Double Angle Formulae
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
Using the double angle formulae, simplify:
(a)\(2\sin 3\theta\cos 3\theta\)
(b)\(\cos^2 2x - \sin^2 2x\)
(c)\(\dfrac{2\tan 5\theta}{1-\tan^2 5\theta}\)
Question 2
Given \(\sin\theta=\dfrac{3}{5}\) and \(\theta\) is acute, find without a calculator:
(a)\(\cos\theta\)
(b)\(\sin 2\theta\)
(c)\(\cos 2\theta\)
(a)
Answer
\(\cos\theta=\dfrac{4}{5}\)
(b)
Formula
\(\sin2\theta=2\cdot\tfrac{3}{5}\cdot\tfrac{4}{5}\)
(c)
Formula
\(\cos2\theta=1-2\sin^2\theta=1-\tfrac{18}{25}\)
Question 3
Prove the following identities:
(a)\(\dfrac{\sin 2\theta}{1+\cos 2\theta}=\tan\theta\)
(b)\(\cos^4\theta - \sin^4\theta = \cos 2\theta\)
(a)
Substitute formulae
\(\dfrac{2\sin\theta\cos\theta}{1+(2\cos^2\theta-1)}=\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}=\dfrac{\sin\theta}{\cos\theta}=\tan\theta\) ✓
(b)
Difference of squares
\((\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta)=1\cdot\cos2\theta=\cos2\theta\) ✓
Question 4
Solve \(\cos 2x = \sin x\) for \(0\leq x\leq2\pi\).
Substitute
\(1-2\sin^2x=\sin x\Rightarrow 2\sin^2x+\sin x-1=0\Rightarrow(2\sin x-1)(\sin x+1)=0\)
Solve
\(\sin x=\tfrac{1}{2}\Rightarrow x=\tfrac{\pi}{6},\tfrac{5\pi}{6}\); \(\sin x=-1\Rightarrow x=\tfrac{3\pi}{2}\)
Answer
\(x=\dfrac{\pi}{6},\;\dfrac{5\pi}{6},\;\dfrac{3\pi}{2}\)
Question 5
Express \(3\sin x + 4\cos x\) in the form \(R\sin(x+\alpha)\), giving exact values of \(R\) and \(\alpha\).
Expand and compare
\(R\sin x\cos\alpha+R\cos x\sin\alpha\Rightarrow R\cos\alpha=3,\;R\sin\alpha=4\)
Find \(R\) and \(\alpha\)
\(R=\sqrt{9+16}=5;\;\alpha=\arctan\tfrac{4}{3}\approx0.927\)
Answer
\(5\sin(x+\arctan\tfrac{4}{3})\)
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