If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given [latex]\tan \theta =-\frac{3}{4}[/latex], such that [latex]\theta [/latex] is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because [latex]\theta [/latex] is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse:

[latex]\begin{align}{\left(-4\right)}^{2}+{\left(3\right)}^{2}&={c}^{2}\\ 16+9&={c}^{2}\\ 25&={c}^{2}\\ c&=5\end{align}[/latex]

Now we can draw a triangle similar to the one shown in Figure 2.

Figure 2

1. Let’s begin by writing the double-angle formula for sine.
   [latex]\sin \left(2\theta \right)=2\sin \theta \cos \theta [/latex]

   We see that we to need to find [latex]\sin \theta [/latex] and [latex]\cos \theta [/latex]. Based on Figure 2, we see that the hypotenuse equals 5, so [latex]\sin \theta =\frac{3}{5}[/latex], and [latex]\cos \theta =−\frac{4}{5}[/latex]. Substitute these values into the equation, and simplify.
   Thus,

   [latex]\begin{align}\sin \left(2\theta \right)&=2\left(\frac{3}{5}\right)\left(−\frac{4}{5}\right) \\ &=−\frac{24}{25} \end{align}[/latex]
2. Write the double-angle formula for cosine.
   [latex]\cos \left(2\theta \right)={\cos }^{2}\theta -{\sin }^{2}\theta [/latex]

   Again, substitute the values of the sine and cosine into the equation, and simplify.

   [latex]\begin{align}\cos \left(2\theta \right)&={\left(−\frac{4}{5}\right)}^{2}−{\left(\frac{3}{5}\right)}^{2} \\ &=\frac{16}{25}−\frac{9}{25} \\ &=\frac{7}{25}\end{align}[/latex]
3. Write the double-angle formula for tangent.
   [latex]\tan \left(2\theta \right)=\frac{2\tan \theta }{1−{\tan }^{2}\theta }[/latex]

   In this formula, we need the tangent, which we were given as [latex]\tan \theta =−\frac{3}{4}[/latex]. Substitute this value into the equation, and simplify.

   [latex]\begin{align}\tan \left(2\theta \right)&=\frac{2\left(-\frac{3}{4}\right)}{1-{\left(-\frac{3}{4}\right)}^{2}} \\ &=\frac{-\frac{3}{2}}{1-\frac{9}{16}} \\ &=-\frac{3}{2}\left(\frac{16}{7}\right) \\ &=-\frac{24}{7} \end{align}[/latex]

[latex]\begin{align}\cos \left(6x\right)&=\cos \left(3x+3x\right) \\ &=\cos 3x\cos 3x-\sin 3x\sin 3x \\ &={\cos }^{2}3x-{\sin }^{2}3x \end{align}[/latex]

Analysis of the Solution

This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.

We will work on the right side of the equal sign and rewrite the expression until it matches the left side.

[latex]\begin{align}{\left(\sin \theta +\cos \theta \right)}^{2}&={\sin }^{2}\theta +2\sin \theta \cos \theta +{\cos }^{2}\theta \\ &=\left({\sin }^{2}\theta +{\cos }^{2}\theta \right)+2\sin \theta \cos \theta \\ &=1+2\sin \theta \cos \theta \\ &=1+\sin \left(2\theta \right)\end{align}[/latex]

Analysis of the Solution

This process is not complicated, as long as we recall the perfect square formula from algebra:

[latex]{\left(a\pm b\right)}^{2}={a}^{2}\pm 2ab+{b}^{2}[/latex]

where [latex]a=\sin \theta [/latex] and [latex]b=\cos \theta [/latex]. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.

In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.

[latex]\begin{align}\tan \left(2\theta \right)&=\frac{2\tan \theta }{1-{\tan }^{2}\theta }&& \text{Double-angle formula} \\ &=\frac{2\tan \theta \left(\frac{1}{\tan \theta }\right)}{\left(1-{\tan }^{2}\theta \right)\left(\frac{1}{\tan \theta }\right)}&& \text{Multiply by a term that results in desired numerator}. \\ &=\frac{2}{\frac{1}{\tan \theta }-\frac{{\tan }^{2}\theta }{\tan \theta }} \\ &=\frac{2}{\cot \theta -\tan \theta }&& \text{Use reciprocal identity for }\frac{1}{\tan \theta }.\end{align}[/latex]

Analysis of the Solution

Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show

[latex]\frac{2\tan \theta }{1-{\tan }^{2}\theta }=\frac{2}{\cot \theta -\tan \theta }[/latex]

Let’s work on the right side.

[latex]\begin{align}\frac{2}{\cot \theta -\tan \theta }&=\frac{2}{\frac{1}{\tan \theta }-\tan \theta }\left(\frac{\tan \theta }{\tan \theta }\right) \\ &=\frac{2\tan \theta }{\frac{1}{\cancel{\tan \theta }}\left(\cancel{\tan \theta }\right)-\tan \theta \left(\tan \theta \right)} \\ &=\frac{2\tan \theta }{1-{\tan }^{2}\theta } \end{align}[/latex]

When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.

We will apply the reduction formula for cosine twice.

[latex]\begin{align}{\cos }^{4}x&={\left({\cos }^{2}x\right)}^{2} \\ &={\left(\frac{1+\cos \left(2x\right)}{2}\right)}^{2}&& {\text{Substitute reduction formula for cos}}^{2}x. \\ &=\frac{1}{4}\left(1+2\cos \left(2x\right)+{\cos }^{2}\left(2x\right)\right) \\ &=\frac{1}{4}+\frac{1}{2}\cos \left(2x\right)+\frac{1}{4}\left(\frac{1+\cos \left(2\left(2x\right)\right)}{2}\right) && {\text{ Substitute reduction formula for cos}}^{2}x. \\ &=\frac{1}{4}+\frac{1}{2}\cos \left(2x\right)+\frac{1}{8}+\frac{1}{8}\cos \left(4x\right) \\ &=\frac{3}{8}+\frac{1}{2}\cos \left(2x\right)+\frac{1}{8}\cos \left(4x\right) \end{align}[/latex]

Analysis of the Solution

The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.

We will work on simplifying the left side of the equation:

[latex]\begin{align}{\sin }^{3}\left(2x\right)&=\left[\sin \left(2x\right)\right]\left[{\sin }^{2}\left(2x\right)\right] \\ &=\sin \left(2x\right)\left[\frac{1-\cos \left(4x\right)}{2}\right]&& \text{Substitute the power-reduction formula}. \\ &=\sin \left(2x\right)\left(\frac{1}{2}\right)\left[1-\cos \left(4x\right)\right] \\ &=\frac{1}{2}\left[\sin \left(2x\right)\right]\left[1-\cos \left(4x\right)\right] \end{align}[/latex]

Analysis of the Solution

Note that in this example, we substituted

for [latex]{\sin }^{2}\left(2x\right)[/latex]. The formula states

We let [latex]\theta =2x[/latex], so [latex]2\theta =4x[/latex].

Since [latex]{15}^{\circ }=\frac{{30}^{\circ }}{2}[/latex], we use the half-angle formula for sine:

[latex]\begin{align} \sin \frac{{30}^{\circ }}{2}&=\sqrt{\frac{1-\cos {30}^{\circ }}{2}} \\ &=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}} \\ &=\sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}} \\ &=\sqrt{\frac{2-\sqrt{3}}{4}} \\ &=\frac{\sqrt{2-\sqrt{3}}}{2} \end{align}[/latex]

Analysis of the Solution

Notice that we used only the positive root because [latex]\sin \left({15}^{\text{o}}\right)[/latex] is positive.

Using the given information, we can draw the triangle shown in Figure 3. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate [latex]\sin \alpha =-\frac{8}{17}[/latex] and [latex]\cos \alpha =-\frac{15}{17}[/latex].

Figure 3

1. Before we start, we must remember that, if [latex]\alpha [/latex] is in quadrant III, then [latex]180^\circ <\alpha <270^\circ [/latex], so [latex]\frac{180^\circ }{2}<\frac{\alpha }{2}<\frac{270^\circ }{2}[/latex]. This means that the terminal side of [latex]\frac{\alpha }{2}[/latex] is in quadrant II, since [latex]90^\circ <\frac{\alpha }{2}<135^\circ [/latex].To find [latex]\sin \frac{\alpha }{2}[/latex], we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in Figure 3 and simplify.
   [latex]\begin{align} \sin \frac{\alpha }{2}&=\pm \sqrt{\frac{1-\cos \alpha }{2}} \\ &=\pm \sqrt{\frac{1-\left(-\frac{15}{17}\right)}{2}} \\ &=\pm \sqrt{\frac{\frac{32}{17}}{2}} \\ &=\pm \sqrt{\frac{32}{17}\cdot \frac{1}{2}} \\ &=\pm \sqrt{\frac{16}{17}} \\ &=\pm \frac{4}{\sqrt{17}} \\ &=\frac{4\sqrt{17}}{17} \end{align}[/latex]

   We choose the positive value of [latex]\sin \frac{\alpha }{2}[/latex] because the angle terminates in quadrant II and sine is positive in quadrant II.

2. To find [latex]\cos \frac{\alpha }{2}[/latex], we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in Figure 3, and simplify.
   [latex]\begin{align} \cos \frac{\alpha }{2}&=\pm \sqrt{\frac{1+\cos \alpha }{2}} \\ &=\pm \sqrt{\frac{1+\left(-\frac{15}{17}\right)}{2}} \\ &=\pm \sqrt{\frac{\frac{2}{17}}{2}} \\ &=\pm \sqrt{\frac{2}{17}\cdot \frac{1}{2}} \\ &=\pm \sqrt{\frac{1}{17}} \\ &=-\frac{\sqrt{17}}{17}\end{align}[/latex]

   We choose the negative value of [latex]\cos \frac{\alpha }{2}[/latex] because the angle is in quadrant II because cosine is negative in quadrant II.

3. To find [latex]\tan \frac{\alpha }{2}[/latex], we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in Figure 3 and simplify.
   [latex]\begin{align} \tan \frac{\alpha }{2}&=\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }} \\ &=\pm \sqrt{\frac{1-\left(-\frac{15}{17}\right)}{1+\left(-\frac{15}{17}\right)}} \\ &=\pm \sqrt{\frac{\frac{32}{17}}{\frac{2}{17}}} \\ &=\pm \sqrt{\frac{32}{2}} \\ &=-\sqrt{16} \\ &=-4 \end{align}[/latex]

   We choose the negative value of [latex]\tan \frac{\alpha }{2}[/latex] because [latex]\frac{\alpha }{2}[/latex] lies in quadrant II, and tangent is negative in quadrant II.

Since the angle for novice competition measures half the steepness of the angle for the high level competition, and [latex]\tan \theta =\frac{5}{3}[/latex] for high competition, we can find [latex]\cos \theta [/latex] from the right triangle and the Pythagorean theorem so that we can use the half-angle identities.

[latex]\begin{gathered}{3}^{2}+{5}^{2}=34 \\ c=\sqrt{34} \end{gathered}[/latex]

Figure 4

We see that [latex]\cos \theta =\frac{3}{\sqrt{34}}=\frac{3\sqrt{34}}{34}[/latex]. We can use the half-angle formula for tangent: [latex]\tan \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}[/latex]. Since [latex]\tan \theta [/latex] is in the first quadrant, so is [latex]\tan \frac{\theta }{2}[/latex]. Thus,

[latex]\begin{align} \tan \frac{\theta }{2}&=\sqrt{\frac{1-\frac{3\sqrt{34}}{34}}{1+\frac{3\sqrt{34}}{34}}} \\ &=\sqrt{\frac{\frac{34 - 3\sqrt{34}}{34}}{\frac{34+3\sqrt{34}}{34}}} \\ &=\sqrt{\frac{34 - 3\sqrt{34}}{34+3\sqrt{34}}} \\ &\approx 0.57\end{align}[/latex]

We can take the inverse tangent to find the angle: [latex]{\tan }^{-1}\left(0.57\right)\approx {29.7}^{\circ }[/latex]. So the angle of the ramp for novice competition is [latex]\approx {29.7}^{\circ }[/latex].