# challenging trig identity please help!?

solve

sin(x) + sin(x + 14π/3) + sin(x - 8π/3) = 0

thanks in advance! please show me your steps

hi Pope, do you mind explaining your answer a bit? :/ im not following the switch to cos and the negative sign out of the bracket thanks!

### 3 Answers

- PopeLv 78 years agoFavorite Answer
LHS = sin(x) + sin(x + 14π/3) + sin(x - 8π/3)

= sin(x) + sin(x + 2π/3) + sin(x - 2π/3)

= sin(x) + sin(x)cos(2π/3) + cos(x)sin(2π/3) + sin(x)cos(-2π/3) + cos(x)sin(-2π/3)

= sin(x) + sin(x)cos(2π/3) + cos(x)sin(2π/3) + sin(x)cos(2π/3) - cos(x)sin(2π/3)

= sin(x) + 2sin(x)cos(2π/3)

= sin(x) + 2sin(x)(-1/2)

= sin(x) - sin(x)

= 0

= RHS

Here are a few identities that may address your followup questions:

sin(A + B) = sin(A)cos(B) + cos(A)sin(B)

sin(-A) = -sin(A)

cos(-A) = cos(A)

- Anonymous8 years ago
sin(x) + sin(x + 14π/3) + sin(x - 8π/3) = 0

First, you substitute the sins to the parentheses.

sin(x)+ sin(x)+ sin(14π/3)+ sin(x)- sin(8π/3) = 0

Next, you add all the single sins. There are 3 single sins with just an x

3sin(x)+ sin(14π/3)- sin(8π/3) = 0

The complex sins with what is in the parentheses have the same denominator. all you do is subtract what is in the parentheses. 14-8=6

3sin(x)+ sin(6π/3)= 0

Subtract sin(6π/3) from both sides

3sin= -sin(6π/3)

Divide both side by 3

sin= sin(2π)

After that, i don't know. (I hope i was able to help you a little)

- 4 years ago
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