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Find the maximum value of y, If $y={{\cos }^{2}}\left( 45{}^\circ +x \right){{\left( \sin x-\cos x \right)}^{2}}$.

Last updated date: 15th Jul 2024
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Hint: Multiply and divide by $\sqrt{2}$ in the bracket of ${{\left( \sin x-\cos x \right)}^{2}}$. Try to convert it to a single sine or cosine function. Also use the fact that $\cos x$ has a maximum value of ‘1’ at x equal to zero degree.

Here, we are given the expression as,
$y={{\cos }^{2}}\left( 45{}^\circ +x \right){{\left( \sin x-\cos x \right)}^{2}}$....................(1)
We have to calculate the maximum value of a given function y as written in equation (1). So first of all we need to simplify the given expression, then we look for the maximum value of ‘y’.
Let us divide and multiply the second bracket${{\left( \sin x-\cos x \right)}^{2}}$by $\sqrt{2}$ .
Now, we can write ‘y’ from equation (1) as;
  & y={{\cos }^{2}}\left( 45+x \right)\times {{\left( \dfrac{\sqrt{2}}{\sqrt{2}}\left( \sin x-\cos x \right) \right)}^{2}} \\
 & y={{\cos }^{2}}\left( 45+x \right)\times {{\left( \sqrt{2} \right)}^{2}}{{\left( \dfrac{1}{\sqrt{2}}\sin x-\dfrac{1}{\sqrt{2}}\cos x \right)}^{2}} \\
We know the value of $\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}$ , so we can replace $\dfrac{1}{\sqrt{2}}$ by $\cos 45{}^\circ $.
And also $\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}$, so we can replace $\dfrac{1}{\sqrt{2}}$by $\sin 45{}^\circ $ in following way in the above equation:
  & y=2{{\cos }^{2}}\left( 45+x \right){{\left( \sin 45\sin x-\cos 45\cos x \right)}^{2}} \\
 & As,{{A}^{2}}={{\left( -A \right)}^{2}}, \\
So we can rewrite ‘y’ as
$y=2{{\cos }^{2}}\left( 45+x \right){{\left( \cos 45\cos x-\sin 45\sin x \right)}^{2}}........\left( 2 \right)$
As we have a trigonometric identity as;
\[\cos \left( A+B \right)=cosAcosB-sinAsinB\] or vice versa is also true.
Therefore, equation (2) can be simplified as
  & y=2{{\cos }^{2}}\left( 45+x \right){{\left( \cos \left( 45+x \right) \right)}^{2}} \\
 & y=2{{\cos }^{4}}\left( 45+x \right).............\left( 3 \right) \\
As we know range of $\cos x$ is [-1,1]
$-1\le \cos x\le 1$
Therefore, $0\le {{\cos }^{2}}x\le 1$
Now $y=2{{\cos }^{4}}\left( 45+x \right)$
Will have a maximum value of 2 by taking the maximum value of \[\cos \left( 45+x \right)\] i.e. 1 at $\left( -45{}^\circ \right)$.
Hence, maximum value of given expression $y={{\cos }^{2}}\left( 45{}^\circ +x \right){{\left( \sin x-\cos x \right)}^{2}}$is 2.

Note: Another approach for the given equation would be that we can apply formula of
\[\cos \left( A+B \right)=cosAcosB-sinAsinB\]with \[\cos \left( 45+x \right)\]and simplifying ${{\left( \sin x-\cos x \right)}^{2}}$
  & y={{\cos }^{2}}\left( 45+x \right){{\left( \sin x-\cos x \right)}^{2}} \\
 & y={{\left( \cos 45\cos x-\sin 45\sin x \right)}^{2}}{{\left( \sin x-\cos x \right)}^{2}} \\
 & y={{\left( \dfrac{1}{\sqrt{2}}\cos x-\dfrac{1}{\sqrt{2}}\sin x \right)}^{2}}{{\left( \sin x-\cos x \right)}^{2}} \\
One can go wrong while relating the maximum value of ${{\cos }^{4}}\left( 45+x \right)$. (45 + x) has no effect on maximum value of ${{\cos }^{4}}\left( 45+x \right)$. As $\cos \theta $ always lies in [-1,1]. So, one can confuse here to get the maximum value of ${{\cos }^{4}}\left( 45+x \right)$.