Definition 3.2.1. Addition and Subtraction Formulas for Cosine.
\begin{align*}
\cos(\alpha+\beta)\amp = \cos\alpha\cos\beta-\sin\alpha\sin\beta\\
\cos(\alpha-\beta)\amp = \cos\alpha\cos\beta+\sin\alpha\sin\beta
\end{align*}
Section 3.2 Addition and Subtraction Formulas
In SectionΒ 1.4, we used right triangles to determine the deviation of a waΚ»a (canoe) from its course based on the angle of deviation. If a waΚ»a sails for 120 nautical miles (NM), we were able to calculate the deviation from its course using right triangles to get the equation:
\begin{equation*}
\mbox{deviation} = (120\mbox{ NM})\cdot\sin(\theta)\text{.}
\end{equation*}
Before setting sail, a voyager studies a table listing the deviation distances corresponding to different houses of deviation. Itβs crucial to understand that while adding angles may yield a third angle, adding their corresponding deviations will not accurately determine the total deviation. In other words:
\begin{equation*}
\sin(\alpha) + \sin(\beta) \neq \sin(\alpha + \beta)
\end{equation*}
for certain angles of deviation \(\alpha\) and \(\beta\text{.}\)
To illustrate this, letβs calculate the deviation distances for 1, 2, and 3 houses respectively. Using the given formula, we have:
\begin{align*}
120 \sin(1 \mbox{ house}) \amp = 120 \sin(11.25^{\circ}) \approx 23.4 \mbox{ NM}\\
120 \sin(2 \mbox{ houses})\amp = 120 \sin(22.5^{\circ}) \approx 45.9 \mbox{ NM}\\
120 \sin(3 \mbox{ houses}) \amp = 120 \sin(33.75^{\circ}) \approx 66.7 \mbox{ NM}
\end{align*}
However,
\begin{equation*}
120 \sin(1 \mbox{ house}) + 120 \sin(2 \mbox{ houses}) \approx 23.4 + 45.9 \mbox{ NM} \approx 69.3 \mbox{ NM}\text{,}
\end{equation*}
which differs from the actual deviation of 66.7 NM when deviating by 3 houses.
These calculations demonstrate that the deviation distances for multiple houses cannot be determined by simply adding individual deviations, highlighting the importance of understanding trigonometric principles for accurate navigation. In this section, we will explore the formulas for the addition and subtraction of angles in trigonometric functions.
Subsection 3.2.1 Addition and Subtraction Formulas for Cosine
First we will derive the addition and subtraction formulas for the cosine function.
Proof.
First we will prove the Subtraction Formula for Cosine
\begin{equation*}
\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\text{.}
\end{equation*}
We begin by considering two points on the unit circle. Point \(P\) is at an angle of \(\beta\) in standard position with coordinates \((\cos\beta,\sin\beta)\) and point \(Q\) is at an angle of \(\alpha\) in standard position with coordinates \((\cos\alpha,\sin\alpha)\text{.}\)
We use the Distance Formula to calculate the distance between \(P\) and \(Q\) to get
\begin{align*}
d(P,Q)\amp = \sqrt{(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2}\\
\amp = \sqrt{\cos^2\alpha-2\cos\alpha\cos\beta+\cos^2\beta+\sin^2\alpha-2\sin\alpha\sin\beta+\sin^2\beta}\\
\amp = \sqrt{(\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta}\text{.}
\end{align*}
By the Pythagorean Identity (DefinitionΒ 1.5.20), \(\cos^2\alpha+\sin^2\alpha=1\) and \(\cos^2\beta+\sin^2\beta=1\text{.}\) Thus the distance becomes
\begin{align*}
d(P,Q) \amp = \sqrt{1+1-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta}\\
\amp = \sqrt{2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta}\text{.}
\end{align*}
Next, consider two additional points on a second unit circle. Point \(A\) has coordinates \((1,0)\) and Point \(B\) is at an angle of \(\alpha-\beta\) in standard position with coordinates \((\cos(\alpha-\beta),\sin(\alpha-\beta))\text{.}\)
The distance between \(A\) and \(B\) is
\begin{align*}
d(A,B)\amp = \sqrt{(\cos(\alpha-\beta)-1)^2+(\sin(\alpha-\beta)-0)^2}\\
\amp = \sqrt{\cos^2(\alpha-\beta)-2\cos(\alpha-\beta)+1+\sin^2(\alpha-\beta)}\\
\amp = \sqrt{(\cos^2(\alpha-\beta)+\sin^2(\alpha-\beta))-2\cos(\alpha-\beta)+1}\\
\amp = \sqrt{1-2\cos(\alpha-\beta)+1}\\
\amp = \sqrt{2-2\cos(\alpha-\beta)}\text{.}
\end{align*}
Note that since \(OP\text{,}\) \(OQ\text{,}\) \(OA\text{,}\) and \(OB\) are line segments from the center to points on the unit circle, they are congruent and have length of 1. Also note that \(\angle POQ=\angle AOB=\alpha-\beta\text{.}\) Since two sides and the included angle of \(\triangle OPQ\) and \(\triangle OAB\) are congruent, we can conclude by the Side-Angle-Side Theorem (SAS) in geometry that the two triangles are congruent. Thus, corresponding sides have the same lengths, giving us \(d(P,Q)=d(A,B)\text{.}\) Substituting our results for \(d(P,Q)\) and \(d(A,B)\) we get
\begin{align*}
d(P,Q)\amp = d(A,B)\\
\sqrt{2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta}\amp = \sqrt{2-2\cos(\alpha-\beta)}\\
2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta\amp = 2-2\cos(\alpha-\beta)\\
-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta\amp = -2\cos(\alpha-\beta)\text{.}
\end{align*}
Dividing both sides by -2 we arrive at the Subtraction Formula for Cosine
\begin{equation*}
\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos(\alpha-\beta)\text{.}
\end{equation*}
To prove the Addition Formula for Cosine, replace \(\beta\) with \(-\beta\) in the Subtraction Formula and use the Even and Odd Trigonometric Properties (DefinitionΒ 1.5.22) where \(\sin(-\beta)=-\sin\beta\) and \(\cos(-\beta)=\cos\beta\) to get
\begin{align*}
\cos\alpha\cos(-\beta)+\sin\alpha\sin(-\beta)\amp = \cos(\alpha-(-\beta))\\
\cos\alpha\cos(\beta)+\sin\alpha(-\sin\beta)\amp = \cos(\alpha+\beta)\\
\cos\alpha\cos\beta-\sin\alpha\sin\beta\amp = \cos(\alpha+\beta)\text{.}
\end{align*}
Example 3.2.2.
Find the exact value of \(\cos105^{\circ}\text{.}\)
Solution.
First note that \(105^{\circ}=60^{\circ}+45^{\circ}\text{.}\) Now, using the Addition Formula for Cosine (DefinitionΒ 3.2.1),
\begin{align*}
\cos105^{\circ}\amp = \cos(60^{\circ}+45^{\circ})\\
\amp = \cos60^{\circ}\cos45^{\circ}-\sin60^{\circ}\sin45^{\circ}\\
\amp = \frac{1}{2}\cdot\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}\\
\amp = \frac{1}{4}\left(\sqrt{2}-\sqrt{6}\right)\text{.}
\end{align*}
Example 3.2.3.
Find the exact value of \(\cos\left(\frac{\pi}{4}-\frac{\pi}{6}\right)\text{.}\)
Solution.
Using the Subtraction Formula for Cosine we get
\begin{align*}
\cos\left(\frac{\pi}{4}-\frac{\pi}{6}\right)\amp = \cos\frac{\pi}{4}\cos\frac{\pi}{6}+\sin\frac{\pi}{4}\sin\frac{\pi}{6}\\
\amp = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}\\
\amp = \frac{1}{4}\left(\sqrt{6}+\sqrt{2}\right)\text{.}
\end{align*}
Example 3.2.4.
Find the exact value of the expression \(\cos25^{\circ}\cos35^{\circ}-\sin25^{\circ}\sin35^{\circ}\text{.}\)
Solution.
Notice this expression is the Addition Formula for Cosine with \(\alpha=25^{\circ}\) and \(\beta=35^{\circ}\text{.}\) So
\begin{equation*}
\cos25^{\circ}\cos35^{\circ}-\sin25^{\circ}\sin35^{\circ}=\cos(25^{\circ}+35^{\circ})=\cos60^{\circ}=\frac{1}{2}\text{.}
\end{equation*}
Subsection 3.2.2 Addition and Subtraction Formulas for Sine
Next we will learn about the addition and subtraction formulas for the sine function.
Definition 3.2.5. Addition and Subtraction Formulas for Sine.
\begin{align*}
\sin(\alpha+\beta)\amp = \sin\alpha\cos\beta+\cos\alpha\sin\beta\\
\sin(\alpha-\beta)\amp = \sin\alpha\cos\beta-\cos\alpha\sin\beta
\end{align*}
We will prove the Addition Formula for Sine in ExampleΒ 3.2.10 and the Subtraction Formula can be established using the Even and Odd Properties (DefinitionΒ 1.5.22).
Example 3.2.6.
Given \(\sin\alpha=-\frac{12}{13}\text{,}\) with \(\frac{3\pi}{2}\lt \alpha\lt 2\pi\) and \(\cos\beta=-\frac{3}{5}\text{,}\) with \(\frac{\pi}{2}\lt \beta\lt \pi\text{,}\) find the exact value of \(\sin(\alpha+\beta)\text{.}\)
Solution.
The Addition Formula for Sine gives us
\begin{equation*}
\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\text{.}
\end{equation*}
At this moment, we do not know the exact values of \(\cos\alpha\) and \(\sin\beta\) but we can compute them.
Given \(\sin\alpha=-\frac{12}{13}\) with \(\frac{3\pi}{2}\lt \alpha\lt 2\pi\) and \(\cos\beta=-\frac{3}{5}\) with \(\frac{\pi}{2}\lt \beta\lt \pi\text{,}\) we can draw the following triangles associated with \(\alpha\) and \(\beta\text{,}\) respectively:
Next, using the Pythagorean Theorem, we solve for the missing sides on the triangles:
\begin{align*}
x^2+(-12)^2\amp = 13^2\\
x+144\amp = 169\\
x^2\amp = 25\\
x\amp = 5\text{.}
\end{align*}
Thus we get
\begin{equation*}
\cos\alpha=\frac{5}{13}\text{.}
\end{equation*}
Similarly,
\begin{align*}
(-3)^2+y^2\amp = 5^2\\
9+y^2\amp = 25\\
y^2\amp = 16\\
y\amp = 4\text{.}
\end{align*}
Thus we get
\begin{equation*}
\sin\beta=\frac{4}{5}\text{.}
\end{equation*}
We now have all the information needed to proceed.
\begin{align*}
\sin(\alpha+\beta)\amp = \sin\alpha\cos\beta+\cos\alpha\sin\beta\\
\amp = \left(-\frac{12}{13}\right)\left(-\frac{3}{5}\right)+\left(\frac{5}{13}\right)\left(\frac{4}{5}\right)\\
\amp = \frac{36}{65}+\frac{20}{65}\\
\amp = \frac{56}{65}\text{.}
\end{align*}
Notice we did not have to know the values of \(\alpha\) or \(\beta\) to do this example.
Subsection 3.2.3 Addition and Subtraction Formulas for Tangent
Now we will learn about the addition and subtraction formulas for the tangent function.
Definition 3.2.7. Addition and Subtraction Formulas for Tangent.
\begin{align*}
\tan(\alpha+\beta)\amp = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\\
\tan(\alpha-\beta)\amp = \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}
\end{align*}
Proof.
Recall that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) as long as \(\cos\theta\neq0\text{.}\) Using this fact and our new formulas for the sum of sine and cosine, we get
\begin{align*}
\tan(\alpha+\beta)\amp = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\\
\amp = \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}\\
\amp = \frac{\dfrac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta}}{\dfrac{\cos\alpha\cos\beta-\sin\alpha\sin\beta}{\cos\alpha\cos\beta}}\\
\amp = \frac{\dfrac{\sin\alpha\cancel{\cos\beta}}{\cos\alpha\cancel{\cos\beta}}+\dfrac{\cancel{\cos\alpha}\sin\beta}{\cancel{\cos\alpha}\cos\beta}}{\dfrac{\cancel{\cos\alpha\cos\beta}}{\cancel{\cos\alpha\cos\beta}}-\dfrac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}}\\
\amp = \frac{\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\sin\beta}{\cos\beta}}{1-\dfrac{\sin\alpha}{\cos\alpha}\dfrac{\sin\beta}{\cos\beta}}\\
\amp = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\text{.}
\end{align*}
The subtraction formula for tangent can be established using the Even and Odd Properties (DefinitionΒ 1.5.22).
Example 3.2.8.
Find the exact value of \(\tan\left(\frac{3\pi}{4}+\frac{\pi}{6}\right)\)
.
Solution.
Using the Addition Formula for Tangent (DefinitionΒ 3.2.7), we get
\begin{align*}
\tan\left(\frac{3\pi}{4}+\frac{\pi}{6}\right) \amp =\frac{\tan\frac{3\pi}{4}+\tan\frac{\pi}{6}}{1-\tan\frac{3\pi}{4}\tan\frac{\pi}{6}}\\
\amp =\frac{(-1)+\frac{\sqrt{3}}{3}}{1-(-1)\cdot\frac{\sqrt{3}}{3}}\\
\amp =\frac{-1+\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}\\
\amp =\frac{\frac{-3+\sqrt{3}}{3}}{\frac{3+\sqrt{3}}{3}}\\
\amp =\frac{-3+\sqrt{3}}{3+\sqrt{3}}\text{.}
\end{align*}
Subsection 3.2.4 Cofunction Identities
Recall the Cofunction Identities (DefinitionΒ 1.4.7):
\begin{align*}
\sin\theta\amp =\cos\left(\frac{\pi}{2}-\theta\right), \amp \cos\theta\amp =\sin\left(\frac{\pi}{2}-\theta\right)\\
\tan\theta\amp =\cot\left(\frac{\pi}{2}-\theta\right), \amp \cot\theta\amp =\tan\left(\frac{\pi}{2}-\theta\right)\\
\sec\theta\amp =\csc\left(\frac{\pi}{2}-\theta\right), \amp \csc\theta\amp =\sec\left(\frac{\pi}{2}-\theta\right)
\end{align*}
Armed with the knowledge of the subtraction formulas, we can prove the Cofunction Identities.
Proof.
We will prove the Cofunction Identity for \(\sin\theta\) in ExampleΒ 3.2.9. The proof for \(\cos\theta\) is given as ExerciseΒ 3.2.7.97. The Cofunction Identities for \(\tan\theta\) and \(\cot\theta\) can be found using the Quotient Identities (DefinitionΒ 1.4.3); for \(\csc\theta\) and \(\sec\theta\) can be found using the Reciprocal Identities (DefinitionΒ 1.4.2).
Example 3.2.9.
Use the Subtraction Formula for Sine to establish the identity \(\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right)\text{.}\)
Solution.
To establish an identity, we will start from one side of the equation and use properties to end up with the expression on the other side of the equation. So,
\begin{align*}
\sin\left(\frac{\pi}{2}-\theta\right) \amp =\sin\left(\frac{\pi}{2}\right)\cos\left(\theta\right)-\cos\left(\frac{\pi}{2}\right)\sin\left(\theta\right)\\
\amp =1\cdot\cos\left(\theta\right)-0\cdot\sin\left(\theta\right)\\
\amp =\cos\left(\theta\right)\text{.}
\end{align*}
Visually, we have \(\cos\theta=\frac{x}{r}=\sin\left(\frac{\pi}{2}-\theta\right)\text{.}\)
Example 3.2.10.
Prove the Addition Formula for Sine
\begin{equation*}
\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\text{.}
\end{equation*}
Solution.
We begin by using the cofunction identity
\begin{align*}
\sin(\alpha+\beta)\amp =\cos\left(\frac{\pi}{2}-(\alpha+\beta)\right)\\
\amp =\cos\left(\left(\frac{\pi}{2}-\alpha\right)-\beta\right)\text{.}
\end{align*}
By the Subtraction Formula for Cosine:
\begin{align*}
=\cos\left(\left(\frac{\pi}{2}-\alpha\right)-\beta\right)\amp =\cos\left(\frac{\pi}{2}-\alpha\right)\cos\beta+\sin\left(\frac{\pi}{2}-\alpha\right)\sin\beta\\
\amp =\sin\alpha\cos\beta+\cos\alpha\sin\beta\text{,}
\end{align*}
in the last step, we use the Cofunction Identity.
Subsection 3.2.5 Sums of Sines and Cosines
Sometimes we may come across functions of the form
\begin{equation*}
a\sin x+b\cos x\text{.}
\end{equation*}
It can often be useful to rewrite this expression as a single trigonometric function.
Definition 3.2.11.
For any real numbers \(a\) and \(b\text{,}\) let \(\theta\) be an angle in standard position where \(P(a,b)\) is a point on the terminal side of \(\theta\text{.}\) Then
\begin{equation*}
a \sin x + b \cos x = \sqrt{a^2 + b^2} \sin(x + \theta)\text{.}
\end{equation*}
Proof.
We begin by considering the triangle formed by angle \(\theta\) and point \(P(a,b)\text{,}\) shown in FigureΒ 3.2.12. By the Pythagorean Theorem, the hypotenuse of this triangle, with base \(a\) and height \(b\text{,}\) is \(\sqrt{a^2+b^2}\text{.}\) According to DefinitionΒ 1.4.1, we have
\begin{equation*}
\cos\theta=\frac{a}{\sqrt{a^2+b^2}}, \quad \sin\theta=\frac{b}{\sqrt{a^2+b^2}}
\end{equation*}
or equivalently:
\begin{equation*}
a = \sqrt{a^2+b^2}\cos\theta, \quad b= \sqrt{a^2+b^2}\sin\theta\text{.}
\end{equation*}
Therefore, using the addition formula for sine, we get
\begin{align*}
a \sin x + b \cos x \amp = \sqrt{a^2+b^2}\cos\theta\sin x + \sqrt{a^2+b^2}\sin\theta\cos x \\
\amp = \sqrt{a^2+b^2}\left(\cos\theta\sin x + \sin\theta\cos x\right) \\
\amp = \sqrt{a^2+b^2}\sin(x+\theta) \text{.}
\end{align*}
Example 3.2.13.
Express
\begin{equation*}
-\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x
\end{equation*}
in terms of sine only.
Solution.
To express the given expression in terms of sine only, we will use DefinitionΒ 3.2.11. Considering the point \(P(a,b)=\left( -\frac{\sqrt{3}}{2},\frac{1}{2}\right)\text{,}\) which lies in Quadrant II, we determine the angle \(\theta\text{.}\) Using either TableΒ 1.5.18 or inverse trigonometric methods where
\begin{equation*}
\tan\theta=\frac{b}{a}=\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}=-\frac{1}{\sqrt{3}}\text{,}
\end{equation*}
we find
\begin{equation*}
\theta=150^{\circ}\text{.}
\end{equation*}
Therefore, by DefinitionΒ 3.2.11, we have:
\begin{align*}
-\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x \amp =\sqrt{\left(-\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{2}\right)^2}\sin(x+150^{\circ}) \\
\amp =\sqrt{\frac{3}{4}+\frac{1}{4}}\sin(x+150^{\circ}) \\
\amp = \sin(x+150^{\circ}) \text{.}
\end{align*}
Subsection 3.2.6 Summary
To review, the addition and subtraction formulas are:
\begin{align*}
\sin(\alpha+\beta)\amp = \sin\alpha\cos\beta+\cos\alpha\sin\beta\\
\sin(\alpha-\beta)\amp = \sin\alpha\cos\beta-\cos\alpha\sin\beta\\
\cos(\alpha+\beta)\amp = \cos\alpha\cos\beta-\sin\alpha\sin\beta\\
\cos(\alpha-\beta)\amp = \cos\alpha\cos\beta+\sin\alpha\sin\beta\\
\tan(\alpha+\beta)\amp = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\\
\tan(\alpha-\beta)\amp = \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\text{.}
\end{align*}
Exercises 3.2.7 Exercises
Exercise Group.
Use the Addition or Subtraction Formula to find the exact value of each expression.
1.
\(\cos(15^{\circ})\)
Answer.
\(\frac{\sqrt{6}+\sqrt{2}}{4}\)
2.
\(\sin(75^{\circ})\)
Answer.
\(\frac{\sqrt{6}+\sqrt{2}}{4}\)
3.
\(\sin(195^{\circ})\)
Answer.
\(\frac{-\sqrt{6}+\sqrt{2}}{4}\)
4.
\(\tan(165^{\circ})\)
Answer.
\(\frac{1-\sqrt{3}}{1+\sqrt{3}}\)
5.
\(\tan(105^{\circ})\)
Answer.
\(\frac{1+\sqrt{3}}{1-\sqrt{3}}\)
6.
\(\cos(255^{\circ})\)
Answer.
\(\frac{-\sqrt{6}+\sqrt{2}}{4}\)
7.
\(\tan\left(\frac{5\pi}{12}\right)\)
Hint.
\(\frac{5\pi}{12}=\frac{\pi}{6}+\frac{\pi}{4}\)
Answer.
\(\frac{3+\sqrt{3}}{3-\sqrt{3}}\)
8.
\(\sin\left(\frac{7\pi}{12}\right)\)
Hint.
\(\frac{7\pi}{12}=\frac{\pi}{3}+\frac{\pi}{4}\)
Answer.
\(\frac{\sqrt{6}+\sqrt{2}}{4}\)
9.
\(\cos\left(\frac{\pi}{12}\right)\)
Hint.
\(\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}\)
Answer.
\(\frac{\sqrt{6}+\sqrt{2}}{4}\)
10.
\(\sin\left(\frac{13\pi}{12}\right)\)
Hint.
\(\frac{13\pi}{12}=\frac{11\pi}{6}-\frac{3\pi}{4}\)
Answer.
\(\frac{\sqrt{2}-\sqrt{6}}{4}\)
11.
\(\cos\left(\frac{17\pi}{12}\right)\)
Hint.
\(\frac{17\pi}{12}=\frac{5\pi}{3}-\frac{\pi}{4}\)
Answer.
\(\frac{\sqrt{2}-\sqrt{6}}{4}\)
12.
\(\tan\left(-\frac{\pi}{12}\right)\)
Hint.
\(-\frac{\pi}{12}=\frac{\pi}{4}-\frac{\pi}{3}\)
Answer.
\(\frac{1-\sqrt{3}}{1+\sqrt{3}}\)
Exercise Group.
Use the Addition or Subtraction Formula to find the exact value of each expression.
13.
\(\sin(172^{\circ})\cos(68^{\circ})+\cos(172^{\circ})\sin(68^{\circ})\)
Answer.
\(-\frac{\sqrt{3}}{2}\)
14.
\(\sin(317^{\circ})\cos(257^{\circ})-\cos(317^{\circ})\sin(257^{\circ})\)
Answer.
\(\frac{\sqrt{3}}{2}\)
15.
\(\cos(337^{\circ})\cos(22^{\circ})+\sin(337^{\circ})\sin(22^{\circ})\)
Answer.
\(\frac{1}{\sqrt{2}}\)
16.
\(\cos(59^{\circ})\cos(211^{\circ})-\sin(59^{\circ})\sin(211^{\circ})\)
Answer.
\(0\)
17.
\(\frac{\tan(85^{\circ})-\tan(25^{\circ})}{1+\tan(85^{\circ})\tan(25^{\circ})}\)
Answer.
\(\sqrt{3}\)
18.
\(\sin\left(\frac{5\pi}{16}\right)\cos\left(\frac{\pi}{16}\right)-\cos\left(\frac{5\pi}{16}\right)\sin\left(\frac{\pi}{16}\right)\)
Answer.
\(\frac{\sqrt{2}}{2}\)
19.
\(\cos\left(\frac{5\pi}{22}\right)\cos\left(\frac{3\pi}{11}\right)-\sin\left(\frac{5\pi}{22}\right)\sin\left(\frac{3\pi}{11}\right)\)
Answer.
\(0\)
20.
\(\frac{\tan\left(\frac{\pi}{7}\right)+\tan\left(\frac{4\pi}{21}\right)}{1-\tan\left(\frac{\pi}{7}\right)\tan\left(\frac{4\pi}{21}\right)}\)
Answer.
\(\sqrt{3}\)
Exercise Group.
Find the exact value of each expression given:
21.
\(\sin(\alpha)\)
Answer.
\(\frac{3}{5}\)
22.
\(\cos(\alpha)\)
Answer.
\(\frac{4}{5}\)
23.
\(\tan(\alpha)\)
Answer.
\(\frac{3}{4}\)
24.
\(\sin(\beta)\)
Answer.
\(\frac{20}{29}\)
25.
\(\cos(\beta)\)
Answer.
\(\frac{21}{29}\)
26.
\(\tan(\beta)\)
Answer.
\(\frac{20}{21}\)
27.
\(\sin(\alpha+\beta)\)
Answer.
\(\frac{3}{5}\cdot\frac{21}{29}+\frac{4}{5}\cdot\frac{20}{29}=\frac{143}{145}\)
28.
\(\cos(\alpha+\beta)\)
Answer.
\(\frac{4}{5}\cdot\frac{21}{29}-\frac{3}{5}\cdot\frac{20}{29}=\frac{24}{145}\)
29.
\(\tan(\alpha+\beta)\)
Answer.
\(\frac{\frac{3}{4}+\frac{20}{21}}{1-\frac{3}{4}\cdot\frac{20}{21}}=\frac{143}{24}\)
30.
\(\sin(\alpha-\beta)\)
Answer.
\(\frac{3}{5}\cdot\frac{21}{29}-\frac{4}{5}\cdot\frac{20}{29}=-\frac{17}{145}\)
31.
\(\cos(\alpha-\beta)\)
Answer.
\(\frac{4}{5}\cdot\frac{21}{29}+\frac{3}{5}\cdot\frac{20}{29}=\frac{144}{145}\)
32.
\(\tan(\alpha-\beta)\)
Answer.
\(\frac{\frac{3}{4}-\frac{20}{21}}{1+\frac{3}{4}\cdot\frac{20}{21}}=-\frac{17}{24}\)
Exercise Group.
Find the exact value of each expression given:
33.
\(\sin(\alpha)\)
Answer.
\(\frac{\sqrt{6}}{\sqrt{10}}=\sqrt{\frac{3}{5}}\)
34.
\(\cos(\alpha)\)
Answer.
\(\frac{2}{\sqrt{10}}\)
35.
\(\tan(\alpha)\)
Answer.
\(\frac{\sqrt{6}}{2}\)
36.
\(\sin(\beta)\)
Answer.
\(\frac{3}{\sqrt{13}}\)
37.
\(\cos(\beta)\)
Answer.
\(\frac{2}{\sqrt{13}}\)
38.
\(\tan(\beta)\)
Answer.
\(\frac{3}{2}\)
39.
\(\sin(\alpha+\beta)\)
Answer.
\(\frac{\sqrt{6}}{\sqrt{10}}\cdot\frac{2}{\sqrt{13}}+\frac{2}{\sqrt{10}}\cdot\frac{3}{\sqrt{13}}=\frac{2\sqrt{6}+6}{\sqrt{130}}\)
40.
\(\cos(\alpha+\beta)\)
Answer.
\(\frac{2}{\sqrt{10}}\cdot\frac{2}{\sqrt{13}}-\frac{\sqrt{6}}{\sqrt{10}}\cdot\frac{3}{\sqrt{13}}=\frac{4-3\sqrt{6}}{\sqrt{130}}\)
41.
\(\tan(\alpha+\beta)\)
Answer.
\(\frac{\frac{\sqrt{6}}{2}+\frac{3}{2}}{1-\frac{\sqrt{6}}{2}\cdot\frac{3}{2}}=\frac{2\sqrt{6}+6}{4-3\sqrt{6}}\)
42.
\(\sin(\alpha-\beta)\)
Answer.
\(\frac{\sqrt{6}}{\sqrt{10}}\cdot\frac{2}{\sqrt{13}}-\frac{2}{\sqrt{10}}\cdot\frac{3}{\sqrt{13}}=\frac{2\sqrt{6}-6}{\sqrt{130}}\)
43.
\(\cos(\alpha-\beta)\)
Answer.
\(\frac{2}{\sqrt{10}}\cdot\frac{2}{\sqrt{13}}+\frac{\sqrt{6}}{\sqrt{10}}\cdot\frac{3}{\sqrt{13}}=\frac{4+3\sqrt{6}}{\sqrt{130}}\)
44.
\(\tan(\alpha-\beta)\)
Answer.
\(\frac{\frac{\sqrt{6}}{2}-\frac{3}{2}}{1+\frac{\sqrt{6}}{2}\cdot\frac{3}{2}}=\frac{2\sqrt{6}-6}{4+3\sqrt{6}}\)
Exercise Group.
Find the exact value of each expression given:
45.
\(\sin(\alpha)\)
Answer.
\(\frac{3}{\sqrt{34}}\)
46.
\(\cos(\alpha)\)
Answer.
\(\frac{5}{\sqrt{34}}\)
47.
\(\tan(\alpha)\)
Answer.
\(\frac{3}{5}\)
48.
\(\sin(\beta)\)
Answer.
\(\frac{\sqrt{15}}{8}\)
49.
\(\cos(\beta)\)
Answer.
\(\frac{7}{8}\)
50.
\(\tan(\beta)\)
Answer.
\(\frac{\sqrt{15}}{7}\)
51.
\(\sin(\alpha+\beta)\)
Answer.
\(\frac{3}{\sqrt{34}}\cdot\frac{7}{8}+\frac{5}{\sqrt{34}}\cdot\frac{\sqrt{15}}{8}=\frac{21+5\sqrt{15}}{8\sqrt{34}}\)
52.
\(\cos(\alpha+\beta)\)
Answer.
\(\frac{5}{\sqrt{34}}\cdot\frac{7}{8}-\frac{3}{\sqrt{34}}\cdot\frac{\sqrt{15}}{8}=\frac{35-3\sqrt{15}}{8\sqrt{34}}\)
53.
\(\tan(\alpha+\beta)\)
Answer.
\(\frac{\frac{3}{5}+\frac{\sqrt{15}}{7}}{1-\frac{3}{5}\cdot\frac{\sqrt{15}}{7}}=\frac{21+5\sqrt{15}}{35-3\sqrt{15}}\)
54.
\(\sin(\alpha-\beta)\)
Answer.
\(\frac{3}{\sqrt{34}}\cdot\frac{7}{8}-\frac{5}{\sqrt{34}}\cdot\frac{\sqrt{15}}{8}=\frac{21-5\sqrt{15}}{8\sqrt{34}}\)
55.
\(\cos(\alpha-\beta)\)
Answer.
\(\frac{5}{\sqrt{34}}\cdot\frac{7}{8}+\frac{3}{\sqrt{34}}\cdot\frac{\sqrt{15}}{8}=\frac{35+3\sqrt{15}}{8\sqrt{34}}\)
56.
\(\tan(\alpha-\beta)\)
Answer.
\(\frac{\frac{3}{5}-\frac{\sqrt{15}}{7}}{1+\frac{3}{5}\cdot\frac{\sqrt{15}}{7}}=\frac{21-5\sqrt{15}}{35+3\sqrt{15}}\)
Exercise Group.
Find the exact of each expression given \(\sin\alpha=\frac{20}{29}\text{,}\) \(0\lt\alpha\lt\frac{\pi}{2}\) and \(\cos\beta=\frac{24}{25}\text{,}\) \(0\lt\beta\lt\frac{\pi}{2}\)
57.
\(\cos(\alpha)\)
Answer.
\(\frac{21}{29}\)
58.
\(\tan(\alpha)\)
Answer.
\(\frac{20}{21}\)
59.
\(\sin(\beta)\)
Answer.
\(\frac{7}{25}\)
60.
\(\tan(\beta)\)
Answer.
\(\frac{7}{24}\)
61.
\(\sin(\alpha+\beta)\)
Answer.
\(\frac{627}{725}\)
62.
\(\sin(\alpha-\beta)\)
Answer.
\(\frac{333}{725}\)
63.
\(\cos(\alpha+\beta)\)
Answer.
\(\frac{364}{725}\)
64.
\(\cos(\alpha-\beta)\)
Answer.
\(\frac{644}{725}\)
65.
\(\tan(\alpha+\beta)\)
Answer.
\(\frac{627}{364}\)
66.
\(\tan(\alpha-\beta)\)
Answer.
\(\frac{333}{644}\)
Exercise Group.
Find the exact value of each expression given \(\tan\alpha=\frac{8}{15}\text{,}\) \(\pi\lt\alpha\lt\frac{3\pi}{2}\) and \(\cos\beta=-\frac{3}{5}\text{,}\) \(\frac{\pi}{2}\lt\beta\lt\pi\)
67.
\(\sin(\alpha)\)
Answer.
\(-\frac{8}{17}\)
68.
\(\cos(\alpha)\)
Answer.
\(-\frac{15}{17}\)
69.
\(\sin(\beta)\)
Answer.
\(\frac{4}{5}\)
70.
\(\tan(\beta)\)
Answer.
\(-\frac{4}{3}\)
71.
\(\sin(\alpha+\beta)\)
Answer.
\(-\frac{36}{85}\)
72.
\(\sin(\alpha-\beta)\)
Answer.
\(\frac{84}{85}\)
73.
\(\cos(\alpha+\beta)\)
Answer.
\(\frac{77}{85}\)
74.
\(\cos(\alpha-\beta)\)
Answer.
\(\frac{13}{85}\)
75.
\(\tan(\alpha+\beta)\)
Answer.
\(-\frac{36}{77}\)
76.
\(\tan(\alpha-\beta)\)
Answer.
\(\frac{84}{13}\)
Exercise Group.
Verify the identity.
77.
\(\sin\left(\theta+\frac{\pi}{2}\right)=\cos\theta\)
78.
\(\sin(\theta-\pi)=-\sin\theta\)
79.
\(\cos(\theta-\pi)=-\cos\theta\)
80.
\(\tan(\theta-\pi)=\tan\theta\)
81.
\(\tan\left(\frac{\pi}{4}-\theta\right)=\frac{1-\tan\theta}{1+\tan\theta}\)
82.
\(\sin\left(\frac{\pi}{2}-\theta\right)=\sin\left(\frac{\pi}{2}+\theta\right)\)
83.
\(\cos\left(\theta+\frac{\pi}{3}\right)=-\sin\left(\theta-\frac{\pi}{6}\right)\)
84.
\(\cos(x+y)\cos(x-y)=\cos^2x-\sin^2y\)
85.
\(\frac{\sin(x+y)}{\cos x\cos y}=\tan x+\tan y\)
86.
\(\cot(x-y)=\frac{\cot x\cot y+1}{\cot y-\cot x}\)
87.
\(\sin(x+y)-\sin(x-y)=2\cos x\sin y\)
88.
\(\cos(x+y)+\cos(x-y)=2\cos x\cos y\)
Exercise Group.
Write each expression in terms on sine only. Round your angles to one decimal.
89.
\(-\frac{\sqrt{2}}{2}\sin x -\frac{\sqrt{2}}{2}\cos x\)
Answer.
\(\sin(x+225^{\circ})\)
90.
\(\frac{\sqrt{3}}{2}\sin x -\frac{1}{2}\cos x\)
Answer.
\(\sin(x+330^{\circ})\)
91.
\(\frac{1}{2}\sin x +\frac{\sqrt{3}}{2}\cos x\)
Answer.
\(\sin(x+60^{\circ})\)
92.
\(-\frac{\sqrt{3}}{2}\sin x -\frac{1}{2}\cos x\)
Answer.
\(\sin(x+210^{\circ})\)
93.
\(3\sin x + 7\cos x\)
Answer.
\(\sqrt{58}\sin(x+66.8^{\circ})\)
94.
\(-5\sin x - 9\cos x\)
Answer.
\(\sqrt{106}\sin(x+240.9^{\circ})\)
95.
\(8\sin x - 2\cos x\)
Answer.
\(\sqrt{68}\sin(x+346.0^{\circ})\)
96.
\(-7\sin x + 4\cos x\)
Answer.
\(\sqrt{65}\sin(x+150.3^{\circ})\)
97.
Use the Subtraction Formula for Cosine to prove the Cofunction Identity for Sine: \(\sin\theta=\cos\left(\frac{\pi}{2}-\theta\right)\text{.}\)
