Question

How do you prove ${\sin ^2x} - {\cos ^2x} = 2{\sin ^2x} - 1$?

Answer 1

This we have to first pick out the formulas which we are going to apply. The first formula which should come to our mind is ${\sin ^2} + {\cos ^2} = 1$. Then we can take any side, and start solving that side and we will at the end prove that both the sides are equal.
Formula used:
${\sin ^2}x + {\cos ^2}x = 1$

Complete step by step solution:
The given equation according to question is:
${\sin ^2}x - {\cos ^2}x = 2{\sin ^2}x - 1$
$\Rightarrow {\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x$
First, we need to pick a side to solve. We can use either the left-hand side or right-hand side. We will choose the right-hand side here because it is easier to solve that side as compared to the left-hand side. Right-hand side is also called RHS in short and Left-hand side is also called LHS. So, RHS= $1 - 2{\sin ^2}x$
Here, we will put the value of $1$ as ${\sin ^2}x + {\cos ^2}x$ according to the trigonometry formula ${\sin ^2}x + {\cos ^2}x = 1$. So, we will get:
$\Rightarrow 1 - 2{\sin ^2}x = {\sin ^2}x + {\cos ^2}x - 2{\sin ^2}x$
Now, we will take the like terms at one side and start grouping so that it will become easy for us to solve. Here, we are taking ${\sin ^2}x$ terms together making a group.
$\Rightarrow 1 - 2{\sin ^2}x = {\cos ^2}x - 2{\sin ^2}x + {\sin ^2}x$
$\Rightarrow 1 - 2{\sin ^2}x = {\cos ^2}x - {\sin ^2}x$
According to question, LHS= ${\cos ^2}x - {\sin ^2}x$
We have proved that RHS= ${\cos ^2}x - {\sin ^2}x$
This means, that LHS=RHS and so, it is proved that ${\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x$ which means ${\sin ^2}x - {\cos ^2}x = 2{\sin ^2}x - 1$ is also proved.

Note: There are various methods to solve this type of problems. There might be various formulas also to solve this type of problem, but we have to think, and choose the correct method and the right formula to solve this type of problem. We can do this by a lot of practice in trigonometry and by learning the formulas.