Chapter 1.8, Problem 22E

a.

To determine

To check the identity

  ${\sin }^2\theta +{\cos }^2\theta =1$

At $\theta =0$

Answer to Problem 22E

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

Explanation of Solution

Given:

  $\theta =0$

Concept Used:

  $sin0=0,\text{ and }\cos \left(0\right)=1$

Calculation:

To check the identity

  ${\sin }^2\theta +{\cos }^2\theta =1$

At $\theta =0$

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

  $\begin{array}{c}{\sin }^{2}0+{\cos }^{2}0={\left(\sin 0\right)}^2+{\left(\cos 0\right)}^2\\ =0^2+1^2\\ =1\\ =\text{ right hand side}\end{array}$

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

  ${\sin }^2\theta +{\cos }^2\theta =1$

At $\theta =\pi /4$

Answer to Problem 22E

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

Explanation of Solution

Given:

  $\theta =\pi /4$

Concept Used:

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

Calculation:

To check the identity

  ${\sin }^2\theta +{\cos }^2\theta =1$

At $\theta =\pi /4$

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

  $\begin{array}{c}{\sin }^2\left(\pi /4\right)+{\cos }^2\left(\pi /4\right)={\left(\sin \left(\pi /4\right)\right)}^2+{\left(\cos \left(\pi /4\right)\right)}^2\\ ={\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2\\ =\frac{1}{2}+\frac{1}{2}\\ =1\\ =\text{ right hand side}\end{array}$

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

  ${\sin }^2\theta +{\cos }^2\theta =1$

At $\theta =\pi /2$

Answer to Problem 22E

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

Explanation of Solution

Given:

  $\theta =\pi /2$

Concept Used:

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

Calculation:

To check the identity

  ${\sin }^2\theta +{\cos }^2\theta =1$

At $\theta =\pi /2$

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

  $\begin{array}{c}{\sin }^2\left(\pi /2\right)+{\cos }^2\left(\pi /2\right)={\left(\sin \left(\pi /2\right)\right)}^2+{\left(\cos \left(\pi /2\right)\right)}^2\\ ={\left(1\right)}^2+{\left(0\right)}^2\\ =1\\ =\text{ right hand side}\end{array}$

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

  ${\sin }^2\theta +{\cos }^2\theta =1$

At $\theta =\pi$

Answer to Problem 22E

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

Explanation of Solution

Given:

  $\theta =\pi$

Concept Used:

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

Calculation:

To check the identity

  ${\sin }^2\theta +{\cos }^2\theta =1$

At $\theta =\pi$

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

  $\begin{array}{c}{\sin }^2\left(\pi \right)+{\cos }^2\left(\pi \right)={\left(\sin \left(\pi \right)\right)}^2+{\left(\cos \left(\pi \right)\right)}^2\\ ={\left(0\right)}^2+{\left(-1\right)}^2\\ =1\\ =\text{ right hand side}\end{array}$

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 $\theta =\pi$