Chapter 1.8, Problem 26E

a.

To determine

To check the identity

  $\sin \left(2\theta \right)=2\sin \theta \cos \theta$

At $\theta =0$

Answer to Problem 26E

The given identity is true at $\theta =0$

Explanation of Solution

Given:

  $\theta =0$

Concept Used:

  $\cos \left(0\right)=1,\text{ and }\sin \left(0\right)=0$

Calculation:

To check the identity

  $\sin \left(2\theta \right)=2\sin \theta \cos \theta$ , at $\theta =0$

Put $\theta =0$ in left hand side of given identity respectively and check whether they are equal, as

  $\begin{array}{c}\sin \left(2\theta \right)=\sin \left(2\left(0\right)\right)\\ =\sin \left(0\right)\\ =0\end{array}$

And;

  $\begin{array}{c}2\sin \theta \cos \theta =2\left(\sin 0\right)\cos \left(0\right)\\ =2\left(0\right)\left(1\right)\\ =0\end{array}$

This, implies that left hand side and right hand side of given identity are true for $\theta =0$ .

Hence, the given identity is true at $\theta =0$

b.

To determine

To check the identity

  $\sin \left(2\theta \right)=2\sin \theta \cos \theta$

At $\theta =\pi /4$

Answer to Problem 26E

The given identity is true at $\theta =\pi /4$

Explanation of Solution

Given:

  $\theta =\pi /4$

Concept Used:

  $sin\left(\frac{\pi }{2}\right)=1,\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}},\text{ and }\sin \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$

Calculation:

To check the identity

  $\sin \left(2\theta \right)=2\sin \theta \cos \theta$ , at $\theta =\pi /4$

Put $\theta =\pi /4$ in left hand side of given identity respectively and check whether they are equal, as

  $\begin{array}{c}\sin \left(2\theta \right)=\sin \left(2\left(\frac{\pi }{4}\right)\right)\\ =\sin \left(\frac{\pi }{2}\right)\\ =1\end{array}$

And;

  $\begin{array}{c}2\sin \theta \cos \theta =2\left(\sin \frac{\pi }{4}\right)\cos \left(\frac{\pi }{4}\right)\\ =2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)\\ =\frac{2}{2}\\ =1\end{array}$

This, implies that left hand side and right hand side of given identity are true for $\theta =\pi /4$ .

Hence, the given identity is true at $\theta =\pi /4$ .

c.

To determine

To check the identity

  $\sin \left(2\theta \right)=2\sin \theta \cos \theta$

At $\theta =\pi /2$

Answer to Problem 26E

The given identity is true at $\theta =\pi /2$

Explanation of Solution

Given:

  $\theta =\pi /2$

Concept Used:

  $sin\left(\frac{\pi }{2}\right)=1,\cos \left(\frac{\pi }{2}\right)=0,\text{ and }\sin \left(\pi \right)=0$

Calculation:

To check the identity

  $\sin \left(2\theta \right)=2\sin \theta \cos \theta$ , at $\theta =\pi /2$

Put $\theta =\pi /2$ in left hand side of given identity respectively and check whether they are equal, as

  $\begin{array}{c}\sin \left(2\theta \right)=\sin \left(2\left(\frac{\pi }{2}\right)\right)\\ =\sin \left(\pi \right)\\ =0\end{array}$

And;

  $\begin{array}{c}2\sin \theta \cos \theta =2\left(\sin \frac{\pi }{2}\right)\cos \left(\frac{\pi }{2}\right)\\ =2\left(1\right)\left(0\right)\\ =0\end{array}$

This, implies that left hand side and right hand side of given identity are true for $\theta =\pi /2$ .

Hence, the given identity is true at $\theta =\pi /2$ .

d.

To determine

To check the identity

  $\sin \left(2\theta \right)=2\sin \theta \cos \theta$

At $\theta =\pi$

Answer to Problem 26E

The given identity is true at $\theta =\pi$

Explanation of Solution

Given:

  $\theta =\pi$

Concept Used:

  $sin\left(\pi \right)=0,\cos \left(\pi \right)=-1,\text{ and }\sin \left(2\pi \right)=0$

Calculation:

To check the identity

  $\sin \left(2\theta \right)=2\sin \theta \cos \theta$ , at $\theta =\pi$

Put $\theta =\pi$ in left hand side of given identity respectively and check whether they are equal, as

  $\begin{array}{c}\sin \left(2\theta \right)=\sin \left(2\pi \right)\\ =0\end{array}$

And;

  $\begin{array}{c}2\sin \theta \cos \theta =2\left(\sin \pi \right)\cos \left(\pi \right)\\ =2\left(0\right)\left(-1\right)\\ =0\end{array}$

This, implies that left hand side and right hand side of given identity are true for $\theta =\pi$ .

Hence, the given identity is true at $\theta =\pi$ .