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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

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

Example Questions

Example Question #191 : Trigonometry

Which of the following systems of trigonometric equations have a solution with an -coordinate of $\frac{\pi}{4}$ ?

Possible Answers:

y=sin(x)

y=cos(3x)

y=sin(x)

y=cos(5x)

y=sin(x)

y=cos(7x)

More than one of these answers has a solutions at $x=\frac{\pi}{4}$ .

Correct answer:

y=sin(x)

y=cos(7x)

Explanation:

The solution to the correct answer would be $\left(\frac{\pi}{4}, \frac{1}{\sqrt2}\right)$ .

For all of the other answers, plugging in $\frac{\pi}{4}$ for the second equation gives a y value of $-\frac{1}{\sqrt2}$ .

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Example Question #2 : Solving Trigonometric Equations

Solve the system for $\theta$ :

$\begin{matrix} y = 6 \sin \theta - 1 \\ y = \sin ^2 \theta + 4 \end{matrix}$

Possible Answers:

no solution

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

$\pi$

$\frac{ \pi }{2 }$

$2 \pi$

Correct answer:

$\frac{ \pi }{2 }$

Explanation:

First, set both equations equal to each other:

$6 \sin \theta - 1 = \sin ^2 \theta + 4$ subtract $6 \sin \theta$ from both sides

$-1 = \sin ^2 \theta - 6 \sin \theta + 4$ add 1 to both sides

$0 = \sin ^2 \theta - 6 \sin \theta + 5$

Now we can solve this as a quadratic equation, where "x" is $\sin \theta$ . Using the quadratic formula:

$\sin \theta = \frac{ 6 \pm \sqrt{ 36 - 4(1)(5)}}{2} = \frac{6 \pm \sqrt{16 }}{2} = \frac{ 6 \pm 4 }{2}$

This gives us 2 potential solutions for $\sin \theta$ :

$\sin \theta = \frac{10}{2} = 5$ the sine of an angle cannot be greater than 1

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

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

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Example Question #3 : Solving Trigonometric Equations

Solve this system for $\theta$ :

$\begin{matrix} y = 5 \sin (\frac{ \theta }{2}) + 3 \\ y = 3 \sin (\frac{\theta }{2} ) + 2 \end{matrix}$

Possible Answers:

$\frac{ 11 \pi }{12} , \frac{ 7 \pi }{12}$

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

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

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

$\frac{ 11 \pi }{3} , \frac{ 7 \pi }{3}$

Correct answer:

$\frac{ 11 \pi }{3} , \frac{ 7 \pi }{3}$

Explanation:

First, set the two equations equal to each other

$5 \sin (\frac{ \theta }{2}) + 3 = 3 \sin (\frac{ \theta }{2} ) + 2$ subtract the sine term from the right

$2 \sin (\frac{ \theta } { 2 }) + 3 = 2$ subtract 3 from both sides

$2 \sin (\frac{ \theta }{2} ) = -1$ divide by 2

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

$\frac{ \theta } {2} = \sin ^{-1} (- \frac{1}{2}) = \frac{ 11 \pi }{6} , \frac{ 7 \pi }{6}$ multiply by 2

$\theta = \frac{ 11 \pi }{3} , \frac{ 7 \pi }{3}$

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Example Question #1 : Solving Trigonometric Equations

Solve this system for $\theta$ :

$y = \sin ^ 2 \theta + \cos \theta$

$y = \cos \theta + \frac{1}{2}$

Possible Answers:

$\frac{ \pi }{4} , \frac{ 3 \pi }{4} ,\frac{ 5 \pi }{4} , \frac{ 7 \pi }{4 }$

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

$\frac{ \pi }{4} , \frac{ 7 \pi }{4}$

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

no solution

Correct answer:

$\frac{ \pi }{4} , \frac{ 3 \pi }{4} ,\frac{ 5 \pi }{4} , \frac{ 7 \pi }{4 }$

Explanation:

Set the two equations equal to each other

$\sin ^2 \theta + \cos \theta = \cos \theta + \frac{ 1}{2}$ subtract cos from both sides

$\sin ^ 2 \theta = \frac{ 1}{2 }$ take the square root of both sides

$\sin \theta = \pm \frac{ 1}{ \sqrt2}$

$\theta = \sin ^{-1} ( \pm \frac{ 1} {\sqrt2}) = \frac{ \pi }{4} , \frac{ 3 \pi }{4} , \frac{ 5 \pi }{4} , \frac{ 7 \pi }{4}$

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Example Question #31 : Trigonometric Equations

Solve this system for $\theta$ :

$\\ y = \cos ^2 \theta + \frac{1}{2} \\ y = \sin ^2 \theta$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Set both equations equal to each other:

$\cos ^2 \theta + \frac{1}{2} = \sin ^2 \theta$ subtract $\frac{1}{2}$ from both sides

$\cos ^2 \theta = \sin ^2 \theta - \frac{1}{2}$ subtract $\sin ^2 \theta$ from both sides

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

We can re-write the left side using a trigonometric identity

$\cos ( 2 \theta ) = -\frac{1}{2 }$ take the inverse cosine

$2 \theta = \cos ^ {-1}\left(-\frac{1}{2} \right)$

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

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

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Example Question #1 : Solving Trigonometric Equations

Solve this system for $\theta$ :

$\\y = 4 \cos ^ 2 \theta + \cos \theta - 5 \\ y = \cos \theta - 2$

Possible Answers:

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

$\frac{ \pi }{3} , \frac{ 11 \pi }{3}$

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

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

no solution

Correct answer:

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

Explanation:

Set the two equations equal to each other:

$4 \cos ^2 \theta + \cos \theta - 5 = \cos \theta - 2$ subtract $\cos \theta$ from both sides

$4 \cos ^ 2 \theta - 5 = -2$ add 5 to both sides

$4 \cos ^ 2 \theta = 3$ divide both sides by 4

$\cos ^ 2 \theta = \frac{ 3}{4}$ take the square root of both sides

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

$\theta = \cos ^ {-1} \left( \pm \frac{ \sqrt 3}{2}\right)$

$\theta = \frac{ \pi }{6} , \frac{ 11 \pi }{6}$

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Example Question #1 : Solving Trigonometric Equations

Solve the following system:

Screen shot 2020 05 21 at 10.30.36 am

Possible Answers:

$\frac{-\pi}{3}$

$\pi$

The system does not have a solution.

$-\pi$

$\frac{\pi}{3}$

Correct answer:

The system does not have a solution.

Explanation:

A number x is a solution if it satisfies both equations.

We note first we can write the first equation in the form :

$cosx\ge \frac{4}{3}$

We know that $cosx\le 1$ for all reals. This means that there is no x that

satisifies the first inequality. This shows that the system cannot satisfy both equations since it does not satisfy one of them. This shows that our system does not have a solution.

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Example Question #1 : Solving Trigonometric Equations

Solve this system for $\theta$ :

$\begin{matrix} y = \sin ^2 \theta + 1 \\ y = \cos \theta + 2 \end{matrix}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

First, set both equations equal to each other:

$\sin ^2 \theta + 1 = \cos \theta + 2$ subtract $\cos \theta$ from both sides

$\sin ^2 \theta - \cos \theta + 1 = 2$

Using a trigonometric identity, we can re-write $\sin^2 \theta$ as $1 - \cos ^2 \theta$ :

$1 - \cos^2 \theta - \cos \theta + 1 =2$ combine like terms

$- \cos ^2 \theta - \cos \theta + 2 = 2$ subtract 2 from both sides

$- \cos^2 \theta - \cos \theta = 0$

We can solve for $\cos \theta$ using the quadratic formula:

$\cos \theta = \frac{ 1 \pm \sqrt{ 1 - 4(-1)(0)}}{2(-1)} = \frac{ 1 \pm 1}{ -2 }$

This gives us 2 possible values for cosine

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

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

$\cos \theta = \frac{ 1 - 1 }{-2 } = \frac{ 0 }{-2} = 0$

$\theta = \cos ^ {-1} (0) = \frac{ \pi }{2} , \frac{ 3 \pi }{2}$

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Example Question #1 : Finding Trigonometric Roots

Which of the following is a solution to the following equation such that $0\leq x<2\pi?$

2sin^2(x)-sin(x)+3=4

Possible Answers:

$\frac{\pi}{6}$

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

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

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

$\frac{5\pi}{4}$

Correct answer:

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

Explanation:

We begin by getting the right side of the equation to equal zero.

2sin^2(x)-sinx-1=0

Next we factor.

(2sin(x)+1)(sin(x)-1)=0

We then set each factor equal to zero and solve.

2sin(x)+1=0 or sin(x)-1=0

$sin(x)=-\frac{1}{2}$ sin(x)=1

We then determine the angles that satisfy each solution within one revolution.

The angles $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ satisfy the first, and $\frac{\pi}{2}$ satisfies the second. Only $\frac{11\pi}{6}$ is among our answer choices.

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Example Question #2 : Finding Trigonometric Roots

Solve the following equation for $0\leq x<2\pi$ .

$\sin 2x=\cos 2x$

Possible Answers:

$x=\frac{\pi}{8}; \frac{5\pi}{8}$

$x=\frac{\pi}{4}; \frac{5\pi}{4}$

$x=\frac{\pi}{4}; \frac{5\pi}{4}; \frac{9\pi}{4}; \frac{13\pi}{4}$

$x=\frac{\pi}{8}; \frac{5\pi}{8}; \frac{9\pi}{8}; \frac{13\pi}{8}$

No solution exists

Correct answer:

$x=\frac{\pi}{8}; \frac{5\pi}{8}; \frac{9\pi}{8}; \frac{13\pi}{8}$

Explanation:

The fastest way to solve this problem is to substitute a new variable. Let u=2x .

The equation now becomes:

$\sin u=\cos u$

So at what angles are the sine and cosine functions equal. This occurs at

$u=\frac{\pi}{4}; \frac{5\pi}{4}; \frac{9\pi}{4}; \frac{13\pi}{4}$

You may be wondering, "Why did you include

$\frac{9\pi}{4}; \frac{13\pi}{4}$ if they're not between and $2\pi$ ?"

The reason is because once we substitute back the original variable, we will have to divide by 2. This dividing by 2 will bring the last two answers within our range.

$2x=\frac{\pi}{4}; 2x=\frac{5\pi}{4}; 2x=\frac{9\pi}{4}; 2x=\frac{13\pi}{4}$

Dividing each answer by 2 gives us

$x=\frac{\pi}{8}; \frac{5\pi}{8}; \frac{9\pi}{8}; \frac{13\pi}{8}$

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #191 : Trigonometry

Example Question #2 : Solving Trigonometric Equations

Example Question #3 : Solving Trigonometric Equations

Example Question #1 : Solving Trigonometric Equations

Example Question #31 : Trigonometric Equations

Example Question #1 : Solving Trigonometric Equations

Example Question #1 : Solving Trigonometric Equations

Example Question #1 : Solving Trigonometric Equations

Example Question #1 : Finding Trigonometric Roots

Example Question #2 : Finding Trigonometric Roots

All Trigonometry Resources

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