Question

The equation of the tangent to the curve $y=2\cos x$ at $x=\frac{\pi }{4}$ is

• A. $y–\sqrt{2}=2\sqrt{2(x–(\frac{\pi }{4})})$
• B. $y–\sqrt{2}=\sqrt{2}(x+\frac{\mathrm{\pi }}{4})$
• C. $y–\sqrt{2}=–\sqrt{2}(x-\frac{\mathrm{\pi }}{4})$
• D. $y–\sqrt{2}=\sqrt{2}(x-\frac{\mathrm{\pi }}{4})$

Answer 1

The correct option is C

$y–\sqrt{2}=–\sqrt{2}(x-\frac{\mathrm{\pi }}{4})$

Explanation for correct answer:

Given $y=2\cos x$ at $x=\frac{\pi }{4}$

$\begin{array}{rcl}\mathrm{At},x& =& \frac{\pi }{4}\\ y& =& 2\cos \frac{\pi }{4}\\ & =& \frac{2}{\sqrt{2}}\\ & =& \sqrt{2}\end{array}$

Differentiating with respect to $x$ we get,

$\begin{array}{rcl}\frac{dy}{dx}& =& -2\sin x\\ & \Rightarrow & {\left(\frac{dy}{dx}\right)}_{x=\frac{\pi }{4}}=-\sqrt{2}\end{array}$

The equation of the tangent at $\left(\frac{\pi }{4},\sqrt{2}\right)$ using the formula $y-y_1=\frac{dy}{dx}(x-x_1)$ is $y–\sqrt{2}=–\sqrt{2}(x-\frac{\mathrm{\pi }}{4})$.

Hence, Option (C) is the correct answer.