Question

The equation of the tangent to the curve $y=2 \cos x$ at $x=\frac{\pi}{4}$ is
A. $y-\sqrt{2}=2 \sqrt{2}\left(x-\frac{\pi}{4}\right)$
B. $y+\sqrt{2}=\sqrt{2}\left(x+\frac{\pi}{4}\right)$
C. $y-\sqrt{2}=-\sqrt{2}\left(x-\frac{\pi}{4}\right)$
D. $y-\sqrt{2}=\sqrt{2}\left(x-\frac{\pi}{4}\right)$

Answer 1

Hint: Given, the curve $y = 2\cos x$ at $x = \dfrac{\pi }{4}$. First, we will find the value of $y$ at $x = \dfrac{\pi }{4}$. Then we will differentiate the equation of the curve $y = 2\cos x$ with respect to $x$. Then, after differentiation we put the value $x = \dfrac{\pi }{4}$ to find the value of slope of tangent. At last, with the help of formula
$(y - {y_0}) = m(x - {x_0})$ we will find the equation of tangent.

Complete step by step solution:
Given, the equation of curve $y = 2\cos x$
At $x = \dfrac{\pi }{4}$
We will find the value of $y$ by putting $x = \dfrac{\pi }{4}$
$y = 2\cos \left( {\dfrac{\pi }{4}} \right)$
We know that $\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$
$y = \dfrac{2}{{\sqrt 2 }}$
After solving above equation
we get
$y = \sqrt 2 $
Now, we will differentiate the given equation.
$y = 2\cos x$
Differentiating with respect to $x$
$\dfrac{{dy}}{{dx}} = 2\dfrac{{d(\cos x)}}{{dx}}$
We know that
$\dfrac{{d(\cos x)}}{{dx}} = - \sin x$
So, $\dfrac{{dy}}{{dx}} = - 2\sin x$
Now, finding slope using $x = \dfrac{\pi }{4}$
${\dfrac{{dy}}{{dx}}_{x = \dfrac{\pi }{4}}} = - 2\sin x$
After putting $x = \dfrac{\pi }{4}$
We know $\sin \dfrac{\pi }{2} = \dfrac{1}{{\sqrt 2 }}$
$ = - 2 \times \dfrac{1}{{\sqrt 2 }}$
$ = - \sqrt 2 $
So, the slope of the tangent is $ - \sqrt 2 $
We know the formula for equation of line $(y - {y_0}) = m(x - {x_0})$
We will use above formula to find equation of tangent
${x_0} = \dfrac{\pi }{4},\,{y_0} = \sqrt 2 ,\,m = - \sqrt 2 $
Putting in formula
$(y - \sqrt 2 ) = - \sqrt 2 (x - \dfrac{\pi }{4})$
So, the equation of tangent is
$y - \sqrt 2 = - \sqrt 2 (x - \dfrac{\pi }{4})$

Option ‘A’ is correct

Note: A method for finding a function's derivative is called differentiation. Mathematicians use a procedure called differentiation to determine a function's instantaneous rate of change based on one of its variables. The most typical illustration is velocity, which is the rate at which a distance changes in relation to time.
While finding the equation of tangent students should attempt each step carefully in order to avoid any complication. They should pay attention while differentiating because the derivative of $\cos x$ is $ - \sin x$, if they forget to put a negative sign they will get the wrong answer. In order to get the correct answer without any errors, they should solve the question step by step.