Double Angle Formulae
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
Using the double angle formulae, simplify:
(a)\(2\sin 3 heta\cos 3 heta\)
(b)\(\cos^2 2x - \sin^2 2x\)
(c)\(\dfrac{2 an 5 heta}{1- an^2 5 heta}\)
Question 2
Given \(\sin heta=\dfrac{3}{5}\) and \( heta\) is acute, find without a calculator:
(a)\(\cos heta\)
(b)\(\sin 2 heta\)
(c)\(\cos 2 heta\)
(a)
Answer
\(\cos heta=\dfrac{4}{5}\)
(b)
Formula
\(\sin2 heta=2\cdot frac{3}{5}\cdot frac{4}{5}\)
(c)
Formula
\(\cos2 heta=1-2\sin^2 heta=1- frac{18}{25}\)
Question 3
Prove the following identities:
(a)\(\dfrac{\sin 2 heta}{1+\cos 2 heta}= an heta\)
(b)\(\cos^4 heta - \sin^4 heta = \cos 2 heta\)
(a)
Substitute formulae
\(\dfrac{2\sin heta\cos heta}{1+(2\cos^2 heta-1)}=\dfrac{2\sin heta\cos heta}{2\cos^2 heta}=\dfrac{\sin heta}{\cos heta}= an heta\) ✓
(b)
Difference of squares
\((\cos^2 heta+\sin^2 heta)(\cos^2 heta-\sin^2 heta)=1\cdot\cos2 heta=\cos2 heta\) ✓
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
\(\sin x= frac{1}{2}\Rightarrow x= frac{\pi}{6}, frac{5\pi}{6}\); \(\sin x=-1\Rightarrow x= frac{3\pi}{2}\)
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 \(R\sin(x+lpha)\), giving exact values of \(R\) and \(lpha\).
Expand and compare
\(R\sin x\coslpha+R\cos x\sinlpha\Rightarrow R\coslpha=3,\;R\sinlpha=4\)
Find \(R\) and \(lpha\)
\(R=\sqrt{9+16}=5;\;lpha=rctan frac{4}{3}pprox0.927\)
Answer
\(5\sin(x+rctan frac{4}{3})\)
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