Simplifying Expressions Involving Trigonometric and Inverse Trigonometric Functions (C)

Identities involving trigonometric functions and their inverses

We are interested in simplifying such hair raising expressions as $\tan(\arcsin(x/3))$ or $\sin(2\arctan(x))$. There is a simple way to deal with all such expressions. Just recall that the value of an inverse trigonometric function is an angle. For example:

• $\arctan(x/3)$ is an angle, call it $\theta$, whose tangent is x/3 and which lies in the interval $(-\pi/2,\pi/2)$. If we assume, for the moment, that x is positive, then there is a right triangle containing $\theta$ where the side opposite $\theta$ has length x and the side adjacent $\theta$ has length 3. Therefore the hypotenuse of this triangle has length $\sqrt{9 + x^2}$, so $\sin(\arctan(x/3)) = \sin(\theta) = x/\sqrt{9+x^2}$. All the domains and ranges have been worked out so that this calculation works even if x < 0, but if you want to check for yourself, place the vertex of $\arctan(x/3)$ at the origin, make one side the positive x axis, and the let the other side go through (3,x). This side will intersect the unit reference circle at $(3/\sqrt{9+x^2},x/\sqrt{9+x^2})$, so the sine of this angle is $x/\sqrt{9+x^2})$.
• Simplify $\sin(2\arccos(x))$. First we employ the double angle identity

  $$\sin(2\arccos(x)) = 2\sin(\arccos(x))\cos(\arccos(x)).$$

  One part is easy: $\cos(\arccos(x)) = x$. As for $\sin(\arccos(x))$ we proceed as in the previous example: There is a right triangle containing $\arccos(x)$ where the hypotenuse of the triangle is of length 1 and whose side adjacent to $\arccos(x)$ has length x. The other side of this right triangle must have length $\sqrt{1-x^2}$ so $\sin(\arccos(x)) = \sqrt{1-x^2}$. Therefore $\sin(2\arccos(x)) = 2x\sqrt{1-x^2}$.