Question

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).\[-p\sqrt{2-{{p}^{2}}}\]

Answer 1

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.