Question

complete the left-hand column of the table below following the steps indicated in the right-hand column to show that sin a sin b =1/2[cos(a-b)-cos(a+b)] is an identity. use the definitions of sum and difference formulas for cosine.

Calculation: Reason:
_____________ Given on the right side of the original equation
_____________ Apply the definitions of the sum and difference identities for
cosine
_____________ Simplify the expressions

Answer 1

Answer:

$\frac{1}{2}[cos(A-B)-cos(A+B)]=sinA\:sinB$

Step-by-step explanation:

Formula used:

$cos(A-B)=cosA\:cosB+sinA\:sinB$

$cos(A+B)=cosA\:cosB-sinA\:sinB$

Now,

$cos(A-B)=cosA\:cosB+sinA\:sinB$....................(1)

$cos(A+B)=cosA\:cosB-sinA\:sinB$.....................(2)

Adding (1) and (2)

$cos(A-B)-cos(A+B)=(cosA\:cosB+sinA\:sinB)-(cosA\:cosB-sinA\:sinB)$

$cos(A-B)-cos(A+B)=cosA\:cosB+sinA\:sinB-cosA\:cosB+sinA\:sinB$

$cos(A-B)-cos(A+B)=2\:sinA\:sinB$

$\implies\:\frac{1}{2}[cos(A-B)-cos(A+B)]=sinA\:sinB$