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# If $\cos 25{}^\circ +\sin 25{}^\circ =p$, then $\cos 50{}^\circ$ is?(a).$\sqrt{2-{{p}^{2}}}$(b).$-\sqrt{2-{{p}^{2}}}$(c).$p\sqrt{2-{{p}^{2}}}$(d).$-p\sqrt{2-{{p}^{2}}}$ Verified
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Hint: Square the equation from both the sides and use the formulae ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, ${{\cos }^{2}}\theta {}^\circ +{{\sin }^{2}}\theta {}^\circ =1$, and $2\times \sin \theta {}^\circ \times \cos \theta {}^\circ =\sin \left( 2\times \theta \right){}^\circ$; you will get the equation for $\sin 50{}^\circ$ then again square the equation and use the formula ${{\sin }^{2}}\theta {}^\circ =1-{{\cos }^{2}}\theta {}^\circ$ and simplify it to get the final answer.

To solve the above equation we will write it down first, Therefore,
$\cos 25{}^\circ +\sin 25{}^\circ =p$
If we square the above equation on both sides we will get,
$\therefore {{\left( \cos 25{}^\circ +\sin 25{}^\circ \right)}^{2}}={{p}^{2}}$ …………………………………………… (1)
To proceed further in the solution we should know the formula given below,
Formula:
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
If we use the above formula in equation (1) we will get,
$\therefore {{\left( \cos 25{}^\circ \right)}^{2}}+2\times \left( \cos 25{}^\circ \right)\times \left( \sin 25{}^\circ \right)+{{\left( \sin 25{}^\circ \right)}^{2}}={{p}^{2}}$
By rearranging the above equation we will get,
$\therefore {{\left( \cos 25{}^\circ \right)}^{2}}+{{\left( \sin 25{}^\circ \right)}^{2}}+2\times \left( \sin 25{}^\circ \right)\times \left( \cos 25{}^\circ \right)={{p}^{2}}$
$\therefore {{\cos }^{2}}25{}^\circ +{{\sin }^{2}}25{}^\circ +2\times \left( \sin 25{}^\circ \right)\times \left( \cos 25{}^\circ \right)={{p}^{2}}$ …………………………………….. (2)
To proceed further in the solution in the solution we should know the formula given below,
Formula:
${{\cos }^{2}}\theta {}^\circ +{{\sin }^{2}}\theta {}^\circ =1$
If we use the above formula in equation (2) we will get,
$\therefore 1+2\times \left( \sin 25{}^\circ \right)\times \left( \cos 25{}^\circ \right)={{p}^{2}}$
To proceed further in the solution in the solution we should know the formula given below,
Formula:
$2\times \sin \theta {}^\circ \times \cos \theta {}^\circ =\sin \left( 2\times \theta \right){}^\circ$
By using above formula we will get,
$\therefore 1+\sin \left( 2\times 25 \right){}^\circ ={{p}^{2}}$
After multiplication we will get,
$\therefore 1+\sin 50{}^\circ ={{p}^{2}}$
If we shift ‘1’ on the right hand side of the equation we will get,
$\therefore \sin 50{}^\circ ={{p}^{2}}-1$
Now, to find the value of $\cos 50{}^\circ$ we will square the above equation on the both sides so that we can use the formula ${{\sin }^{2}}\theta {}^\circ =1-{{\cos }^{2}}\theta {}^\circ$ and simplify it.
$\therefore {{\left( \sin 50{}^\circ \right)}^{2}}={{\left( {{p}^{2}}-1 \right)}^{2}}$
By using the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ in the above equation we will get,
$\therefore {{\sin }^{2}}50{}^\circ ={{\left( {{p}^{2}} \right)}^{2}}-2\times {{p}^{2}}\times 1+{{1}^{2}}$
If we do further simplification in the solution we will get,
$\therefore {{\sin }^{2}}50{}^\circ ={{p}^{4}}-2{{p}^{2}}+1$
If we use the formula ${{\sin }^{2}}\theta {}^\circ =1-{{\cos }^{2}}\theta {}^\circ$ in the above equation we will get,
$\therefore 1-{{\cos }^{2}}50{}^\circ ={{p}^{4}}-2{{p}^{2}}+1$
If we multiply the above equation by ‘-1’ we will get,
$\therefore -1\times \left( 1-{{\cos }^{2}}50{}^\circ \right)=-1\times \left( {{p}^{4}}-2{{p}^{2}}+1 \right)$
$\therefore -1+{{\cos }^{2}}50{}^\circ =-{{p}^{4}}+2{{p}^{2}}-1$
If we shift ‘-1’ on the right hand side of the equation we will get,
$\therefore {{\cos }^{2}}50{}^\circ =-{{p}^{4}}+2{{p}^{2}}-1+1$
$\therefore {{\cos }^{2}}50{}^\circ =-{{p}^{4}}+2{{p}^{2}}$
If we take ${{p}^{2}}$ common from the above equation we will get,
$\therefore {{\cos }^{2}}50{}^\circ ={{p}^{2}}\left( -{{p}^{2}}+2 \right)$
By rearranging the above equation we will get,
$\therefore {{\cos }^{2}}50{}^\circ ={{p}^{2}}\left( 2-{{p}^{2}} \right)$
As we have to find the value of $\cos 50{}^\circ$ therefore we will simply take the square roots on both sides of the equation, therefore we will get,
$\therefore \sqrt{{{\cos }^{2}}50{}^\circ }=\sqrt{{{p}^{2}}\left( 2-{{p}^{2}} \right)}$
As the square root of ${{\cos }^{2}}50{}^\circ$ is $\cos 50{}^\circ$ and he square root of ${{p}^{2}}$is p therefore the above equation will become,
$\therefore \cos 50{}^\circ =p\sqrt{\left( 2-{{p}^{2}} \right)}$
Therefore, if $\cos 25{}^\circ +\sin 25{}^\circ =p$ the value of $\cos 50{}^\circ$ is $p\sqrt{\left( 2-{{p}^{2}} \right)}$.
Therefore the correct answer is option (c).

Note: You can solve the step $1-{{\cos }^{2}}50{}^\circ ={{p}^{4}}-2{{p}^{2}}+1$ without multiplying it by ‘-1’ but then you have to be very careful about the signs.
Last updated date: 24th Sep 2023
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