Question

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

• A. ${\tan }^{-1}\left(2\sqrt{2}\right)$
• B. ${\tan }^{-1}\left(3\sqrt{2}\right)$
• C. ${\tan }^{-1}\left(3\sqrt{3}\right)$
• D. ${\tan }^{-1}\left(5\sqrt{2}\right)$

Answer 1

Answer: (A)

${\tan }^{-1}\left(2\sqrt{2}\right)$

C₁ : $y=\sin x$, C₂ : $y=\cos x$.
Equating the $y$' s,
$\sin x=\cos x$ ∴ $x=\frac{\pi }{4}$
∴ curves intersect each other at the point P : $x=\frac{\pi }{4}$.
Now, differentiating w.r.t. $x$,

C₁ gives : $\frac{dy}{dx}=\cos x$

C₂ gives : $\frac{dy}{dx}=-\sin x$.

Hence slopes $m₁andm₂ofC₁andC₂atP:$x = \dfrac{π}{4}$arem₁$= \cos \dfrac{π}{4} = \dfrac{1}{√2}$

m₂ =$-\sin \frac{\pi }{4}=-\frac{1}{\surd 2}$.

If $\theta$ is the acute angle between them at $P$, then

$tan\theta =\frac{|(m₁-m₂)|}{|(1+m₁m₂)|}$
$=\frac{|\left(\right(\frac{1}{\surd 2})-(-\frac{1}{\surd 2}\left)\right)|}{|(1+(\frac{1}{\surd 2}\left)\right(-\frac{1}{\surd 2}\left)\right)|}$
$=\frac{|2\left(\frac{1}{\surd 2}\right)|}{|(\frac{(2-1)}{\left(2\right)}|}$
$=2\surd 2$
∴ $\theta =\tan ֿ¹\left(2\surd 2\right)$