3.4.3: Simplifying Trigonometric Expressions with Double-Angle Identities
- Page ID
- 4223
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.
\(\begin{aligned} &\cos x+2\sin ^2\dfrac{x}{2} \\ &\cos x+2\left(\sqrt{\dfrac{1−\cos x}{2}}\right)^2 \\ &\cos x+2\cdot \dfrac{1−\cos x}{2} \\ &\cos x+1−\cos x \\ &1 \end{aligned}\)
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.
\(\begin{aligned}\dfrac{\tan 2x}{\dfrac{\tan x}{1+\tan x}}&=\dfrac{2\tan x}{1−\tan ^2x}\cdot \dfrac{1+\tan x}{\tan x} \\&=\dfrac{2\tan x}{(1+\tan x)(1−\tan x)}\cdot \dfrac{1+\tan x}{\tan x}=\dfrac{2}{1−\tan x}\end{aligned}\)
Simplify \(\dfrac{\sin 2x}{\sin x}\).
Solution
\(\dfrac{\sin 2x}{\sin x}=\dfrac{2\sin x\cos x}{\sin x}=2\cos x\)
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.
- \(\sqrt{2+2\cos x} \left(\cos \dfrac{x}{2}\right)\)
- \(\dfrac{\cos 2x}{\cos ^2x}\)
- \(\tan 2x(1+\tan x)\)
- \(\cos 2x−3\sin ^2x\)
- \(\dfrac{1+\cos 2x}{\cot x}\)
- \((1+\cos x)^2 \tan \dfrac{x}{2}\)
Verify the following identities.
- \(\cot \dfrac{x}{2}=\dfrac{\sin x}{1−\cos x}\)
- \(\dfrac{\sin x}{1+\cos x}=\dfrac{1−\cos x}{\sin x}\)
- \(\dfrac{\sin 2x}{1+\cos 2x}=\tan x\)
- \((\sin x+\cos x)^2=1+\sin 2x\)
- \(\sin x\tan \dfrac{x}{2}+2\cos x=2\cos ^2 \dfrac{x}{2}\)
- \(\cot x+\tan x=2\csc 2x\)
- \(\cos 3x=4\cos ^3x−3\cos x\)
- \(\cos 3x=\cos ^3x−3\sin ^2x\cos x\)
- \(\sin 2x−\tan x=\tan x\cos 2x\)
- \(\cos ^4x−\sin ^4x=\cos 2x\)
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 14.16.
Additional Resources
Practice: Simplifying Trigonometric Expressions with Double-Angle Identities