Question

The maximum value of $(5\sin x-12\cos x)(5\cos x+12\sin x)$ is
(A) $\dfrac{169}{2}$
(B) $169$
(C) $13$
(D) $85$

Answer 1

We are given a trigonometric expression and we are asked to find the maximum value of the expression. We will first multiply the terms of the two brackets with each other and then on reducing the expression we will get a new expression which is, $-\dfrac{119}{2}\sin 2x-60\cos 2x$. This expression is of the form $a\sin x+b\cos x$, and so the maximum value of these type of expression is obtained using the formula, $\sqrt{{{a}^{2}}+{{b}^{2}}}$. We will substitute the values from the expression we obtained in this formula, which will look like, $\sqrt{{{\left( -\dfrac{119}{2} \right)}^{2}}+{{\left( -60 \right)}^{2}}}$. On solving the above expression, we will get the maximum value of the given trigonometric expression.

Complete step-by-step solution:
According to the given question, we are given a trigonometric expression and we are asked to find the maximum value of the expression.
The expression we have is,
$(5\sin x-12\cos x)(5\cos x+12\sin x)$
We will now open up the brackets and multiply the terms in the bracket and we get,
$\Rightarrow 25\sin x\cos x+60{{\sin }^{2}}x-60{{\cos }^{2}}x-144\sin x\cos x$
We will rearrange the above terms such that the similar terms come together and we get,
$\Rightarrow 25\sin x\cos x-144\sin x\cos x-60{{\cos }^{2}}x+60{{\sin }^{2}}x$
Subtracting the terms, we get,
$\Rightarrow -119\sin x\cos x-60\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)$
We know that the expression, ${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$, so the expression we get is,
$\Rightarrow -\dfrac{119}{2}\left( 2\sin x\cos x \right)-60\left( \cos 2x \right)$
Now, we know that $2\sin x\cos x=\sin 2x$, so we have,
$\Rightarrow -\dfrac{119}{2}\sin 2x-60\cos 2x$
If we observe carefully, the above expression is of the form, $a\sin x+b\cos x$. And, so the maximum value for such type of expression is obtained using the formula, $\sqrt{{{a}^{2}}+{{b}^{2}}}$.
The maximum value for the expression we obtained is,
$Maximum\\_value=\sqrt{{{\left( -\dfrac{119}{2} \right)}^{2}}+{{\left( -60 \right)}^{2}}}$
Squaring up the terms, we get,
$\Rightarrow \sqrt{\left( \dfrac{14161}{4} \right)+\left( 3600 \right)}$
$\Rightarrow \sqrt{\dfrac{14161}{4}+3600}$
Taking the LCM and simplifying the expression further, we get,
$\Rightarrow \sqrt{\dfrac{14161+14400}{4}}$
$\Rightarrow \sqrt{\dfrac{28561}{4}}$
We can see that the numerator and denominator have numbers which have perfect square roots, so we get,
$\Rightarrow \dfrac{169}{2}$
Therefore, the correct answer is (A) $\dfrac{169}{2}$.

Note: While opening up the brackets, the terms should be orderly and correctly multiplied and make sure no term is repeated or is wrongly multiplied. Also, while getting the expression in the form, $a\sin x+b\cos x$, the signs should be correctly written. And also the substitution of values in the formula, $\sqrt{{{a}^{2}}+{{b}^{2}}}$ should also be done in a step – wise manner and even though the terms will get squared still the sign should be intact.