Prove sin(α+β)sin(α-β)=cos^2β-cos^2α
To prove the given identity:
Step 1: Start with the left-hand side (LHS) of the equation:
LHS = sin(α + β)sin(α - β)
Step 2: Use the trigonometric identity for the product of trigonometric functions:
sin(x)sin(y) = (1/2)[cos(x - y) - cos(x + y)]
Applying this identity to LHS, we have:
LHS = (1/2)[cos((α + β) - (α - β)) - cos((α + β) + (α - β))]
Simplifying further:
LHS = (1/2)[cos(2β) - cos(2α)]
Step 3: Now let's look at the right-hand side (RHS) of the equation:
RHS = cos^2β - cos^2α
Step 4: Recall the identity for squared cosine functions:
cos^2(x) = (1/2)(1 + cos(2x))
Using this identity for both terms on the RHS, we can rewrite it as:
RHS = (1/2)(1 + cos(2β)) - (1/2)(1 + cos(2α))
Simplifying further:
RHS = (1/2)(cos(2β) - cos(2α))
Step 5: Compare LHS and RHS:
We observe that both LHS and RHS simplify to the same expression:
(1/2)(cos(2β) - cos(2α))
Therefore, we conclude that LHS = RHS, and the identity sin(α+β)sin(α-β) = cos^2β - cos^2α is proved.