trig question

[ydoaPs]

[math]sin(\frac{\pi}{6}+x)=\frac{1}{2}(cosx+\sqrt{3}sinx)[/math]

 

how do i start this one? am i supposed to distribute the right side or work with the left?

[swansont]

I'd start with the changing the sin(a+b) term using the angle sum identity.

 

Squaring both sides might end up simplifying things

[ydoaPs]

[math]sin(\frac{\pi}{6}+x)=\frac{1}{2}(cosx+\sqrt{3}sinx)[/math]

[math]sin\frac{\pi}{6}cosx+cos\frac{pi}{6}sinx=\frac{1}{2}(cosx+\sqrt{3}sinx)[/math]

[math]\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx=\frac{1}{2}(cosx+\sqrt{3}sinx)[/math]

[math]\frac{1}{2}(cosx+\sqrt{3}sinx)=\frac{1}{2}(cosx+\sqrt{3}sinx)[/math]

 

wow, i'm stupid. i made that way harder than it was. thanx for the help.