Chapter 5.4, Problem 53E

To determine

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

Answer to Problem 53E

The algebraic form of the trigonometric expression $\sin \left(\arcsin x+\arccos x\right)$ is $1$ .

Explanation of Solution

Given information:

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

Formula used:

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

Calculation:

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

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

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

Trigonometric and its inverse trigonometric function cancel each other.

Apply it,

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

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,

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

Thus, the algebraic form of the trigonometric expression $\sin \left(\arcsin x+\arccos x\right)$ is $1$ .