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Trigonometric Identities

+2 votes
I seem to be confused when it comes to solving thisidentity: (sin a + sin b)/(csc a + csc b) = sin a sin b I've tried about 3 different methods to verify thisidentity and I still can't seem to solve it.
asked Feb 14, 2013 in TRIGONOMETRY by payton Apprentice

1 Answer

+3 votes

LHS [sin(a) + sin(b)]/[csc(a) + csc(b)]

Consider: csc(a) + csc(b)

Reciprocal identities: cscθ = 1/sinθ.

= 1/sin(a) + 1/sin(b)

Rewrite the expression with common denominator.

= [sin(b) + sin(a)]/sin(a)sin(b).

1/[csc(a) + csc(b)] = [sin(a)sin(b)]/[sin(b) + sin(a)]

Therefore [sin(a) + sin(b)]/[csc(a) + csc(b)] = [sin(a) + sin(b)]{[sin(a)sin(b)]/[sin(b) + sin(a)]}

[sin(a) + sin(b)]/[csc(a) + csc(b)] = sin(a)sin(b).

answered Feb 15, 2013 by britally Apprentice

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