Q80 / 160 · TS EAMCET 2003 · Mathematics

The angle between the curves $y=\sin x$ and $y=\cos x$ is

  • Correct answer
    A
    $\tan ^{-1}(2 \sqrt{2})$
  • B
    $\tan ^{-1}(3 \sqrt{2})$
  • C
    $\tan ^{-1}(3 \sqrt{3})$
  • D
    $\tan ^{-1}(5 \sqrt{2})$

Solution

Given that, $y=\sin x$ ...(i) and $y=\cos x$ ...(ii) From Eqs. (i) and (ii), $\sin x=\cos x \Rightarrow \tan x=1 \Rightarrow x=\frac{\pi}{4}$ Slope $\quad m_1=\frac{d y}{d x}=\cos x$ $m_2=\frac{d y}{d x}=-\sin x$ $\therefore \quad m_1=\left(\frac{d y}{d x}\right)_{\frac{\pi}{4}}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$ and $\quad m_2=\left(\frac{d y}{d x}\right)_{\pi / 4}=-\frac{1}{\sqrt{2}}$ $\therefore \quad \tan \theta=\frac{m_1-m_2}{1+m_1 m_2}$ $\Rightarrow \quad \tan \theta=\frac{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}}{1-\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}}=\frac{2}{\sqrt{2}} \cdot 2=2 \sqrt{2}$ $\therefore \quad \theta=\tan ^{-1}(2 \sqrt{2})$

Hence, the correct answer is option: 'A'.

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