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: 14th Mar 2023
•
Total views: 305.4k
•
Views today: 4.85k
Answer
305.4k+ views
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;
\[\begin{align}
& y={{\cos }^{2}}\left( 45+x \right)\times {{\left( \dfrac{\sqrt{2}}{\sqrt{2}}\left( \sin x-\cos x \right) \right)}^{2}} \\
& y={{\cos }^{2}}\left( 45+x \right)\times {{\left( \sqrt{2} \right)}^{2}}{{\left( \dfrac{1}{\sqrt{2}}\sin x-\dfrac{1}{\sqrt{2}}\cos x \right)}^{2}} \\
\end{align}\]
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:
$\begin{align}
& y=2{{\cos }^{2}}\left( 45+x \right){{\left( \sin 45\sin x-\cos 45\cos x \right)}^{2}} \\
& As,{{A}^{2}}={{\left( -A \right)}^{2}}, \\
\end{align}$
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
\[\begin{align}
& y=2{{\cos }^{2}}\left( 45+x \right){{\left( \cos \left( 45+x \right) \right)}^{2}} \\
& y=2{{\cos }^{4}}\left( 45+x \right).............\left( 3 \right) \\
\end{align}\]
As we know range of $\cos x$ is [-1,1]
Or
$-1\le \cos x\le 1$
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 ${{\left( \sin x-\cos x \right)}^{2}}$
$\begin{align}
& y={{\cos }^{2}}\left( 45+x \right){{\left( \sin x-\cos x \right)}^{2}} \\
& y={{\left( \cos 45\cos x-\sin 45\sin x \right)}^{2}}{{\left( \sin x-\cos x \right)}^{2}} \\
& y={{\left( \dfrac{1}{\sqrt{2}}\cos x-\dfrac{1}{\sqrt{2}}\sin x \right)}^{2}}{{\left( \sin x-\cos x \right)}^{2}} \\
\end{align}$
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 ${{\cos }^{4}}\left( 45+x \right)$.
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;
\[\begin{align}
& y={{\cos }^{2}}\left( 45+x \right)\times {{\left( \dfrac{\sqrt{2}}{\sqrt{2}}\left( \sin x-\cos x \right) \right)}^{2}} \\
& y={{\cos }^{2}}\left( 45+x \right)\times {{\left( \sqrt{2} \right)}^{2}}{{\left( \dfrac{1}{\sqrt{2}}\sin x-\dfrac{1}{\sqrt{2}}\cos x \right)}^{2}} \\
\end{align}\]
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:
$\begin{align}
& y=2{{\cos }^{2}}\left( 45+x \right){{\left( \sin 45\sin x-\cos 45\cos x \right)}^{2}} \\
& As,{{A}^{2}}={{\left( -A \right)}^{2}}, \\
\end{align}$
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
\[\begin{align}
& y=2{{\cos }^{2}}\left( 45+x \right){{\left( \cos \left( 45+x \right) \right)}^{2}} \\
& y=2{{\cos }^{4}}\left( 45+x \right).............\left( 3 \right) \\
\end{align}\]
As we know range of $\cos x$ is [-1,1]
Or
$-1\le \cos x\le 1$
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 ${{\left( \sin x-\cos x \right)}^{2}}$
$\begin{align}
& y={{\cos }^{2}}\left( 45+x \right){{\left( \sin x-\cos x \right)}^{2}} \\
& y={{\left( \cos 45\cos x-\sin 45\sin x \right)}^{2}}{{\left( \sin x-\cos x \right)}^{2}} \\
& y={{\left( \dfrac{1}{\sqrt{2}}\cos x-\dfrac{1}{\sqrt{2}}\sin x \right)}^{2}}{{\left( \sin x-\cos x \right)}^{2}} \\
\end{align}$
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 ${{\cos }^{4}}\left( 45+x \right)$.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
