Evaluating Trig Functions

a. This is our reference angle. \({\frac{\pi }{3}}\) is in the first quadrant, where the \(x\)-coordinate (and therefore cosine) is positive.So, \(\cos({\frac{\pi }{3}}) = {\frac{1}{2}}\text{.}\)

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b. \({\frac{2\pi }{3}}\) is in the second quadrant, where the \(x\)-coordinate (and therefore cosine) is negative.So, \(\cos({\frac{2\pi }{3}}) = {-\frac{1}{2}}\text{.}\)

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c. \({\frac{4\pi }{3}}\) is in the third quadrant, where the \(x\)-coordinate (and therefore cosine) is negative.So, \(\cos({\frac{4\pi }{3}}) = {-\frac{1}{2}}\text{.}\)

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d. \({\frac{5\pi }{3}}\) is in the fourth quadrant, where the \(x\)-coordinate (and therefore cosine) is positive.So, \(\cos({\frac{5\pi }{3}}) = {\frac{1}{2}}\text{.}\)

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e. \({\frac{7\pi }{3}}\) is coterminal with \({\frac{\pi }{3}}\text{,}\) and both are in the first quadrant, where the \(x\)-coordinate (and therefore cosine) is positive.So, \(\cos({\frac{7\pi }{3}}) = {\frac{1}{2}}\text{.}\)

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f. \({\frac{-2\pi }{3}}\) is coterminal with \({\frac{4\pi }{3}}\text{,}\) and both are in the third quadrant, where the \(x\)-coordinate (and therefore cosine) is negative.So, \(\cos({\frac{-2\pi }{3}}) = {-\frac{1}{2}}\text{.}\)

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g. \({\frac{-4\pi }{3}}\) is coterminal with \({\frac{2\pi }{3}}\text{,}\) and both are in the second quadrant, where the \(x\)-coordinate (and therefore cosine) is negative.So, \(\cos({\frac{-4\pi }{3}}) = {-\frac{1}{2}}\text{.}\)

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h. \({\frac{-5\pi }{3}}\) is coterminal with \({\frac{\pi }{3}}\text{,}\) and both are in the first quadrant, where the \(x\)-coordinate (and therefore cosine) is positive.So, \(\cos({\frac{-5\pi }{3}}) = {\frac{1}{2}}\text{.}\)

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2.

Without using a calculator, state the exact value of the following trig functions for the specified angle.

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\(\displaystyle{ \sin\left({\frac{\pi }{3}}\right) = }\)
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\(\displaystyle{ \sin\left({\frac{2\pi }{3}}\right) = }\)
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\(\displaystyle{ \sin\left({\frac{4\pi }{3}}\right) = }\)
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\(\displaystyle{ \sin\left({\frac{5\pi }{3}}\right) = }\)
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\(\displaystyle{ \sin\left({\frac{7\pi }{3}}\right) = }\)
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\(\displaystyle{ \sin\left({\frac{-2\pi }{3}}\right) = }\)
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\(\displaystyle{ \sin\left({\frac{-4\pi }{3}}\right) = }\)
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\(\displaystyle{ \sin\left({\frac{-5\pi }{3}}\right) = }\)
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Hint.

Do you notice anything about all the angles in this problem?
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What is the reference angle for each?
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And what does that tell you about the sine of each angle?
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Answer 1.

\(\frac{\sqrt{3}}{2}\)

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Answer 2.

\(\frac{\sqrt{3}}{2}\)

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Answer 3.

\(-\frac{\sqrt{3}}{2}\)

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Answer 4.

\(-\frac{\sqrt{3}}{2}\)

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Answer 5.

\(\frac{\sqrt{3}}{2}\)

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Answer 6.

\(-\frac{\sqrt{3}}{2}\)

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Answer 7.

\(\frac{\sqrt{3}}{2}\)

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Answer 8.

\(\frac{\sqrt{3}}{2}\)

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Solution.

Right off the bat, you should recognize that all angles appearing in this problem have \(\displaystyle{\frac{\pi }{3}}\) as their reference angle.

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This means that the only difference in your answers will be whether your answer will be positive or negative.

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The answer, of course, depends on the quadrant where \(\theta\) lives.

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From our unit circle, we know that \(\sin\left(\displaystyle{\frac{\pi }{3}}\right) = \displaystyle{\frac{\sqrt{3}}{2}}\text{.}\)

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\(\displaystyle{\frac{\pi }{3}}\) corresponds to the point \(\left(\displaystyle{\frac{1}{2}},{\frac{\sqrt{3}}{2}}\right)\) on the unit circle;

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and \(\sin(\theta)\) corresponds to the \(y\)-coordinate.

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a. This is our reference angle. \(\displaystyle{\frac{\pi }{3}}\) is in the first quadrant, where the \(y\)-coordinate (and therefore sine) is positive.So, \(\sin\left(\displaystyle{\frac{\pi }{3}}\right) = \displaystyle{\frac{\sqrt{3}}{2}}\text{.}\)

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b. \(\displaystyle{\frac{2\pi }{3}}\) is in the second quadrant, where the \(y\)-coordinate (and therefore sine) is positive.So, \(\sin\left(\displaystyle{\frac{2\pi }{3}}\right) = \displaystyle{\frac{\sqrt{3}}{2}}\text{.}\)

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c. \(\displaystyle{\frac{4\pi }{3}}\) is in the third quadrant, where the \(y\)-coordinate (and therefore sine) is negative.So, \(\sin\left(\displaystyle{\frac{4\pi }{3}}\right) = \displaystyle{-\frac{\sqrt{3}}{2}}\text{.}\)

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d. \(\displaystyle{\frac{5\pi }{3}}\) is in the fourth quadrant, where the \(y\)-coordinate (and therefore sine) is negative.So, \(\sin\left(\displaystyle{\frac{5\pi }{3}}\right) = \displaystyle{-\frac{\sqrt{3}}{2}}\text{.}\)

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e. \(\displaystyle{\frac{7\pi }{3}}\) is coterminal with \(\displaystyle{\frac{\pi }{3}}\text{,}\) and both are in the first quadrant, where the \(y\)-coordinate (and therefore sine) is positive.So, \(\sin\left(\displaystyle{\frac{7\pi }{3}}\right) = \displaystyle{\frac{\sqrt{3}}{2}}\text{.}\)

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f. \(\displaystyle{\frac{-2\pi }{3}}\) is coterminal with \({\frac{4\pi }{3}}\text{,}\) and both are in the third quadrant, where the \(y\)-coordinate (and therefore sine) is negative.So, \(\sin\left(\displaystyle{\frac{-2\pi }{3}}\right) = \displaystyle{-\frac{\sqrt{3}}{2}}\text{.}\)

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g. \(\displaystyle{\frac{-4\pi }{3}}\) is coterminal with \(\displaystyle{\frac{2\pi }{3}}\text{,}\) and both are in the second quadrant, where the \(y\)-coordinate (and therefore sine) is positive.So, \(\sin({\frac{-4\pi }{3}}) = {\frac{\sqrt{3}}{2}}\text{.}\)

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h. \(\displaystyle{\frac{-5\pi }{3}}\) is coterminal with \({\frac{\pi }{3}}\text{,}\) and both are in the first quadrant, where the \(y\)-coordinate (and therefore sine) is positive.So, \(\sin({\frac{-5\pi }{3}}) = {\frac{\sqrt{3}}{2}}\text{.}\)

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3.

Without using a calculator, state the exact value of the following trig functions for the specified angle.

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\(\displaystyle{ \tan\left({\frac{\pi }{4}}\right) = }\)
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\(\displaystyle{ \tan\left({\frac{3\pi }{4}}\right) = }\)
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\(\displaystyle{ \tan\left({\frac{5\pi }{4}}\right) = }\)
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\(\displaystyle{ \tan\left({\frac{7\pi }{4}}\right) = }\)
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\(\displaystyle{ \tan\left({\frac{9\pi }{4}}\right) = }\)
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\(\displaystyle{ \tan\left({\frac{-3\pi }{4}}\right) = }\)
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\(\displaystyle{ \tan\left({\frac{-5\pi }{4}}\right) = }\)
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\(\displaystyle{ \tan\left({\frac{-7\pi }{4}}\right) = }\)
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Hint.

Do you notice anything about all the angles in this problem?

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What is the reference angle for each?

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And what does that tell you about the tangent of each angle?

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Answer 1.

\(1\)

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Answer 2.

\(-1\)

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Answer 3.

\(1\)

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Answer 4.

\(-1\)

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Answer 5.

\(1\)

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Answer 6.

\(1\)

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Answer 7.

\(-1\)

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Answer 8.

\(1\)

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Solution.

Right off the bat, you should recognize that all angles appearing in this problem have \({\frac{\pi }{4}}\) as their reference angle.

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This means that the only difference in your answers will be whether your answer will be positive or negative.

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The answer, of course, depends on the quadrant where \(\theta\) lives.

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From our unit circle, we know that \(\tan({\frac{\pi }{4}}) = \frac{{\frac{\sqrt{2}}{2}}}{{\frac{\sqrt{2}}{2}}} = {1}\text{.}\)

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\({\frac{\pi }{4}}\) corresponds to the point \(({\frac{\sqrt{2}}{2}},{\frac{\sqrt{2}}{2}})\) on the unit circle;

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and \(\tan(\theta)\) corresponds to \(\frac{\sin(\theta)}{\cos(\theta)}\) or \(\frac{y}{x}\text{.}\)

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a. This is our reference angle. \({\frac{\pi }{4}}\) is in the first quadrant, where \(x\) is positive and \(y\) is positive, so \(\frac{y}{x}\) (and therefore tangent) is positive.So, \(\tan({\frac{\pi }{4}}) = {1}\text{.}\)

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b. \({\frac{3\pi }{4}}\) is in the second quadrant, where \(x\) is negative and \(y\) is positive, so \(\frac{y}{x}\) (and therefore tangent) is negative.So, \(\tan({\frac{3\pi }{4}}) = {-1}\text{.}\)

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c. \({\frac{5\pi }{4}}\) is in the third quadrant, where \(x\) is negative and \(y\) is negative, so \(\frac{y}{x}\) (and therefore tangent) is positive.So, \(\tan({\frac{5\pi }{4}}) = {1}\text{.}\)

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d. \({\frac{7\pi }{4}}\) is in the fourth quadrant, where \(x\) is positive and \(y\) is negative, so \(\frac{y}{x}\) (and therefore tangent) is negative.So, \(\tan({\frac{7\pi }{4}}) = {-1}\text{.}\)

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e. \({\frac{9\pi }{4}}\) is coterminal with \({\frac{\pi }{4}}\text{,}\) and both are in the first quadrant, where \(x\) is positive and \(y\) is positive, so \(\frac{y}{x}\) (and therefore tangent) is positive.So, \(\tan({\frac{9\pi }{4}}) = {1}\text{.}\)

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f. \({\frac{-3\pi }{4}}\) is coterminal with \({\frac{5\pi }{4}}\text{,}\) and both are in the third quadrant, where \(x\) is negative and \(y\) is negative, so \(\frac{y}{x}\) (and therefore tangent) is positive.So, \(\tan({\frac{-3\pi }{4}}) = {1}\text{.}\)

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g. \({\frac{-5\pi }{4}}\) is coterminal with \({\frac{3\pi }{4}}\text{,}\) and both are in the second quadrant, where \(x\) is negative and \(y\) is positive, so \(\frac{y}{x}\) (and therefore tangent) is negative.So, \(\tan({\frac{-5\pi }{4}}) = {-1}\text{.}\)

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h. \({\frac{-7\pi }{4}}\) is coterminal with \({\frac{\pi }{4}}\text{,}\) and both are in the first quadrant, where \(x\) is positive and \(y\) is positive, so \(\frac{y}{x}\) (and therefore tangent) is positive.So, \(\tan({\frac{-7\pi }{4}}) = {1}\text{.}\)

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4.

If \(P(t) = (\cos t, \sin t)\) has coordinates (0.486,0.874), find the coordinates of

a. \(P(t+ \pi)\)

\(x\) = \(y\) =

b. \(P(- t)\)

\(x\) = \(y\) =

c. \(P(t- \pi)\)

\(x\) = \(y\) =

c. \(P(- t- \pi)\)

\(x\) = \(y\) =

Answer 1.

\(-0.486\)

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Answer 2.

\(-0.874\)

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Answer 3.

\(0.486\)

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Answer 4.

\(-0.874\)

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Answer 5.

\(-0.486\)

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Answer 6.

\(-0.874\)

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Answer 7.

\(-0.486\)

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Answer 8.

\(0.874\)

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5.

From the information given, find the quadrant in which the terminal point determined by \(t\) lies. Input I, II, III, or IV.

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(a) \(\sin (t)\lt 0\) and \(\cos (t)\lt 0\text{,}\) quadrant ;

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(b) \(\sin (t)>0\) and \(\cos (t)\lt 0\text{,}\) quadrant ;

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(c) \(\sin (t)>0\) and \(\cos (t)>0\text{,}\) quadrant ;

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(d) \(\sin (t)\lt 0\) and \(\cos (t)>0\text{,}\) quadrant ;

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Answer 1.

III

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Answer 2.

II

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Answer 3.

I

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Answer 4.

IV

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6.

Evaluate the following expressions.

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1.	\(\sin\mathopen{}\left(\frac{7\pi }{6}\right)\)	\(=\)
2.	\(\cos\mathopen{}\left(\frac{7\pi }{6}\right)\)	\(=\)
3.	\(\tan\mathopen{}\left(\frac{7\pi }{6}\right)\)	\(=\)
4.	\(\sec\mathopen{}\left(\frac{7\pi }{6}\right)\)	\(=\)

Remark: You are not allowed to use decimals in your answer.

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Answer 1.

\(-0.5\)

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Answer 2.

\(-0.866025\)

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Answer 3.

\(0.57735\)

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Answer 4.

\(-1.1547\)

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7.

Refer to the right triangle in the figure. Click on the picture to see it more clearly.

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If , \(BC=9\) and the angle \(\alpha=60 ^\circ\text{,}\) find any missing angles or sides. Give your answer to at least 3 decimal digits.

AB =

AC =

\(\beta\)=

Answer 1.

\(5.1961524370658\)

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Answer 2.

\(10.3923048525928\)

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Answer 3.

\(30\)

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8.

\(\tan (-\pi) =\)

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\(\cot (-\frac{\pi}{2}) =\)

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\(\sec (\frac{\pi}{3}) =\)

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\(\csc (-\frac{\pi}{6}) =\)

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Answer 1.

\(-1\)

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Answer 2.

\(0.707106781186548\)

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Answer 3.

\(0\)

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Answer 4.

\(0\)

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Answer 5.

\(2\)

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Answer 6.

\(-2\)

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9.

Questions 8-16:

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Find the exact value of each without using a calculator:

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a) \(\tan{ \left( \frac{\pi}{6} \right) }\) =

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b) \(\tan{ \left( \frac{\pi}{4} \right) }\) =

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c) \(\cot{ \left( \frac{7 \pi}{6} \right) }\) =

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d) \(\sec{ \left( \frac{5 \pi}{4} \right) }\) =

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e) \(\csc{ \left( \frac{2 \pi}{3} \right) }\) =

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Answer 1.

\(0.577350269189626\)

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Answer 2.

\(1\)

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Answer 3.

\(1.73205080756888\)

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Answer 4.

\(-1.4142135623731\)

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Answer 5.

\(1.15470053837925\)

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Solution.

SOLUTION\(\tan{ \left( \frac{\pi}{6} \right) } = \frac{1}{\sqrt{3}}\)\(\tan{ \left( \frac{\pi}{4} \right) } = 1\)\(\cot{ \left( \frac{7 \pi}{6} \right) } = \sqrt{3};\)\(\sec{ \left( \frac{5 \pi}{4} \right) } = - \sqrt{2};\)\(\csc{ \left( \frac{2 \pi}{3} \right) } = \frac{2}{\sqrt{3}};\)

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10.

Given \(\cot(\alpha)=\sqrt 3\) and \(0\lt \alpha\lt \pi/2\text{,}\) find the exact values of the remaining five trigonometric functions.

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Note: You are not allowed to use decimals in your answer.

\(\sin(\alpha)\) = .

\(\cos(\alpha)\) = .

\(\tan(\alpha)\) = .

\(\csc(\alpha)\) = .

\(\sec(\alpha)\) = .

Answer 1.

\(0.5\)

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Answer 2.

\(0.866025403784439\)

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Answer 3.

\(0.577350269189626\)

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Answer 4.

\(2\)

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Answer 5.

\(1.15470053837925\)

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11.

If \(\cos( \theta ) = - \frac {5} {8}\) and \(\theta\) is in quadrant II , then find

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(a) \(\tan( \theta ) \cot( \theta ) =\)

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(b) \(\csc( \theta ) \tan( \theta ) =\)

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(c) \(\sin ^2 ( \theta ) + \cos ^2 ( \theta ) =\)

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Answer 1.

\(1\)

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Answer 2.

\(-1.6\)

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Answer 3.

\(1\)

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12.

Evaluate the following expressions. The answer must be given as a fraction, NO DECIMALS. If the answer involves a square root it should be entered as sqrt. For instance, the square root of 2 should be written as sqrt(2).

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If \(\tan( \theta ) = - \frac {6} {5}\) and \(\sin ( \theta ) \lt 0\text{,}\) then find

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(a) \(\sin( \theta ) =\)

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(b) \(\cos( \theta ) =\)

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(c) \(\sec( \theta ) =\)

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(d) \(\csc( \theta ) =\)

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(e) \(\cot( \theta ) =\)