Given \( \sin\theta = 3/5 \) and \( \theta \) is acute, find \( \cos\theta \) and \( \tan\theta \).
\( \cos\theta = 4/5 \), \( \tan\theta = 3/4 \).
Trigonometry
12 exam-style questions on ratios, identities, equations and graphs. Tap Show answer for full working.
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Solve \( \sin x = 1/2 \) for \( 0^\circ \leq x \leq 360^\circ \).
\( x = 30^\circ \) or \( x = 150^\circ \).
Prove that \( \dfrac{\sin\theta}{\cos\theta} = \tan\theta \).
By definition, \( \tan\theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{\sin\theta}{\cos\theta} \).
Find the exact value of \( \cos 60^\circ + \sin 30^\circ \).
\( 1/2 + 1/2 = 1 \).
Solve \( 2\cos x = 1 \) for \( 0 \leq x \leq 2\pi \).
\( \cos x = 1/2 \) ⇒ \( x = \pi/3 \) or \( x = 5\pi/3 \).
Prove the identity \( 1 + \tan^2\theta = \sec^2\theta \).
Divide \( \sin^2\theta + \cos^2\theta = 1 \) by \( \cos^2\theta \): \( \tan^2\theta + 1 = \sec^2\theta \).
Solve \( \tan 2x = 1 \) for \( 0^\circ \leq x \leq 180^\circ \).
\( 2x = 45^\circ, 225^\circ \) (within \( 0 \) to \( 360 \)) ⇒ \( x = 22.5^\circ \) or \( 112.5^\circ \).
Express \( 3\sin\theta + 4\cos\theta \) in the form \( R\sin(\theta + \alpha) \) where \( R>0 \) and \( \alpha \) is acute.
\( R=\sqrt{9+16}=5 \). \( \cos\alpha=3/5 \), \( \sin\alpha=4/5 \), \( \alpha=\tan^{-1}(4/3) \). So \( 5\sin(\theta+\alpha) \).
Solve \( 2\sin^2 x - \sin x - 1 = 0 \) for \( 0 \leq x < 2\pi \).
Let \( u=\sin x \): \( 2u^2 - u - 1=0 \) ⇒ \( (2u+1)(u-1)=0 \) ⇒ \( u=-1/2 \) or \( 1 \). \( \sin x=1 \) ⇒ \( x=\pi/2 \); \( \sin x=-1/2 \) ⇒ \( x=7\pi/6 \) or \( 11\pi/6 \).
Prove that \( \dfrac{\cos\theta}{1+\sin\theta} + \dfrac{1+\sin\theta}{\cos\theta} = 2\sec\theta \).
LHS \( = \dfrac{\cos^2\theta + (1+\sin\theta)^2}{\cos\theta(1+\sin\theta)} = \dfrac{\cos^2\theta + 1 + 2\sin\theta + \sin^2\theta}{\cos\theta(1+\sin\theta)} = \dfrac{2 + 2\sin\theta}{\cos\theta(1+\sin\theta)} = \dfrac{2(1+\sin\theta)}{\cos\theta(1+\sin\theta)} = \dfrac{2}{\cos\theta} = 2\sec\theta \).
Solve \( \cos 2x = \sin x \) for \( 0 \leq x \leq 2\pi \).
\( 1-2\sin^2 x = \sin x \) ⇒ \( 2\sin^2 x + \sin x - 1 = 0 \) ⇒ \( (2\sin x - 1)(\sin x + 1)=0 \). \( \sin x=1/2 \) or \( -1 \). Solutions: \( x=\pi/6, 5\pi/6, 3\pi/2 \).
The triangle ABC has sides \( a=5 \), \( b=7 \), \( C=60^\circ \). Find the area of the triangle and the length of side \( c \).
Area \( = \dfrac12 ab\sin C = \dfrac12\cdot5\cdot7\cdot\sin60 = \dfrac{35\sqrt{3}}{4} \). Cosine rule: \( c^2 = 25+49-2\cdot5\cdot7\cdot\cos60 = 74-35=39 \) ⇒ \( c=\sqrt{39} \).