# A triangle has sides $3$, $5$, and $7$. Express $\cos(y)+\sin(y)$, where $y$ is the largest angle in the triangle.

A triangle has the side lengths of $3$, $5$, and $7$. Express $\cos(y)+\sin(y)$, where $y$ is the largest angle in the triangle.

I have tried to apply pythagoras theorm, trying to express the other two angles in some way, split the triangle into smaller triangles, but all without success.

The largest angle is the one "against" the side of length $7$. Use the law of cosine to determine the cosine of the angle. Then use the equality:
$$sin^2x+cos^2x=1$$
Note: You'll have $2$ options for the value of the sine, but it is always non-negative on $[0,\pi]$ so you should take the non-negative root.
Let $a=7, b=5, c=3$ The largest angle is the angle against the $a$ side. Let's call it $y$. Then from cosine law: $$a^2 =b^2+c^2-2bccos(y)\Rightarrow 49=25+9-30cos(y)\Rightarrow 30 cos(y)=-15\Rightarrow cos(y)=-\frac{1}{2}$$ Then you use the following identity: $$sin^2(y)+ cos^2(y)=1\Rightarrow sin^2(y)=1- cos^2(y) =1-\frac{1}{4}=\frac{3}{4}\Rightarrow sin(y)=\frac{\sqrt3}{2}$$ Hence: $$cos(y)+sin(y)=-\frac{1}{2}+ \frac{\sqrt3}{2}=\frac{\sqrt3 -1}{2}$$