Question

sin(45+A)+sin(45-A)=√2cosA
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Answer 1

Answer:

LHS = sin(45+A)sin(45-A)

= sin[90-(45-A)]sin(45-A)

= cos(45-A)sin(45-A)

\* By complementary angles:

sin(90-A) = cosA */

= \frac{1}{2}\times 2sin(45-A)cos(45-A)21×2sin(45−A)cos(45−A)

= \frac{1}{2}\times sin[2(45-A)]21×sin[2(45−A)]

/* 2sinAcosA = sin2A */

= \frac{1}{2}sin(90-2A)21sin(90−2A)

= \frac{1}{2}cos2A21cos2A

/* sin(90-A) = cosA */

= RHS

Therefore,

sin(45+A) sin (45-A)

= 1/2 cos2A