Question

If A + B + C = 90°, then sin 2A + sin 2B + sin 2C = 
1. 2cos A cos B cos C
2. 4 cos A cos B cos C
3. 2sin \(\rm A\over2\) cos \(\rm B\over2\) cos\(\rm C\over2\)
4. 4cos \(\rm A\over2\) cos \(\rm B\over2\) cos\(\rm C\over2\)

Answer 1

Correct Answer - Option 2 : 4 cos A cos B cos C

Concept:

sin a + sin b = 2 sin (\(\rm a+b\over 2\)) cos (\(\rm a-b\over 2\))

cos a + cos b = 2 cos (\(\rm a+b\over 2\)) cos (\(\rm a-b\over 2\))

sin 2a = 2 sin a cos a

Calculation:

S = sin 2A + sin 2B + sin 2C

S = [2 sin (\(\rm 2A+2B\over 2\)) cos (\(\rm 2A-2B\over 2\))] + 2 sin C cos C

S = 2[sin (A + B) cos (A - B) + sin C cos C]

Given: A + B + C = 90°

So, A + B = 90 - C and C = 90 - (A + B)

S = 2 [sin (90 - C) cos (A - B) + sin (90 - (A + B) cos C]

S = 2 [cos C cos (A - B) + cos (A + B) cos C]

S = 2 cos C [cos (A - B) + cos (A + B)]

S = 2 cos C [2 cos (\(\rm A+B+A-B\over 2\)) cos (\(\rm A+B-(A-B)\over 2\))]

S = 4 cos A cos B cos C