# Find $\sin(x+y)$, given $\tan x$ and $\cos y$

Given that $\tan x= -2$ and $\cos y= 1/2$ where $x$ and $y$ are in the 4th and 1st quadrants respectively. Find, without evaluating angles $x$ and $y$,

a) $\sin (x+y)$

Here is what i have done so far..

For (X)

a² + b² = c²

1² + (-2)² = c²

1 +4 = c²

√5 = c²

For (Y)

a² + b² = c²

1² + b² = 2²

1 + b² = 2²

√b² = √3

√5 = c²

From here

sin (x+y)

sinx cosy + cosx sin y

= (-2/√5)(1/2) + (1/√5)(√3/2)

= (-2/2√5) + (1√3/2√5)

= -2+√3/√5 x √5/√5

= -2√5+√15/2√25

i got lost at this point

• You are actually done at $$\frac{\sqrt3-2}{2\sqrt5}$$ Optionally we can rationalize the denominator to get $$\frac{\sqrt{15}-2\sqrt5}{10}$$ – lab bhattacharjee Jul 29 '14 at 13:22
• how did you get the 10 in the denominator? – user159676 Jul 29 '14 at 13:25
• I assume by multiply by $\sqrt{5}$ – Buzi Jul 29 '14 at 13:27
• @user159676, What is $$\sqrt{25}?$$ – lab bhattacharjee Jul 29 '14 at 13:28
• @ lab bhattacharjee √25 = 5² – user159676 Jul 29 '14 at 13:31

$$1+\tan^2 x=\sec^2 x$$ so $\sec^2 x=5$ this means that $\cos x =\frac{1}{\sqrt{5}}$ since $x$ is fourth quadrent.
$$\sin x =\tan x \cos x = -\frac{2}{\sqrt{5}}$$
Also $\sin^2 y +\cos ^2 y=1$ gives $\sin y=\frac{\sqrt{3}}{2}$
So $$\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)=-\frac{2}{\sqrt{5}}\frac{1}{2} +\frac{1}{\sqrt{5}}\frac{\sqrt{3}}{2}=\frac{\sqrt{3}-2}{2\sqrt{5}}$$