Find all acute angles $θ$ (in degrees) satisfying $sin θ = cos 2 θ$ .

hard 4 marks Class 10 · Mathematics · Introduction to Trigonometry

Answer

3 0^{\circ}

The correct answer is $3 0^{\circ}$ — Check: $cos 2 θ = cos 6 0^{\circ} = \frac{1}{2} = sin 3 0^{\circ}$ , confirming the solution.

Step-by-step solution

Rewrite $cos 2 θ$ using the identity $cos 2 θ = 1 - 2 sin^{2} θ$ .
Set up the equation: $sin θ = 1 - 2 sin^{2} θ$ .
Bring all terms to one side: $2 sin^{2} θ + sin θ - 1 = 0$ .
Treat $sin θ$ as a variable $x$ and solve the quadratic $2 x^{2} + x - 1 = 0$ .
Factor or use the quadratic formula: $x = \frac{- 1 \pm 1 + 8}{4} = \frac{- 1 \pm 3}{4}$ .
Thus $x = \frac{1}{2}$ or $x = - 1$ . Since $θ$ is acute, $sin θ$ must be positive, so $sin θ = \frac{1}{2}$ .
The acute angle whose sine is $\frac{1}{2}$ is $θ = 3 0^{\circ}$ .
Check: $cos 2 θ = cos 6 0^{\circ} = \frac{1}{2} = sin 3 0^{\circ}$ , confirming the solution.

Common mistake: Using

cos 2 θ = 2 cos^{2} θ - 1

and not converting to a single

sin θ

term, leading to a more complicated equation.

Concept tested: Trigonometric identities and solving equations