Question

If $y = {\cos ^{ - 1}}\left[ {\dfrac{{\left( {3\cos x - 4\sin x} \right)}}{5}} \right]$ then $\dfrac{{dy}}{{dx}}$ equals:
$\left( 1 \right)0$
$\left( 2 \right)1$
$\left( 3 \right) - 1$
$\left( 4 \right)\dfrac{1}{2}$

Answer 1

To solve this question, the standard trigonometric identity formulae and basic algebraic
operations have to be kept in mind. Trigonometric identities are equations involving functions which are true for all the angles for which the particular functions are defined. For example: $\cos \theta = \dfrac{1}{{\sec \theta }}$ and $\sin \theta = \dfrac{1}{{\cos ec\theta }}$ , are trigonometric identities since they hold for all values of $\theta$ except for values where $\sin \theta$ and $\cos \theta$ are not defined. The equation $\tan \theta = \cot \theta$ is a trigonometric equation but it
is not a trigonometric identity as it does not hold true for all the values of $\theta$ .

Complete step by step answer:
According to the given question;
$\Rightarrow y = {\cos ^{ - 1}}\left( {\dfrac{{3\cos x - 4\sin x}}{5}} \right)$
It can also be written as;
$\Rightarrow y = \left( {\dfrac{3}{5}\cos x - \dfrac{4}{5}\sin x} \right){\text{ }}......\left( 1 \right)$
Let $\dfrac{3}{5} = \cos \theta$
We know that, $\cos \theta = \dfrac{{{\text{Adjacent}}}}{{{\text{Hypotenuse}}}}$
$\Rightarrow \cos \theta = \dfrac{b}{h}$
Here, $b = 3{\text{ and }}h = 5$ ;
According to the Pythagoras theorem; ${h^2} = {p^2} + {b^2}$
Therefore perpendicular will be ${\text{p = }}\sqrt {{{\text{h}}^2} - {{\text{b}}^2}} {\text{ }}$
$\therefore {\text{P }} \Rightarrow \sqrt {{5^2} - {3^2}} {\text{ = 4}}$
$\because \sin \theta = \dfrac{{{\text{Perpendicular }}}}{{{\text{hypotenuse}}}}$
$\Rightarrow \sin \theta = \dfrac{p}{h}$
$\Rightarrow \sin \theta = \dfrac{4}{5}$
Therefore, $\cos \theta = \dfrac{3}{5}{\text{ and sin}}\theta {\text{ = }}\dfrac{4}{5}$ .
Put the values of $\cos \theta {\text{ and sin}}\theta$ in equation $\left( 1 \right)$, we get;
$\Rightarrow y = {\cos ^{ - 1}}\left( {\cos \theta \cos x - \sin \theta \sin x} \right){\text{ }}......\left( 2 \right)$
We know the standard formula for ; $\cos \left( {{\text{A}} + {\text{B}}} \right) = \cos {\text{A cosB}} - \sin {\text{A}}\sin {\text{B}}$
Using the above formula in equation $\left( 1 \right)$ , we get;
$\Rightarrow y = {\cos ^{ - 1}}\left( {\cos \left( {\theta + x} \right)} \right){\text{ }}......\left( 3 \right)$ $\left( {{\text{Here , A = }}\theta {\text{ and B = x}}} \right)$
We know that the cosine function and its inverse function cancels each other;
Therefore equation $\left( 3 \right)$, reduces to;
$\Rightarrow y = \theta + x{\text{ }}......\left( 4 \right)$
$\because \cos \theta = \dfrac{3}{5}{\text{ }}\therefore \theta {\text{ = co}}{{\text{s}}^{ - 1}}\dfrac{3}{5}$
Now, put the value of $\theta$ , in equation $\left( 4 \right)$ we get;
$\Rightarrow y = {\cos ^{ - 1}}\dfrac{3}{5} + x$
( $\because {\cos ^{ - 1}}\dfrac{3}{5}$ has a constant angle therefore it’s differentiation will be 0 )
Differentiating the above equation w.r.t. $x$ ;
$\Rightarrow \dfrac{{dy}}{{dx}} = 0 + \dfrac{d}{{dx}}\left( x \right)$
$\Rightarrow \dfrac{{dy}}{{dx}} = 1$
Therefore, the correct answer for this question is $\dfrac{{dy}}{{dx}} = 1$ .

So, the correct answer is “Option 2”.

Note: We have used the standard formula of cosine of the difference and sum of two angles, i.e.
$\left( 1 \right)\cos \left( {{\text{A}} + {\text{B}}} \right) = \cos {\text{A cosB}} - \sin {\text{A}}\sin {\text{B}}$ , this formula holds true for all values of angles of ${\text{A and B}}$ , whether positive, negative or zero. $\left( 2 \right)\cos \left( {{\text{A}} - {\text{B}}} \right) = \cos {\text{A cosB + }}\sin {\text{A}}\sin {\text{B}}$ . Similarly the formula for sine of the difference and sum of two angles is $\left( 3 \right)\sin \left( {{\text{A}} + {\text{B}}} \right) = \sin {\text{A cosB + cosA}}\sin {\text{B}}$ and $\left( 4 \right)\sin \left( {{\text{A}} - {\text{B}}} \right) = \sin {\text{A cosB}} - {\text{cosA}}\sin {\text{B}}$ .