Convert \( 150^\circ \) to radians.
\( \dfrac{5\pi}{6} \).
Radians & Further Trigonometry
10 exam-style questions on radians, arc length, sector area and further trig identities/equations. Tap Show answer for full working.
3 Easy4 Medium3 Hard
A sector of a circle of radius 6 cm has angle \( \dfrac{\pi}{3} \) radians. Find the arc length and the area of the sector.
Arc \( = r\theta = 6\cdot\pi/3=2\pi \) cm. Area \( =\dfrac12 r^2\theta=\dfrac12\cdot36\cdot\pi/3=6\pi \) cm².
Solve \( \sin\theta = \dfrac{1}{2} \) for \( 0 \leq \theta \leq 2\pi \).
\( \theta=\pi/6 \) or \( 5\pi/6 \).
Prove that \( \sin 2\theta = 2\sin\theta\cos\theta \).
Using compound angle: \( \sin(A+B)=\sin A\cos B+\cos A\sin B \) with \( A=B=\theta \).
Solve \( 2\cos^2\theta - \cos\theta - 1 = 0 \) for \( 0 \leq \theta < 2\pi \).
\( (2\cos\theta+1)(\cos\theta-1)=0 \) ⇒ \( \cos\theta=-1/2 \) or \( 1 \). \( \theta=0, 2\pi/3, 4\pi/3 \).
Express \( 3\sin\theta + 4\cos\theta \) in the form \( R\sin(\theta+\alpha) \) where \( R>0 \) and \( 0<\alpha<\pi/2 \).
\( R=5 \), \( \tan\alpha=4/3 \). \( 5\sin(\theta+\alpha) \).
Find the exact value of \( \sin\dfrac{\pi}{12} \).
\( \sin(\pi/3 - \pi/4)=\sin\pi/3\cos\pi/4-\cos\pi/3\sin\pi/4=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{6}-\sqrt{2}}{4} \).
Solve \( \tan 2\theta = \sqrt{3} \) for \( 0 \leq \theta \leq \pi \).
\( 2\theta=\pi/3, 4\pi/3 \) (in \( [0,2\pi] \)) ⇒ \( \theta=\pi/6 \) or \( 2\pi/3 \). (Also \( 2\theta=\pi/3+2\pi \) out of range for \( \theta\leq\pi \).)
A sector of a circle of radius \( r \) has perimeter 20 cm. Show that the area \( A \) of the sector is given by \( A=10r - r^2 \) and find the maximum area.
Perimeter \( 2r+r\theta=20 \) ⇒ \( \theta=(20-2r)/r \). \( A=\dfrac12 r^2\theta=\dfrac12 r(20-2r)=10r-r^2 \). \( \dfrac{dA}{dr}=10-2r=0 \) ⇒ \( r=5 \). Max area \( 25 \) cm².
Prove that \( \dfrac{\sin 2\theta}{1+\cos 2\theta} = \tan\theta \) and hence solve \( \dfrac{\sin 2\theta}{1+\cos 2\theta} = \sqrt{3} \) for \( 0<\theta<\dfrac{\pi}{2} \).
LHS \( =\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}=\tan\theta \). So \( \tan\theta=\sqrt{3} \) ⇒ \( \theta=\pi/3 \).