Trig: Past Paper Style (Non-GDC)
IB Mathematics AA · SL & HL · Shadow Worksheet
Practice
Question 1
Without a calculator, find the exact value of:
(a)\(\sin\dfrac{5\pi}{6}\)
(b)\(\cos\dfrac{7\pi}{4}\)
(c)\(\tan\dfrac{4\pi}{3}\)
(b)
Answer
\(\tfrac{\sqrt{2}}{2}\)
Question 2
Prove that \(\dfrac{\sin\theta+\sin 2\theta}{1+\cos\theta+\cos 2\theta}=\tan\theta\).
Substitute double angle formulae
Numerator: \(\sin\theta+2\sin\theta\cos\theta=\sin\theta(1+2\cos\theta)\)
Denominator: \(1+\cos\theta+2\cos^2\theta-1=\cos\theta+2\cos^2\theta=\cos\theta(1+2\cos\theta)\)
Simplify
\(\dfrac{\sin\theta(1+2\cos\theta)}{\cos\theta(1+2\cos\theta)}=\tan\theta\) ✓
Question 3
The angle \(\alpha\) satisfies \(\cos\alpha=\dfrac{1}{3}\) and \(0<\alpha<\dfrac{\pi}{2}\). Without a calculator find \(\sin 2\alpha\) and \(\cos 2\alpha\).
Find \(\sin\alpha\)
\(\sin\alpha=\sqrt{1-\tfrac{1}{9}}=\dfrac{2\sqrt{2}}{3}\)
Double angles
\(\sin2\alpha=2\cdot\tfrac{2\sqrt2}{3}\cdot\tfrac{1}{3}=\dfrac{4\sqrt2}{9}\)
Answer
\(\sin2\alpha=\dfrac{4\sqrt{2}}{9};\quad\cos2\alpha=2\cdot\tfrac{1}{9}-1=-\dfrac{7}{9}\)
Question 4
Prove that \(\sin^2\!\left(\dfrac{\pi}{4}+x\right)-\sin^2\!\left(\dfrac{\pi}{4}-x\right)=\sin 2x\).
Use \(\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)\)
\(\sin\!\left(\tfrac{\pi}{2}\right)\sin(2x)=1\cdot\sin2x=\sin2x\) ✓
Generated by Mathski · mathski.io · IB Mathematics AA Shadow Worksheets