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1 Pure Mathematics

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Question 73
a.

Prove that

sin⁡2θ1+cos⁡2θ≡tan⁡θ\displaystyle \frac{\sin 2\theta}{1 + \cos 2\theta} \equiv \tan\theta1+cos2θsin2θ​≡tanθ

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\theta1+cos2θsin2θ​=2sinθ for 0°≤θ≤360° 0° \leq \theta \leq 360°\,0°≤θ≤360°

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

Step 1: tan⁡θ=2sin⁡θ\tan\theta = 2\sin\thetatanθ=2sinθ

Step 2: sin⁡θcos⁡θ=2sin⁡θ\dfrac{\sin\theta}{\cos\theta} = 2\sin\thetacosθsinθ​=2sinθ

Step 3: 1cos⁡θ=2\dfrac{1}{\cos\theta} = 2cosθ1​=2

Step 4: cos⁡θ=12\cos\theta = \dfrac{1}{2}cosθ=21​

Step 5: θ=60°\theta = 60°θ=60°, 300°300°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 0°≤θ≤360°0° \leq \theta \leq 360°0°≤θ≤360°.

[2]
Markscheme

1 Pure Mathematics Questions

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