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1.5 Trigonometry

1.5 Trigonometry

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Question 38

A student is asked to solve the equation sin⁡2x=1\sin^2 x = 1sin2x=1 for 0°⩽x⩽360°0° \leqslant x \leqslant 360°0°⩽x⩽360°.

The student writes:

Step 1: sin⁡x=1\sin x = 1sinx=1

Step 2: x=90°x = 90°x=90°

a.

Explain the error the student has made in Step 1.

[1]
b.

State the correct solutions of sin⁡2x=1\sin^2 x = 1sin2x=1 for 0°⩽x⩽360°0° \leqslant x \leqslant 360°0°⩽x⩽360°.

[2]
c.

A second student solves the equation sin⁡xcos⁡x=sin⁡x\sin x\cos x = \sin xsinxcosx=sinx for 0°⩽x<360° 0° \leqslant x < 360°\,0°⩽x<360° by dividing both sides by sin⁡x\sin xsinx, obtaining cos⁡x=1\cos x = 1cosx=1 and hence x=0°x = 0°x=0°. Explain what is wrong with this method, and give the complete solution set.

[3]
Markscheme

1.5 Trigonometry Questions

  1. A Level
  2. /Maths
  3. /1.5 Trigonometry

97 exam-style questions on WJEC A Level Maths 1.5 Trigonometry, covering 1.5.1 Trigonometry, 1.5.2 Trigonometry, 1.5.3 Trigonometry, 1.5.4 Trigonometry, and 1.5.5 Trigonometry. Each one has a worked solution and a mark scheme showing where the marks go.

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