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Question: Show that \[{{\cos }^{2}}A-{{\sin }^{2}}A=2{{\cos }^{2}}A-1\]....

Show that cos⁡2A−sin⁡2A=2cos⁡2A−1{{\cos }^{2}}A-{{\sin }^{2}}A=2{{\cos }^{2}}A-1.

Explanation

Solution

In this problem, we have to prove the given trigonometric expression. Here we can first take the left-hand side part and simplify and solve it to get the right-hand side part. We can replace the sine part in the left-hand side as sin⁡2A=1−cos⁡2A{{\sin }^{2}}A=1-{{\cos }^{2}}A, we can then simplify it by multiplying the signs and adding the similar terms to get the right-hand side part.

Complete step-by-step solution:
Here we have to prove the given trigonometric expression.
The given trigonometric expression is cos⁡2A−sin⁡2A=2cos⁡2A−1{{\cos }^{2}}A-{{\sin }^{2}}A=2{{\cos }^{2}}A-1.
We can first take the left-hand side part and simplify and solve it to get the right-hand side part.
LHS = cos⁡2A−sin⁡2A{{\cos }^{2}}A-{{\sin }^{2}}A
We can now take the sine part as we know that sin⁡2A=1−cos⁡2A{{\sin }^{2}}A=1-{{\cos }^{2}}A.
We can now replace the above formula for sine part in the left-hand side, we get
LHS = cos⁡2A−(1−cos⁡2A){{\cos }^{2}}A-\left( 1-{{\cos }^{2}}A \right)
Here, we can see that we have similar terms, we can first multiply the signs inside and outside the bracket, we get
LHS = cos⁡2A−1+cos⁡2A{{\cos }^{2}}A-1+{{\cos }^{2}}A
We can see that, we have similar terms with similar sign, so we can add them, we get
LHS = 2cos⁡2A−12{{\cos }^{2}}A-1
We can see that,
LHS = RHS.
Therefore, cos⁡2A−sin⁡2A=2cos⁡2A−1{{\cos }^{2}}A-{{\sin }^{2}}A=2{{\cos }^{2}}A-1
Hence proved.

Note: We should also remember some of the trigonometric formulas and identities to be substituted and solved for the both sides. Here we have used the formula sin⁡2A=1−cos⁡2A{{\sin }^{2}}A=1-{{\cos }^{2}}A, as we have only cosine terms in the right-hand side. We have to choose the formula, in order to prove for the other side.