Question

All maths aryabhatt are fake.....
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Koi answer nahi de raha h

Question is.....

Sin^2 A/2 + sin^2 B/2 + sin^2 C/2
=1-2cos A/2 sin B/2 sin C/2
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Ab to de do

Answer 1

Hey here is your answer_________________

Complete question is:

In triangle ABC, show that sin2 A/2 * sin2B/2 * sin2 C/2 = 1 - 2 * sin A/2 * sin

B/2 * sin C/2

Solution:

In a triangle ABC,

A + B + C = 180

LHS     

   sin2 A/2 + sin2 B/2 + sin2 C/2

=  (1/2) * (2sin2 A/2 + 2sin2 B/2 + 2sin2 C/2)

=  (1/2) * (1 - cos A + 1 - cos B + 1 - cos C)

=  (1/2) * {3 -(cos A + cos B + cos C)}

=  (1/2) * [3 - {2cos (A + B)/2 * cos(A - B)/2 + cos C)}]

=  (1/2) * [3 - {2cos(90 - C/2) * cos(A - B)/2 + cos C}]

=  (1/2) * [3 - {2sin C/2 * cos(A - B)/2 + 1 - 2sin2 C/2}]

=  (1/2) * [3 - 1 - 2sin C/2 * {cos(A + B)/2 -sin C/2)}]

=  (1/2) * [2 - 2sin C/2 * {cos(A - B)/2 - sin(90- (A+B)/2}]   {since sin(90 - θ) = cos θ)

=  (1/2) * {2 - sin C/2 * cos(A - B)/2 - cos(A + B)/2}

=  (1/2) * {2 - 2sin C/2* (2sin A/2 * sin B/2)}

=  (1/2) * (2 - 4sin A/2 * sin B/2 * sin C/2)

=  1 - 2*sin A/2 * sin B/2 * sin C/2

=  RHS   

So,
sin2 A/2 * sin2B/2 * sin2 C/2 = 1 - 2 * sin A/2 * sin B/2 * sin C/2


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HOPE THIS ANSWER WILL HELP U...........