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)\( an\dfrac{4\pi}{3}\)
(b)
Answer
\( frac{\sqrt{2}}{2}\)
Question 2
Prove that \(\dfrac{\sin heta+\sin 2 heta}{1+\cos heta+\cos 2 heta}= an heta\).
Substitute double angle formulae
Numerator: \(\sin heta+2\sin heta\cos heta=\sin heta(1+2\cos heta)\)
Denominator: \(1+\cos heta+2\cos^2 heta-1=\cos heta+2\cos^2 heta=\cos heta(1+2\cos heta)\)
Simplify
\(\dfrac{\sin heta(1+2\cos heta)}{\cos heta(1+2\cos heta)}= an heta\) ✓
Question 3
The angle \(lpha\) satisfies \(\coslpha=\dfrac{1}{3}\) and \(0<lpha<\dfrac{\pi}{2}\). Without a calculator find \(\sin 2lpha\) and \(\cos 2lpha\).
Find \(\sinlpha\)
\(\sinlpha=\sqrt{1- frac{1}{9}}=\dfrac{2\sqrt{2}}{3}\)
Double angles
\(\sin2lpha=2\cdot frac{2\sqrt2}{3}\cdot frac{1}{3}=\dfrac{4\sqrt2}{9}\)
Answer
\(\sin2lpha=\dfrac{4\sqrt{2}}{9};\quad\cos2lpha=2\cdot frac{1}{9}-1=-\dfrac{7}{9}\)
Question 4
Prove that \(\sin^2\!\left(\dfrac{\pi}{4}+x
ight)-\sin^2\!\left(\dfrac{\pi}{4}-x
ight)=\sin 2x\).
Use \(\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)\)
\(\sin\!\left( frac{\pi}{2}
ight)\sin(2x)=1\cdot\sin2x=\sin2x\) ✓
Generated by Mathski · mathski.io · IB Mathematics AA Shadow Worksheets