Chapter 5.2, Problem 71E

Solution

To prove: The identity ${\sin }^{2}A+{\sin }^{2}B=1$ .

Given information:

The given identity is ${\sin }^{2}A+{\sin }^{2}B=1$ .

Formula used:

The trigonometric identity is ${\sin }^{2}x+{\cos }^{2}x=1$ .

The co-function is given as $\sin \left(90°-x\right)=\cos x$ .

Proof:

It is known that if two angles are complementary, then the sum of angles is $90°$ .

Add the angles $A$ and $B$ , and equate it to $90°$ and solve the resulting equation for angle $B$ .

  $\begin{array}{c}A+B=90°\\ B=90°-A\end{array}$

Substitute $B=90°-A$ into the left-hand side of the given equation and simplify the left-hand side of the given equation using the co-function, $\sin \left(90°-x\right)=\cos x$ and the trigonometric identity.

  $\begin{array}{c}{\sin }^{2}A+{\sin }^{2}B={\sin }^{2}A+{\sin }^2\left(90°-A\right)\\ ={\sin }^{2}A+{\cos }^{2}A\\ =1\end{array}$

Here, the left-hand side is equal to the right-hand side.

Hence, it is proved that ${\sin }^{2}A+{\sin }^{2}B=1$ .