Chapter 5.4, Problem 38E

Solution

(a)

To prove: The identity ${\sin }^{2}u=\frac{1-\cos 2u}{2}$ .

Given information:

The identity is ${\tan }^{2}u=\frac{1-\cos 2u}{1-\cos 2u}$ .

Formula used:

Use the double angle identities of cosine.

  $\begin{array}{c}\cos 2u=1-2{\sin }^{2}u\\ \cos 2u=2{\cos }^{2}u-1\end{array}$

Proof:

Simplify the left hand side of the given identity by using quotient identity and double angle identities of cosine.

  $\begin{array}{c}{\tan }^{2}u=\frac{{\sin }^{2}u}{{\cos }^{2}u}\\ =\frac{\frac{1-\cos 2u}{2}}{\frac{1+\cos 2u}{2}}\\ =\frac{1-\cos 2u}{2}\cdot \frac{2}{1+\cos 2u}\\ =\frac{1-\cos 2u}{1+\cos 2u}\end{array}$

Therefore, it is proved that ${\tan }^{2}u=\frac{1-\cos 2u}{1-\cos 2u}$ .

(b)

To explain: The reason why the identity in the previous part does not imply that $u=\sqrt{\frac{1-\cos 2u}{1-\cos 2u}}$ .

The identity in the previous part does not imply that $u=\sqrt{\frac{1-\cos 2u}{1-\cos 2u}}$ because the formula is false.

Calculation:

From the square root property, it is known that $x^2=a$ implies that $x=\pm \sqrt{a}$ , which means that $\tan u$ must be positive or negative.

Another reasoning is that when $u$ is an angle in QII or QIV, the formula is false since the left side will be negative and the right side is positive.

Therefore, the identity in the previous part does not imply that $u=\sqrt{\frac{1-\cos 2u}{1-\cos 2u}}$ because the formula is false.