Question

i) sin^ 2 A+sin^ 2 (60^ 0 +A)+sin^ 2 (60^ 0 -A ) = 3÷ 2​

Answer 1

Answer:

Step-by-step explanation:

L.H.S = sin^ 2 A+sin^ 2 (60^ 0 +A)+sin^ 2 (60^ 0 -A )

=sin2^A+1/2(1−cos^2(60+A))+1/2(1−cos^2(60−A))

=sin2^A+1−1/2(cos(120+2A)+cos(120−2A))

=sin2^A+1−1/2×2cos((120+2A)+(120−2A))/2.cos((120+2A)-(120−2A))/2

=sin2^A+1−cos120.cos2A

=sin2^A+1+1/2.(1−2sin2^A)

=sin2^A+1+1/2−sin2^A

=3/2 = R.H.S

Used formulas:

sin^2A=1/2(1−cos2A)

cosC+cosD=2.cos(C+D)/2.cos(C−D)/2