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Precalculus JumpStart

Mathematics Department

Exercises 11.5 Exercises: Trig III

Exercises: Identities.

These exercises emphasize use of some important trigonometric identities. Review Appendix A of Anton (Handout). Here’s a summary of summary important identities. Don’t feel obligated to memorize the Addition Formulas.
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Table 11.5.1.
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Ratio Identities
\begin{equation*} \tan \theta = \frac{\sin \theta}{\cos \theta} \end{equation*}
\begin{equation*} \cot \theta = \frac{\cos \theta}{\sin \theta} \end{equation*}
Reciprocal Identities
\begin{equation*} \csc \theta = \frac{1}{\sin \theta} \end{equation*}
\begin{equation*} \sec \theta = \frac{1}{\cos \theta} \end{equation*}
\begin{equation*} \cot \theta = \frac{1}{\tan \theta} \end{equation*}
Pythagorean Identity
\begin{equation*} \sin^2 (\theta) + \cos^2 (\theta) = 1 \end{equation*}
Opposite Angle Identities
\begin{equation*} \cos(-\theta) = \cos(\theta) \end{equation*}
\begin{equation*} \sin(-\theta) = - \sin(\theta) \end{equation*}
Addition Formulas
\begin{equation*} \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \end{equation*}
\begin{equation*} \cos(\alpha+\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \end{equation*}

1.

Use the the appropriate identities to simplify the following expression as much as possible.
\begin{equation*} \frac{\tan \theta \, \csc^3 \theta}{\cot^4 \theta \, \sin^5 \theta\, (1-\sin^2 \theta)} \end{equation*}
Your final answer should not include any ratios.
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Answer.
Reciprocal identities will be helpful. Also, \(1-\sin^2 \theta = \cos^2\theta\text{,}\) by the Pythagorean Identity.
\begin{equation*} \frac{\tan \theta \, \csc^3 \theta}{\cot^4 \theta \, \sin^5 \theta\, (1-\sin^2 \theta)} = \csc^3 \theta \sec^7 \theta \end{equation*}
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2.

Simplify the following expression.
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\begin{equation*} \frac{1}{1-\sin(x)}-\frac{1}{1+\sin(x)} \end{equation*}
Answer.
\begin{equation*} 2\sec(x)\tan(x) \end{equation*}
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3.

Suppose \(\cos(\theta) = 0.3\) and \(\theta\) is in Quadrant IV. Find the sine of the \(\theta\) using the Pythagorean Identity.
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Answer.
\begin{equation*} \sin(\theta) = - \sqrt{1-(0.3)^2} \approx - 0.9539 \end{equation*}
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4.

Starting with the identity \(\sin^2 \theta + \cos^2 \theta = 1\text{,}\) derive two similar identities involving tangent and cotangent, respectively.
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Hint.
Divide both sides by \(\cos^2\theta\) to see that
\begin{equation*} \tan^2 \theta + 1 = \sec^2 \theta \end{equation*}
Divide both sides by \(\sin^2\theta\) to see that
\begin{equation*} 1+\cot^2 \theta = \csc^2 \theta \end{equation*}
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5.

Use the Addition Formulas and the Opposite Angle Identities to derive the Difference Formulas
\begin{equation*} \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \quad \text{and} \quad \cos(\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta. \end{equation*}
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Hint.
Write the difference of angles as a sum \(\sin(\alpha - \beta) = \sin(\alpha +(-\beta))\text{.}\)
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6.

Use the Difference Formulas above and the Special Triangles to find the exact values of \(\sin(15^\circ)\) and \(\cos(15^\circ)\text{.}\)
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Hint.
\(15^\circ = 45^\circ - 30^\circ\) and the exact values of these can be found from the special triangles.
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7.

Use the Addition Formulas to derive the following Double Angle Identities
\begin{equation*} \sin (2\theta) = 2\sin\theta \cos\theta \quad \text{and} \quad \cos (2\theta) = \cos^2 \theta - \sin^2 \theta. \end{equation*}
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Hint.
Write the double angle as a sum \(\sin (2\theta) = \sin (\theta + \theta)\text{.}\)
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8.

Use the Double Angle Identity for Cosine above to derive the Half-Angle Identities
\begin{equation*} \cos(x/2) = \pm \sqrt{\frac{1+\cos x}{2}} \quad \text{and} \quad \sin(x/2) = \pm \sqrt{\frac{1-\cos x}{2}}. \end{equation*}
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Solution.
I’ll do one. From the double angle identity:
\begin{align*} \cos(2\theta) &= \cos^2 \theta - \sin^2 \theta\\ &= 1-\sin^2 \theta - \sin^2 \theta\\ \cos(2\theta)&= 1-2\sin^2 \theta \end{align*}
Solve this last equation for \(\sin^2 \theta\text{,}\)
\begin{equation*} \sin^2 \theta = \frac{1-\cos(2\theta)}{2} \end{equation*}
Thus,
\begin{equation*} \sin \theta =\pm\sqrt{\frac{1-\cos(2\theta)}{2}}, \end{equation*}
where the sign is determined by the quadrant \(\theta\) is in. Let \(\theta = x/2\text{,}\) then \(2\theta = x\) and
\begin{equation*} \sin (x/2) =\pm\sqrt{\frac{1-\cos(x)}{2}}. \end{equation*}
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