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ho to derive formula of sin2asin2bsin2c

mohitsoni , 10 Years ago
Grade 11
anser 1 Answers
Nishant Vora

Last Activity: 10 Years ago

sin2A sin2B sin2C
=1/2 ( 2 sin2A sin2B sin2C)
=1/ ( cos(2A-2B) – cos(2A + 2B)) cos2C
= ½ (( cos(2A-2B) – cos(2pi – 2C)) cos2C )
= ½ (( cos(2A-2B) – cos(2C)) cos2C )
= ½ (( cos(2A-2B)cos2C – cos22C )
= ½ (( cos(2A-2B)cos(2A + 2B) – cos22C )
= ½ ( cos4A + cos4B – (cos4A +1)/2)
= ¼ ( cos4A + cos4B + cos4A -1 )

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