Chapter 5.4, Problem 37E

Solution

(a)

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

Given information:

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

Formula used:

Use the double angle identities of cosine.

  $\cos 2u={\cos }^{2}u-{\sin }^{2}u$

Use the Pythagorean identity.

  ${\sin }^{2}u+{\cos }^{2}u=1$

Proof:

Simplify the right-hand side of the given identity by using double angle identities of cosine and Pythagorean identity.

  $\begin{array}{c}\frac{1-\cos 2u}{2}=\frac{1-\left({\cos }^{2}u-{\sin }^{2}u\right)}{2}\\ =\frac{{\cos }^{2}u+{\sin }^{2}u-{\cos }^{2}u+{\sin }^{2}u}{2}\\ =\frac{2{\sin }^{2}u}{2}\\ ={\sin }^{2}u\end{array}$

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

(b)

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

Given information:

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

Calculation:

Simplify the right hand side of the given identity by using double angle identities of cosine and Pythagorean identity.

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

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