Answer

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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.

Complete step-by-step 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.

Complete step-by-step 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.

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