1. ## Trig help

If cos x + sin x =2/3, find the value of sin(2x)

I've got that the solution is: -5/9, but how do you get this?

2. Hello, white_cap!

If the answer is .$\displaystyle \text{- }\frac{5}{9}$, there must be typo . . .

If $\displaystyle \cos x + \sin x \:=\:\frac{2}{3}$, .find the value of $\displaystyle {\color{red}\sin}(2x)$

Square both sides: .$\displaystyle (\cos x + \sin x)^2 \:=\:\left(\frac{2}{3}\right)^2\quad\Rightarrow\qu ad \cos^2\!x + 2\cos x\sin x + \sin^2\!x \;=\;\frac{4}{9}$

And we have: .$\displaystyle \underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} + 2\sin x\cos x \;=\;\frac{4}{9}$

Then: .$\displaystyle 1 + \underbrace{2\sin x\cos x}_{\text{This is }\sin(2x)} \:=\:\frac{4}{9}$

Therefore: .$\displaystyle 1 + \sin(2x) \:=\:\frac{4}{9}\quad\Rightarrow\quad \sin(2x) \;=\;-\frac{5}{9}$

3. Thanks so much, you're right it should have read:

if cosx + sinx = 2/3 find the value of sin2x

sorry about that, and thanks for providing the awnser AND correcting my mistake