Question

How do you prove $1 - 2\sin a.\cos a \div 2 = {\sin ^2}(45^\circ - a)$?

Answer 1

Hint: Here we have shown that the left side of the equation is equal to the right side of the equation by considering one side of the equation and simplifying it with the help of trigonometric formulas. In this question we will use the formula $\sin (x - y) = \sin x\cos y - \sin y\cos x$.

Complete step by step answer:
We will solve this problem by considering the right side of the equation and simplify it to get the result which is equal to the left side of the equation.
Considering left side of the equation i.e., ${\sin ^2}(45^\circ - a)$
We can write ${\sin ^2}(45^\circ - a)$$ = \sin (45^\circ - a)\sin (45^\circ - a)$
Apply the formula $\sin (x - y) = \sin x\cos y - \sin y\cos x$ in the above equation.
We have
$\sin (45^\circ - a)\sin (45^\circ - a) = (\sin 45^\circ \cos a - \sin a\cos 45^\circ )(\sin 45^\circ \cos a - \sin a\cos 45^\circ )$
We know that $\sin 45^\circ = \cos 45^\circ = \dfrac{1}{{\sqrt 2 }} = \dfrac{{1 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }} = \dfrac{{\sqrt 2 }}{2}$
So,
$ \Rightarrow \left( {\dfrac{{\sqrt 2 }}{2}\cos a - \dfrac{{\sqrt 2 }}{2}\sin a} \right)\left( {\dfrac{{\sqrt 2 }}{2}\cos a - \dfrac{{\sqrt 2 }}{2}\sin a} \right)$
Taking $\dfrac{{\sqrt 2 }}{2}$ as a common factor from both the expressions. We get,
$ \Rightarrow \dfrac{{\sqrt 2 }}{2}\left( {\cos a - \sin a} \right)\dfrac{{\sqrt 2 }}{2}\left( {\cos a - \sin a} \right)$
Multiplying the square roots and trigonometric expressions. We get,
$ \Rightarrow \dfrac{2}{4}({\cos ^2}a - 2\sin a\cos a + {\sin ^2}a)$
Simplifying and rearranging the above equation. We get,
$ \Rightarrow \dfrac{1}{2}({\sin ^2}a + {\cos ^2}a - 2\sin a\cos a)$
We know that ${\sin ^2}a + {\cos ^2}a = 1$
Therefore,
${\sin ^2}(45^\circ - a) = \dfrac{{1 - 2\sin a\cos a}}{2}$
Hence the right side of the equation is equal to the left side of the equation.
Hence proved.

Note:
In these types of problems in which we have to equal both sides of the equation, first check by solving which side of the equation we can get our desired result. One should remember all trigonometric formulas before solving these types of problems. Some students are confused between the different identities of $\sin $. They both are different. One is the difference of angles of $\sin $i.e., $\sin (x - y)$ and the other one is the difference of two $\sin $ angles i.e., $\sin x - \sin y$.