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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.

Thanks in advance.

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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:

$$sin^2x+cos^2x=1$$

to get the sine of the angle.

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.

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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: $$cos(y)+sin(y)=-\frac{1}{2}+ \frac{\sqrt3}{2}=\frac{\sqrt3 -1}{2} $$

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