Evaluating $\sin\left(\arccos\frac12+\arccos\frac{7}{25}\right)$

Question

Evaluate $$\sin\left(\arccos\frac12+\arccos\frac{7}{25}\right)$$

I know that $\arccos\frac12$ is $60^\circ$. I don't know how to continue.

Answer 1

$$\sin\left(\arccos\frac12+\arccos\frac{7}{25}\right)=$$

$$\sin(x+y) = \sin x \cos y + \cos x \sin y$$

Where $$x=\arccos\frac12$$ and $$y= \arccos\frac{7}{25}$$

Note that $$ \cos x =\frac {1}{2}$$ and $$ \cos y =\frac {7}{25}$$

We can find $$ \sin x= {\sqrt 3}/2 $$ and $$\sin y = 24/{25}$$

upon substitution we get $$\sin\left(\arccos\frac12+\arccos\frac{7}{25}\right)=$$

$$ {7\sqrt 3}/50 +24/{50} = \frac { 7\sqrt 3 + 24}{50} $$

Answer 2

Hint:

let $$ \cos \alpha= \frac{1}{2} \qquad \cos \beta= \frac{7}{25} $$

so yo u have to evaluate $\sin(\alpha+\beta)$.

Use the addition formula and, given $\cos \beta=\frac{7}{25}$ find $\sin \beta=\sqrt{1-\cos^2 \beta}$