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?


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 $$\cos \arcsin \frac{1}{5} = \frac{2\sqrt{6}}{5}.$$

Of course, we could have arrived at this conclusion more directly and more generally, but it is illustrative to think of the inverse trigonometric functions as representing angles, whereas the trigonometric functions themselves take as their arguments an angle and give you a ratio.


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