How did the solution arrive at this step in this inverse Trigonometry step?

Question

Step : $\arcsin(\sin 10^{\circ}) - \arccos(\cos 10^{\circ}) = 3\pi-10-(4\pi-10) = -\pi$

I do not understand how the $\pi$ is coming. Shouldn't it be just $10°-10° = 0$?

Answer 1

You are correct. As stated, the correct answer to the question is $0^\circ$.

Expressed in words, the expression $\arcsin(\sin10^\circ)$ means "The angle between $-90^\circ$ and $90^\circ$ whose sine is the same as $\sin10^\circ$." The only such angle in that range is $10^\circ$.

The expression $\arccos(\cos10^\circ)$ means "The angle between $0^\circ$ and $180^\circ$ whose cosine is the same as $\cos10^\circ$." The only such angle in that range is $10^\circ$.

Therefore, the value of the expression is $10^\circ-10^\circ=0^\circ$.

Note that the only way the given answer could be correct would be in the context of a specially defined, non-standard definition of the inverse sine and cosine functions. You should check to see if that is the case for this exercise.

Answer 2

In inverse trig function we need always consider the co-terminal angle contributions, two solutions per rotation of the radius vector. Using $\theta $ instead of$ 10 ^{\circ}$, we have all angles as

Inverse sine

$$\theta,\pi-\theta,\theta,2\pi+\theta,3\pi-\theta,.. $$

Inverse cosine $$\theta,-\theta,2\pi+\theta,2 \pi-\theta,...$$

Subtract and the difference is always $ k\pi,$ where $k$ need not always be zero.