# What is the exact value of $\cos\left(\arccos\dfrac{1}{7}+\arcsin\dfrac{1}{5}\right)$

What is the exact value of $\cos\left(\arccos\dfrac{1}{7}+\arcsin\dfrac{1}{5}\right)$?

I wasn't sure if I was doing this correctly or using the fastest method, but I wrote $x=\arccos\dfrac{1}{7}, y=\arcsin\dfrac{1}{5}$

and used $\cos(x+y)=\cos x\cos y-\sin x \sin y$

However, I still had $\cos(\arcsin y)$ so do I have to use $\cos x=\sqrt{1-sin^2x}$? Is there a faster way to doing this?

• WA gives $\frac{2\sqrt{3}(\sqrt{2}-2)}{35}$. – pisoir Dec 10 '14 at 9:45
• – lab bhattacharjee Dec 10 '14 at 9:46

Note that the expression $$\arcsin \frac{1}{5}$$ represents an angle whose sine is $\frac{1}{5}$. If we think of a right triangle with one leg with length $1$ and a hypotenuse of $5$, then the other leg by the Pythagorean theorem must have length $$\sqrt{5^2 - 1^2} = \sqrt{24} = 2\sqrt{6}.$$ Since the sine of the angle is $\frac{1}{5}$, this means the leg that is adjacent to the angle is $2\sqrt{6}$, and the opposite leg is $1$. So the cosine of this angle is obviously $$\cos \arcsin \frac{1}{5} = \frac{2\sqrt{6}}{5}.$$