tan^2 x}\cdot \frac{1 + \tan x}{tan x} \\ =\frac{2\tan x}{(1 + \tan x)(1 - \tan x)}\cdot \frac{1 + \tan x}{tan x} =\frac{2}{1-\tan x}\end{align*}

Example 2

Simplify \begin{align*}\frac{\sin 2x}{\sin x}\end{align*}.

\begin{align*}\frac{\sin 2x}{\sin x}=\frac{2 \sin x \cos x}{\sin x}=2 \cos x\end{align*}

Example 3

Verify \begin{align*}\cos x+2 \cos^2 \frac{x}{2}=1+ 2 \cos x\end{align*}.


\begin{align*}\cos x+2 \cos^2 \frac{x}{2}&=1+2 \cos x \\ \cos x+2 \sqrt{\frac{1+ \cos x}{2}}^2&= \\ \cos x+1 + \cos x&= \\ 1+2 \cos x&=\end{align*}

Review

Simplify the following expressions.

  1. \begin{align*}\sqrt{2+2 \cos x} \left(\cos \frac{x}{2}\right)\end{align*}
  2. \begin{align*}\frac{\cos 2x}{\cos^2x}\end{align*}
  3. \begin{align*}\tan 2x(1+ \tan x)\end{align*}
  4. \begin{align*}\cos 2x- 3 \sin^2x\end{align*}
  5. \begin{align*}\frac{1+\cos 2x}{\cot x}\end{align*}
  6. \begin{align*}(1+ \cos x)^2 \tan \frac{x}{2}\end{align*}

Verify the following identities.

  1. \begin{align*}\cot \frac{x}{2}=\frac{\sin x}{1- \cos x}\end{align*}
  2. \begin{align*}\frac{\sin x}{1+ \cos x}=\frac{1- \cos x}{\sin x}\end{align*}
  3. \begin{align*}\frac{\sin 2x}{1+ \cos 2x}= \tan x\end{align*}
  4. \begin{align*}(\sin x+ \cos x)^2=1+ \sin 2x\end{align*}
  5. \begin{align*}\sin x \tan \frac{x}{2}+2 \cos x=2 \cos^2 \frac{x}{2}\end{align*}
  6. \begin{align*}\cot x+ \tan x=2 \csc 2x\end{align*}
  7. \begin{align*}\cos 3x=4 \cos^3x-3 \cos x\end{align*}
  8. \begin{align*}\cos 3x=\cos^3x-3 \sin^2x \cos x\end{align*}
  9. \begin{align*}\sin 2x-\tan x=\tan x \cos 2x\end{align*}
  10. \begin{align*}\cos^4x-\sin^4x=\cos 2x\end{align*}

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.16. 

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