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TRIGONOMETRIC FUNCTIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTRIGONOMETRIC FUNCTIONS1 Mark
MCQ
Choose the correct answer. If $\sin\theta=\frac{-4}{5}$ and $\theta$ lies in third quadrant then the value of $\cos\frac{\theta}{2}$ is:
  • A
    $\frac{1}{5}$
  • B
    $-\frac{1}{\sqrt{10}}$
  • ✓
    $-\frac{1}{\sqrt{5}}$
  • D
    $\frac{1}{\sqrt{10}}$
✓

Answer

Correct option: C.
$-\frac{1}{\sqrt{5}}$
Given that, $\sin\theta=-\frac{4}{5},\theta$ lies in third quadrant
$\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\Big(\frac{-4}{5}\Big)^2}$
$=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\pm\frac{3}{5}$
$\therefore\cos\theta=-\frac{3}{5},\theta$ lies in third quadrant
$\cos\theta=2\cos^2\frac{\theta}{2}-1\Big[\because\pi<\theta\frac{3\pi}{2},\therefore\frac{\pi}{2}<\frac{\theta}{2}<\frac{3\pi}{4}\Big]$
$\Rightarrow\frac{-3}{5}=2\cos^2\frac{\theta}{2}-1$
$\Rightarrow2\cos^2\frac{\theta}{2}=1-\frac{3}{5}=\frac{2}{5}\Rightarrow\cos^2\frac{\theta}{2}=\frac{2}{5\times2}=\frac{1}{5}$
$\Rightarrow\cos\frac{\theta}{2}=\pm\frac{1}{\sqrt5}$
$\Rightarrow\cos\frac{\theta}{2}=-\frac{1}{\sqrt5}\Big[\because\frac{\pi}{2}<\frac{\theta}{2}<\frac{3\pi}{4}\Big]$
Hence, the correct option is (c).

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