N2022 P1 Q4

N2022 P1 Q4

Junior College 2
7 marks
Free
  1. Show that \(\frac{\mathrm{d}}{\mathrm{d}x}(\cot x)=-\mathrm{cosec}^2x\).[2]
  2. Show that \(\sin 2x\tan x=2\sin^2x\).[1]
  3. Hence, find the exact value of \(\int_{\pi/18}^{\pi/9}\mathrm{cosec}\,6x\cot 3x\,\mathrm{d}x\).[4]

Default solution

  1. \(\frac{\mathrm{d}}{\mathrm{d}x}(\cot x)=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\cos x}{\sin x}\right)=\frac{-\sin^2x-\cos^2x}{\sin^2x}\)
    \(\therefore\ \frac{\mathrm{d}}{\mathrm{d}x}(\cot x)=-\mathrm{cosec}^2x\) (shown)
  2. \(\sin 2x\tan x=(2\sin x\cos x)\frac{\sin x}{\cos x}\)
    \(\therefore\ \sin 2x\tan x=2\sin^2x\) (shown)
  3. \(\mathrm{cosec}\,6x\cot 3x=\frac{1}{\sin 6x\tan 3x}=\frac12\mathrm{cosec}^2(3x)\) (using part (b))
    \(\int_{\pi/18}^{\pi/9}\mathrm{cosec}\,6x\cot 3x\,\mathrm{d}x=-\frac16[\cot(3x)]_{\pi/18}^{\pi/9}\) (using part (a))
    \(=-\frac16\left(\cot\frac\pi3-\cot\frac\pi6\right)=\frac{1}{3\sqrt3}\)
Answer:(a) \(\frac{\mathrm{d}}{\mathrm{d}x}(\cot x)=-\mathrm{cosec}^2x\) (b) \(\sin 2x\tan x=2\sin^2x\) (c) \(\frac{1}{3\sqrt3}\)

Need help? Join our JC Math tuition classes.

Learn more