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# How do you prove $\cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A$ ?

Last updated date: 08th Aug 2024
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Hint:
The above-given question is of trigonometric identities. So, we will use a trigonometric formula like cos(A + B) = cos A cos B - sin A sin B, and then we will put B = A in the equation to get 2A on the left side of the identity, and then we will get the required result.

We can see that the above-given question is of trigonometric identity and so we will use trigonometric formulas to prove the above result $\cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A$ .
We have to prove that $\cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A$ and from trigonometric formula of cos(A + B) we know that:
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$
So, when we put B = A in the above equation we will get:
$\cos \left( A+A \right)=\cos A\cos A-\sin A\sin A$
After we add the angle and express the terms on the RHS as squares, we will get the final result as
$\Rightarrow \cos \left( 2A \right)={{\cos }^{2}}A-{{\sin }^{2}}A$
Hence proved.
Hence, this is our required proof and solution to the above question.

Note:
Students are required to note that we can also solve the above question by making LHS (Left Hand Side) i.e. cos 2A equal to RHS (Right Hand Side) i.e. ${{\cos }^{2}}A-{{\sin }^{2}}A$.
Since, LHS = cos 2A and RHS = ${{\cos }^{2}}A-{{\sin }^{2}}A$
We can also write it as:
LHS = $\cos 2A=\cos \left( A+A \right)$
Now, we know that $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ :
So, we can write $\cos 2A=\cos \left( A+A \right)=\cos A\cos A-\sin A\sin A$
$\Rightarrow \cos \left( 2A \right)={{\cos }^{2}}A-{{\sin }^{2}}A$ = RHS
Hence, LHS = RHS.
Hence, proved.