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Trigonometry : Trigonometry

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6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #241 : Trigonometry

Solve for $\theta$ : $10 \cos ^2 ( \theta + 1 ) + 19 \cos (\theta + 1 ) - 15 = 0$

Possible Answers:

52.13^o, 305.87^o

54.13^o, 234.13^o

54.13^o, 307.876^o

52.13^o, 160.13^o

52.13^o, 215.87^o

Correct answer:

52.13^o, 305.87^o

Explanation:

First, solve for $\cos ( \theta + 1 )$ by using the quadratic formula:

$\cos (\theta + 1 ) = \frac{ - 19 \pm \sqrt{ 361 - 4 (10)(-15)}}{20 } = \frac{ -19 \pm 31 }{20}$

This gives two solutions:

$\cos ( \theta + 1) = \frac{ - 19 - 31 }{ 20 } = \frac{ -50 }{20 } = -2.5$ this is outside of the range for cosine, so that does not work as a solution

$\cos ( \theta + 1 ) = \frac{ -19 + 31 }{ 20 } = \frac{ 12}{20 } = 0.6$

To solve for theta, take the inverse of cosine of both sides:

$\theta + 1 = \cos ^ {-1} (0.6)$

$\theta \approx 53.13^o$ according to the calculator. That's just one potential value, though. The other angle that would have a cosine of positive 0.6 would be 53.13 degrees below the x-axis in quadrant IV, so subtract from 360:

360^o - 53.13^o = 306. 87^o

That gives us two values for $\theta + 1$ , so to get theta we have to subtract 1:

$\theta = 52.13^o , 305.87^o$

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Example Question #241 : Trigonometry

Solve for $\theta$ : $3 \cos ^2 \theta - 5 \cos \theta - 2 = 0$

Possible Answers:

109.47^o, 250.53^o

289.47^o, 109.47^o

109.47^o, 218.94^o

70.53^o , 109.47^o

199.47^o, 109.47^o

Correct answer:

109.47^o, 250.53^o

Explanation:

First solve for $\cos \theta$ using the quadratic formula:

$\cos \theta = \frac{ 5 \pm \sqrt{25 - 4(3)(-2)}}{6} = \frac{5 \pm 7 }{6}$

One answer is $\cos \theta = \frac{ 5 + 7 }{6 } = \frac{ 12} {6 } = 2$ this is outside the range for cosine, so it does not work as a solution

The other answer is $\cos \theta = \frac{ 5 - 7 }{6 } = \frac{ -2}{6 } = \frac{ -1}{3 }$

To solve for theta, take the inverse cosine using a calculator:

$\theta = \cos ^ {-1} (-\frac{1}{3} ) \approx 109.47$

This is just one answer for theta, in quadrant II. Cosine is also negative in quadrant III, so we want to find the angle there with the same cosine. This would be -109.47 , or 360^o - 109.47^o = 250.53^o

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Example Question #242 : Trigonometry

Which is not a solution for $\theta$ : $\tan ^2 (3 \theta ) - 5 \tan (3 \theta ) - 14 = 0$

Possible Answers:

38.86^o

27.29^o

87.29^o

92.71^o

98.86^o

Correct answer:

92.71^o

Explanation:

To solve, use the quadratic formula:

$\tan ( 3 \theta ) = \frac{ 5 \pm \sqrt{5^2 - 4(1)(-14)}}{2(1)} = \frac{ 5 \pm 9 }{ 2 }$

This gives two solutions.

The first is:

$\tan ( 3 \theta ) = \frac{ 5 + 9 }{ 2 } = \frac{ 14}{2} = 7$

Using a calculator gives us $3 \theta = \tan ^ {-1} (7 ) \approx 81.87$

This is just one potential value, the one in quadrant I. Tangent is also positive in quadrant III, and we can get this angle by adding 180: 180^o + 81. 87^o = 261.87^o

The second solution from the quadratic formula is:

$\tan ( 3 \theta ) = \frac{ 5 - 9 }{2 } = \frac{ -4 }{2} = -2$

Using a calculator gives us $3 \theta = \tan ^{-1 } (-2 ) \approx -63.43$ , which we can add to 360 to get as a positive value, 296.57^o .

This is just one potential value, the one in quadrant IV. Tangent is also negative in quadrant II, and we can get this angle by subtracting 180:

296.57^o - 180^o = 116.57^o

Dividing all four of these angles by 3 gives us $\theta = 27.29^o, 38.86^o, 87.29^o, 98.86^o$

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Example Question #81 : Trigonometric Equations

Solve for $\theta$ :

$y = 2 \cos ^2 (3 \theta ) + 5 \cos ( 3 \theta ) - 3$

Possible Answers:

$\frac{ 2 \pi }{ 9 } , \frac{ 4 \pi }{ 9 }$

no solution

$2 \pi$

$\frac{ \pi }{ 9 } , \frac{ 5 \pi }{9 }$

$\pi$

Correct answer:

$\frac{ \pi }{ 9 } , \frac{ 5 \pi }{9 }$

Explanation:

Use the quadratic formula to solve for $\cos (3 \theta )$ :

$\cos (3 \theta ) = \frac{ -5 \pm \sqrt{25 - 4(2)(-3)}}{2(2)} = \frac{ -5 \pm \sqrt{49}}{4} = \frac{ -5 \pm 7 }{4}$

One possible solution is:

$\cos (3 \theta ) = \frac{- 5 - 7 }{4} = \frac{-12}{4} = -3$ this is outside of the possible range for cosine

The other solution is:

$\cos (3 \theta ) = \frac{ -5 + 7 }{4} = \frac{ 2}{4} =\frac{ 1}{2}$

$3 \theta = \cos ^ {-1 } (\frac{1}{2} ) = \frac{ \pi }{3} , \frac{ 5 \pi }{3}$ divide by 3

$\theta = \frac{ \pi }{ 9 } , \frac{ 5 \pi }{ 9 }$

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Example Question #241 : Trigonometry

Solve for $\theta$ :

$0 = 4 \sin ^ 4 \theta + 9 \sin ^ 2 \theta - 9$

Possible Answers:

$\frac{ \pi }{6} , \frac{ 5 \pi }{6}$

$\frac{ \pi }{3} , \frac{ 5 \pi }{3}$

$\frac{ \pi }{3} , \frac{ 2 \pi }{3}$

$\frac{ \pi }{6} , \frac{ 5 \pi }{6} , \frac{ 7 \pi }{6} , \frac{ 11 \pi }{6}$

$\frac{ \pi }{3} , \frac{ 2 \pi }{3} , \frac{ 4 \pi }{3} , \frac{ 5 \pi }{3}$

Correct answer:

$\frac{ \pi }{3} , \frac{ 2 \pi }{3} , \frac{ 4 \pi }{3} , \frac{ 5 \pi }{3}$

Explanation:

Solve using the quadratic formula:

$\sin ^ 2 \theta = \frac{ -9 \pm \sqrt{ 81 - 4(4)(-9)}}{2(4)} = \frac{ -9 \pm \sqrt{225}}{8 } = \frac{ -9 \pm 15}{8}$

One possible answer is:

$\sin ^ 2 \theta = \frac{ - 9 + 15 }{ 8 } = \frac{ 6 }{ 8 } = \frac{ 3}{4}$ take the square root

$\sin \theta = \pm \sqrt{ \frac{ 3}{4} } = \pm \frac{ \sqrt3}{2}$

$\theta = \sin ^ {-1} ( \pm \frac{ \sqrt3}{2} )$

$\theta = \frac{ \pi }{3} , \frac{2 \pi }{3} , \frac{ 4 \pi }{3} , \frac{ 5 \pi }{3}$

The other would be:

$\sin ^ 2 \theta = \frac{ -9 - 15 }{8 } = \frac{ - 24}{8 } = -3$ this is outside of the range for sine

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Example Question #51 : Solving Trigonometric Equations

Solve for $\theta$ : $5 = 2 \cos ^2 \theta - 5 \cos \theta + 2$

Possible Answers:

$\theta = \frac{2 \pi }{3} , \frac{4 \pi }{3}$

$\theta = \frac{ \pi }{3}$

$\theta = \frac{ \pi }{6} , \frac{ 5 \pi }{6}$

$\theta = \frac{ \pi }{3} , \frac{ 2 \pi }{3}$

$\theta = \frac{ 2 \pi }{3}$

Correct answer:

$\theta = \frac{2 \pi }{3} , \frac{4 \pi }{3}$

Explanation:

Subtracting 5 from both sides gives the quadratic equation $0 = 2 \cos ^2 \theta - 5 \cos \theta - 3$

Using the quadratic formula gives:

$\cos \theta = \frac{ 5 \pm \sqrt{25 - 4(2)(-3)}}{2(2)} = \frac{ 5 \pm 7}{4} = \frac{ -1}{2} , 3$

The cosine cannot be 3 because that's greater than 1.

$\cos \theta = - \frac{1}{2}$

$\theta = \cos ^{-1} (-\frac{1}{2}) = \frac{2 \pi }{3} , \frac{ 4 \pi }{3}$

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Example Question #241 : Trigonometry

Which is not a solution for $\theta$ for $10 \sin ^2 \theta + \sin \theta - 3 = 0$ ?

Possible Answers:

150^o

143.13^o

60^o

323.13^o

30^o

Correct answer:

60^o

Explanation:

Using the quadratic formula gives:

$\sin \theta = \frac{ -1 \pm \sqrt{ 1 - 4(10)(-3)}}{2(10)} = \frac{-1 \pm 11}{20 } = \frac{1}{2} , -\frac{3}{5}$

$\theta = \sin ^{-1} (\frac{ 1}{2} ) = 30^o, 150^o$ or

$\theta = \sin ^{-1} (-\frac{3}{5}) \approx 323.13^o, 143.13^o$

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Example Question #83 : Trigonometric Equations

Solve for $\theta$ : $0=2 \sin ^2 \theta - 3 \sin \theta + 1$

Possible Answers:

$\frac{\pi }{2} , \frac{3 \pi }{2} , \frac{\pi }{6} , \frac{5 \pi }{6}$

$\frac{\pi }{2} , \frac{\pi }{6} , \frac{5 \pi }{6}$

$\frac{\pi }{2} , \frac{\pi }{3}$

$\frac{\pi}{2} , \frac{\pi }{6} , \frac{11 \pi }{6}$

$\frac{\pi }{2} , \frac{\pi }{6}$

Correct answer:

$\frac{\pi }{2} , \frac{\pi }{6} , \frac{5 \pi }{6}$

Explanation:

Solve using the quadratic formula:

$\sin \theta = \frac{ 3 \pm \sqrt{9 - 4(2)}}{4(1)} = \frac{ 3 \pm \sqrt1}{4} = \frac{1}{2} \enspace or \enspace 1$

$\sin \theta = \frac{1}{2}$

$\theta = \sin ^{-1} (\frac{1}{2} ) = \frac{\pi }{6} , \frac{ 5 \pi }{6}$

$\sin \theta = 1$

$\theta = \sin ^{-1}(1)$

$\theta = \frac{\pi }{2}$

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Example Question #84 : Trigonometric Equations

Find the roots for $y = \sin^2 x + 2 \sin x - 5$

Possible Answers:

No solution

60.56 ^o , 119.45^o

26.71^o, 153.29^o

90^o, 270^o

90^o

Correct answer:

No solution

Explanation:

To solve, use the quadratic formula:

$\sin \theta = \frac{ -2 \pm \sqrt{4 + 20 }}{2 } = \frac{-2 \pm \sqrt{24}}{2} = - \pm \sqrt{6}$

Both $-1 - \sqrt{6}$ and $-1+\sqrt{6}$ are outside of the range of the sine function, $-1 \leq \sin x \leq 1$ so there is no solution.

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Example Question #81 : Trigonometric Equations

Solve for :

$0 = 2 \cos ^2 x + \cos x - 2$

Possible Answers:

141.33^o, 321.33^o

38.67^o, 141.33^o

16.31^o, 343.69^o

38.67^o, 321.33^o

38.67^o, 16.31^o

Correct answer:

38.67^o, 321.33^o

Explanation:

Solve using the quadratic formula:

$\cos x = \frac{-1 \pm \sqrt{1 + 16}}{4 } = \frac{-1 \pm \sqrt{17}}{4}$

$\frac{ -1 - \sqrt{17}}{4} \approx -1.28$ , outside the range for cosine.

$\frac{-1 + \sqrt{17}}{4} \approx 0.78$

$x = \cos ^{-1} (0.78 ) = 38.67^o$ according to a calculator.

The other angle with a cosine of 0.78 would be 360 ^o - 38.67^o = 321.33^o .

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #241 : Trigonometry

Example Question #241 : Trigonometry

Example Question #242 : Trigonometry

Example Question #81 : Trigonometric Equations

Example Question #241 : Trigonometry

Example Question #51 : Solving Trigonometric Equations

Example Question #241 : Trigonometry

Example Question #83 : Trigonometric Equations

Example Question #84 : Trigonometric Equations

Example Question #81 : Trigonometric Equations

All Trigonometry Resources

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