prove sin(α+β)sin(α-β)=cos^2β-cos^2α
please help... this is the only one I
didn't understand out of all my homework...
sin(a+b)sin(a-b)
(sina cosb + cosa sinb)(sinacosb - cosa sinb)
sin^2acos^2b + sinacosasinbcosb - sinacosasinbcosb - cos^2asin^2b)
(1-cos^2a)cos^2b - (1-cos^2b)cos^2a
cos^2b - cos^2acos^2b - cos^2a + cos^2acos^2b
cos^2b - cos^2a
To prove the identity sin(α+β)sin(α-β) = cos²β - cos²α, we'll start with the left side of the equation and simplify it step by step to show that it is equal to the right side.
Step 1: Start with the left side of the equation:
sin(α+β)sin(α-β)
Step 2: Apply the product-to-sum formula to expand the expression:
= (1/2)[cos((α+β)-(α-β)) - cos((α+β)+(α-β))]
Step 3: Simplify the expression inside the brackets:
= (1/2)[cos(2β) - cos(2α)]
Step 4: Use the trigonometric identities:
cos(2β) = 1 - 2sin²β
cos(2α) = 1 - 2sin²α
Step 5: Substitute the identities back into the expression:
= (1/2)[1 - 2sin²β - 1 + 2sin²α]
Step 6: Simplify the expression further:
= (1/2)(2sin²α - 2sin²β)
Step 7: Factor out a 2 from the expression:
= 2(sin²α - sin²β)
Step 8: Apply the identity sin²θ = 1 - cos²θ:
= 2[(1 - cos²α) - (1 - cos²β)]
Step 9: Simplify the expression further:
= 2(cos²β - cos²α)
Step 10: Finally, we end up with the right side of the equation:
= cos²β - cos²α
Therefore, we have shown that sin(α+β)sin(α-β) = cos²β - cos²α, and the identity is proven.