Use the fact that 180° is equivalent to [latex]\pi[/latex] radians as a conversion factor: [latex]1=\dfrac{\pi \, \text{rad}}{180^{\circ}}=\dfrac{180^{\circ}}{\pi \, \text{rad}}[/latex].

1. [latex]225^{\circ}=225^{\circ}·\dfrac{\pi }{180^{\circ}}=\dfrac{5\pi }{4}\text{ rad}[/latex]
2. [latex]\dfrac{5\pi }{3}\text{ rad}[/latex]= [latex]\dfrac{5\pi }{3}·\dfrac{180^{\circ}}{\pi }=300^{\circ}[/latex]

1. On the unit circle, the angle [latex]\theta =\large\frac{2\pi}{3}[/latex] corresponds to the point [latex]\Big(-\large\frac{1}{2}, \frac{\sqrt{3}}{2}\Big)[/latex]. Therefore, [latex]\sin \Big(\large\frac{2\pi}{3}\Big) \normalsize = y = \large\frac{\sqrt{3}}{2}[/latex].
   

   Figure 4. Unit circle depiction of [latex]\theta[/latex]

2. An angle [latex]\theta =-\large\frac{5\pi}{6}[/latex] corresponds to a revolution in the negative direction, as shown. Therefore, [latex]\cos \Big(-\large\frac{5\pi}{6}\Big)[/latex] [latex]=x=-\large\frac{\sqrt{3}}{2}[/latex].
   

   Figure 5. Unit circle depiction of [latex]\theta[/latex]

3. An angle [latex]\theta =\large\frac{15\pi}{4} \normalsize = 2\pi +\large\frac{7\pi}{4}[/latex]. Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of [latex]\large\frac{7\pi}{4}[/latex] corresponds to the point [latex]\Big(\large\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\Big)[/latex], we can conclude that [latex]\tan \Big(\large\frac{15\pi}{4}\Big)\normalsize =\large\frac{y}{x}[/latex] [latex]=-1[/latex].
   

   Figure 6. Unit circle depiction of [latex]\theta[/latex]

Let [latex]x[/latex] denote the length of the ramp. In the following image, we see that [latex]x[/latex] needs to satisfy the equation [latex]\sin(10^{\circ})=\dfrac{4}{x}[/latex]. Solving this equation for [latex]x[/latex], we see that [latex]x=\frac{4}{ \sin(10^{\circ})} \approx 23.035[/latex] ft.

Figure 8. Sketch of the ramp and staircase.

[latex]1+\cos(2\theta)=\cos \theta[/latex]

if and only if

[latex]1+2\cos^2 \theta -1=\cos \theta[/latex],

which is true if and only if

[latex]2\cos^2 \theta -\cos \theta =0[/latex].

To solve this equation, it is important to note that we need to factor the left-hand side and not divide both sides of the equation by [latex]\cos \theta[/latex]. The problem with dividing by [latex]\cos \theta[/latex] is that it is possible that [latex]\cos \theta[/latex] is zero. In fact, if we did divide both sides of the equation by [latex]\cos \theta[/latex], we would miss some of the solutions of the original equation. Factoring the left-hand side of the equation, we see that [latex]\theta[/latex] is a solution of this equation if and only if

[latex]\cos \theta (2\cos \theta -1)=0[/latex]

Since [latex]\cos \theta =0[/latex] when

[latex]\theta =\large \frac{\pi }{2}, \, \frac{\pi}{2} \normalsize \pm \pi, \, \large \frac{\pi}{2} \normalsize \pm 2\pi, \cdots[/latex],

and [latex]\cos \theta =1/2[/latex] when

[latex]\theta =\large \frac{\pi}{3}, \, \frac{\pi}{3} \normalsize \pm 2\pi, \cdots[/latex], or [latex]\theta =-\large \frac{\pi}{3}, \, -\frac{\pi}{3} \normalsize \pm 2\pi, \cdots[/latex],

we conclude that the set of solutions to this equation is

[latex]\theta =\large \frac{\pi}{2} \normalsize +n\pi, \, \theta =\large \frac{\pi}{3} \normalsize +2n\pi[/latex], and [latex]\theta =-\large \frac{\pi}{3}\normalsize +2n\pi, \, n=0, \pm 1, \pm 2,\cdots[/latex]

b. Using the double-angle formula for [latex]\sin(2\theta)[/latex] and the reciprocal identity for [latex]\tan(\theta)[/latex], the equation can be written as

[latex]2\sin \theta \cos \theta =\large \frac{\sin\theta}{\cos \theta}[/latex]

To solve this equation, we multiply both sides by [latex]\cos \theta[/latex] to eliminate the denominator, and say that if [latex]\theta[/latex] satisfies this equation, then [latex]\theta[/latex] satisfies the equation

[latex]2\sin \theta \cos^2 \theta -\sin \theta =0[/latex]

However, we need to be a little careful here. Even if [latex]\theta[/latex] satisfies this new equation, it may not satisfy the original equation because, to satisfy the original equation, we would need to be able to divide both sides of the equation by [latex]\cos \theta[/latex]. However, if [latex]\cos \theta =0[/latex], we cannot divide both sides of the equation by [latex]\cos \theta[/latex]. Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check for extraneous solutions. Returning to the equation, it is important that we factor [latex]\sin \theta[/latex] out of both terms on the left-hand side instead of dividing both sides of the equation by [latex]\sin \theta[/latex]. Factoring the left-hand side of the equation, we can rewrite this equation as

[latex]\sin \theta (2\cos^2 \theta -1)=0[/latex]

Therefore, the solutions are given by the angles [latex]\theta[/latex] such that [latex]\sin \theta =0[/latex] or [latex]\cos^2 \theta =1/2[/latex]. The solutions of the first equation are [latex]\theta =0, \pm \pi, \pm 2\pi, \cdots[/latex]. The solutions of the second equation are [latex]\theta =\pi /4, \, (\pi/4) \pm (\pi/2), \, (\pi/4) \pm \pi, \cdots[/latex]. After checking for extraneous solutions, the set of solutions to the equation is

[latex]\theta =n\pi[/latex] and [latex]\theta =\large \frac{\pi}{4}+\frac{n\pi}{2}, \, \normalsize n=0, \pm 1, \pm 2, \cdots[/latex].

We start with the identity

Dividing both sides of this equation by [latex]\cos^2 \theta[/latex], we obtain

Since [latex]\sin \theta / \cos \theta =\tan \theta[/latex] and [latex]1 / \cos \theta =\sec \theta[/latex], we conclude that