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sin3y/siny=1+2cos2y prove LHS=RHS?
trignometry sum
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- Anonymous1 decade agoFavorite Answer
sin3y/siny
= sin(y+2y) / siny
= ( sinycos2y + cosysin2y) / siny
Using sin2y = 2sinycosy
= [ sinycos2y + 2siny(cosy)^2 ] / siny
= siny[ cos2y + 2(cosy)^2 ] / siny
= cos2y + 2(cosy)^2
= cos2y + 2(cosy)^2 - 1 + 1
Using cos2y = 2(cosy)^2 - 1
= cos2y + cos2y + 1
= 2cos2y + 1
- 1 decade ago
sin3y/siny = 1 + 2cos2y
LHS
= sin3y/siny
= (3 siny - 4 (siny)^3)/siny
= 3 - 4(siny)^2
= 3 - 2(1 - cos2y)
= 1 + 2cos2y
= RHS
- MadhukarLv 71 decade ago
sin3y
= 3siny - 4sin^3 y
= siny (3 - 4sin^2 y)
= siny (4 - 4sin^2 y - 1)
= siny (4cos^2 y - 1)
= siny [2(1 + cos2y) - 1]
= siny (1 + 2cos2y)
=> sin3y/siny = 1 + 2cos2y
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