3.4 Trigonometry (A-level only)

Prove that

$sin⁡2θ1+cos⁡2θ≡tan⁡θ\displaystyle \frac{\sin 2\theta}{1 + \cos 2\theta} \equiv \tan\theta$

A student is attempting to solve the equation

$sin⁡2θ1+cos⁡2θ=2sin⁡θ\displaystyle \frac{\sin 2\theta}{1 + \cos 2\theta} = 2\sin\theta$ for $\leq \theta \leq 360°\,$

They use the result from part (a), and write the following incorrect solution.

Step 1: $tan⁡θ=2sin⁡θ\tan\theta = 2\sin\theta$

Step 2: $sin⁡θcos⁡θ=2sin⁡θ\dfrac{\sin\theta}{\cos\theta} = 2\sin\theta$

Step 3: $1cos⁡θ=2\dfrac{1}{\cos\theta} = 2$

Step 4: $cos⁡θ=12\cos\theta = \dfrac{1}{2}$

Step 5: $θ=60°\theta = 60°$ , $300°$

[3]

b(i).

Explain the error the student has made between Step 2 and Step 3.

[2]

b(ii).

State the complete set of solutions of the equation for $\leq \theta \leq 360°$ .

[2]