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# sin3y/siny=1+2cos2y prove LHS=RHS?

trignometry sum

### 3 Answers

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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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