Chapter 5.4, Problem 55E

To determine

To find: The algebraic form of the given trigonometric expression.

Answer to Problem 55E

The algebraic form of the trigonometric expression $\cos \left(\arccos x+\arcsin x\right)$ is $0$ .

Explanation of Solution

Given information:

The trigonometric expression $\cos \left(\arccos x+\arcsin x\right)$ .

Formula used:

Sum rule for cosine function is $\cos \left(x+y\right)=\cos x\cos y-\sin x\sin y$ .

Calculation:

Consider trigonometric expression $\cos \left(\arccos x+\arcsin x\right)$ .

The above expression is of the form $\cos \left(v+u\right)$ , where $u=\arcsin x$ and $v=\arccos x$ .

Recall sum rule for cosine function is $\cos \left(x+y\right)=\cos x\cos y-\sin x\sin y$ .

Trigonometric and its inverse trigonometric function cancel each other.

Apply it,

  $\begin{array}{c}\cos \left(\arccos x+\arcsin x\right)=cos\left(\arccos x\right)\cos \left(\arcsin x\right)-\sin \left(\arccos x\right)\sin \left(\arcsin x\right)\\ =x\cdot \cos \left(\arcsin x\right)+\sin \left(\arccos x\right)\cdot x\end{array}$

For $u=\arcsin x$ .

Recall that $\sin x=\frac{\text{perpendicular}}{\text{hypotenuse}}$

Construct a right angle triangle with perpendicular p as x and hypotenuse h as 1.

Since, $h^2=p^2+b^2$ , where b is base of triangle.

So, base b of triangle is $\sqrt{1-x^2}$ .

  

For $v=\arccos x$ .

Recall that $\cos x=\frac{\text{base}}{\text{hypotenuse}}$

Construct a right angle triangle with base b as x and hypotenuse h as 1.

Since, $h^2=p^2+b^2$ , where p is perpendicular of triangle.

So, perpendicular p of triangle is $\sqrt{1-x^2}$ .

  

The expression is reduced to,

  $\begin{array}{c}\cos \left(\arccos x+\arcsin x\right)=x\cos \left(\arccos \frac{\sqrt{1-x^2}}{1}\right)-x\sin \left(\arcsin \frac{\sqrt{1-x^2}}{1}\right)\\ =x\sqrt{1-x^2}-x\sqrt{1-x^2}\\ =0\end{array}$

Thus, the algebraic form of the trigonometric expression $\cos \left(\arccos x+\arcsin x\right)$ is $0$ .