Trigonometry : Setting Up Trigonometric Equations

Study concepts, example questions & explanations for Trigonometry

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Example Questions

Example Question #1 : Understanding Trigonometric Equations

A triangle has sides \displaystyle A, \displaystyle B, \displaystyle C of lengths \displaystyle 8, \displaystyle 15, \displaystyle 17 respectively. The angle opposite each side is called \displaystyle a, \displaystyle b, \displaystyle c, respectively. The sine of which angle and the cosine and which different angle will both yield \displaystyle \frac{8}{17}

In the answer, list the sine first and the cosine second.

Possible Answers:

\displaystyle c, a

\displaystyle b, c

\displaystyle b, a

\displaystyle a, b

\displaystyle a, c

Correct answer:

\displaystyle a, b

Explanation:

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This is the figure being described in the problem, and as sine is opposite over hypotenuse and cosine adjacent over hypotenuse, the sine of \displaystyle a and the cosine of \displaystyle b will yield the correct answer.

\displaystyle \sin(a)=\frac{opp}{hyp}=\frac{8}{17}

\displaystyle \cos(b)=\frac{adj}{hyp}=\frac{8}{17}

Example Question #2 : Setting Up Trigonometric Equations

Solve the equation over the domain \displaystyle 0 < x < 360 (answer in degrees).

\displaystyle 3sin(x)-2 = 1

Possible Answers:

\displaystyle 90

\displaystyle 270

\displaystyle 0, 360

\displaystyle 180

Correct answer:

\displaystyle 90

Explanation:

Rearrange algebraically so that,

\displaystyle 3sin(x)=3

 \displaystyle sinx = 1.

Over the interval 0 to 360 degrees, sinx = 1 when x equals 90 degrees.

Example Question #2 : Setting Up Trigonometric Equations

Solve each equation over the domain \displaystyle 0 < x < 360 (answer in degrees). 

\displaystyle tan(x-15) = 1

Possible Answers:

\displaystyle 60, 150

\displaystyle 60, 240

\displaystyle 45, 225

\displaystyle 315, 135

Correct answer:

\displaystyle 60, 240

Explanation:

First, think of the angle values for which \displaystyle tanx = 1. (This is the equivalent of taking the arctan.)

\displaystyle x-15=tan^{-1}(1)

\displaystyle x=tan^{-1}+15.

The angles for which this is true are 45 degrees and 225 degrees.

We set x-15 equal to those two angles and solve for x, giving us 60 and 240. 

Example Question #1 : Trigonometric Equations

Solve each equation over the domain \displaystyle 0 < x < 360 (answer in degrees).

\displaystyle sec(x+7)-2 = 0

Possible Answers:

\displaystyle 23, 143

\displaystyle 53, 293

\displaystyle 30, 150

\displaystyle 60, 300

Correct answer:

\displaystyle 53, 293

Explanation:

Rearrange the equation so that,

 \displaystyle sec(x+7) = 2.

Recall the angles over the interval 0 to 360 degrees for which sec is equal to 2.

These are 60 and 300 degrees.

Set x+7 equal to these angle measures and then find that x equals 53 and 293. 

Example Question #3 : Setting Up Trigonometric Equations

Solve each equation over the interval \displaystyle 0 < x <  \displaystyle 2\pi

\displaystyle 4cos^2(x) - 2 = 0

Possible Answers:

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

\displaystyle \frac{\pi}{6}, \frac{7\pi}{6}

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

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

Correct answer:

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

Explanation:

Rearrange the equation so that,

 \displaystyle cos^2x = \frac{1}{2}.

Take the square of both sides and then recall the angle measures for which,

\displaystyle cos(x)={\frac{1}{\sqrt2}}.

These measures over the interval \displaystyle 0< x< 2 \pi

 \displaystyle x=\frac{\pi}{4}; \frac{7\pi}{4}.

Example Question #1 : Setting Up Trigonometric Equations

Solve each quation over the interval \displaystyle 0 < x < 2\pi

\displaystyle sin^2(x) - 1 = 0

Possible Answers:

\displaystyle 0, \frac{\pi}{2}

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

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

\displaystyle 0, \: 2\pi

Correct answer:

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

Explanation:

Rearrange the equation so that,

\displaystyle sin^2x =1.

Take the square of both sides and find the angles for which

\displaystyle sin(x)=\pm1.

These two angles are \displaystyle \frac{\pi}{2} and \displaystyle \frac{3\pi}{2}.

Example Question #3 : Setting Up Trigonometric Equations

Solve for \displaystyle x using trigonometric ratios. 

Set up trig 1

Possible Answers:

\displaystyle \frac{1}{\sqrt2}

\displaystyle 6\sqrt2

\displaystyle \sqrt{12}

\displaystyle 3

Correct answer:

\displaystyle 6\sqrt2

Explanation:

To solve for x, first set up a trigonometric equation using the information provided in the diagram. The two side lengths given are the hypotenuse, x, and the side opposite the given angle, 6. We can set up our equation like this:

\displaystyle sin\left(\frac{\pi}{4}\right)= \frac{6}{x}

The sine of \displaystyle \frac{\pi}{4} is \displaystyle \frac{1}{\sqrt2}, so we can substitue that in:

\displaystyle \frac{1}{\sqrt2}=\frac{6}{x}

cross multiplying gives us \displaystyle x=6\sqrt2.

Example Question #4 : Setting Up Trigonometric Equations

Solve for \displaystyle x using trigonometric ratios.

Set up trig 2

Possible Answers:

\displaystyle \frac{3\sqrt3}{2}

\displaystyle \frac{3}{2}

\displaystyle 3\sqrt3

\displaystyle \frac{3}{\sqrt3}

Correct answer:

\displaystyle 3\sqrt3

Explanation:

To solve for x, first set up a trigonometric equation using the information provided in the diagram. The two side lengths given are x, the side opposite the angle, and 3, the side adjacent to the angle. This means we'll be using tangent. Set up the equation like this:

\displaystyle tan\left(\frac{\pi}{3}\right)=\frac{x}{3}

We can't just know the tangent by using the unit circle, but we can easily figure it out using sine and cosine. Tangent can be evaluated as sine over cosine.

The sine of \displaystyle \frac{\pi}{3} is \displaystyle \frac{\sqrt3}{2}, and the cosine is \displaystyle \frac{1}{2}. Find the tangent by dividing:

\displaystyle tan\left(\frac{\pi}{3}\right)=\frac{\sqrt3}{2}\div \frac{1}{2}

\displaystyle tan\left(\frac{\pi}{3}\right)=\frac{\sqrt3}{2}\cdot \frac{2}{1}

\displaystyle tan\left(\frac{\pi}{3}\right)=\sqrt3

Now we can substitute that value into the original equation we set up:

\displaystyle \sqrt3 = \frac{x}{3} multiply both sides by 3

\displaystyle 3\sqrt3=x

 

Example Question #2 : Trigonometric Equations

Solve for \displaystyle x using trigonometric ratios.

Set up trig 4

Possible Answers:

\displaystyle 2

\displaystyle \frac{1}{\sqrt3}

\displaystyle \frac{2}{\sqrt3}

\displaystyle \frac{1}{2}

Correct answer:

\displaystyle 2

Explanation:

Start by setting up the trigonometric ratio using the angles and sides given. We have the side length adjacent to the angle, 1, and the hypotenuse, x, so we will be using cosine:

\displaystyle cos\left(\frac{\pi}{3}\right)=\frac{1}{x}

The cosine of \displaystyle \frac{\pi}{3} is \displaystyle \frac{1}{2}, so this makes our equation now:

\displaystyle \frac{1}{2}=\frac{1}{x}. x must equal 2.

Example Question #3 : Setting Up Trigonometric Equations

Which of these sine functions fulfills the following criteria?

  • Range of \displaystyle [-1, 1]
  • Period of \displaystyle 2 \pi
  • \displaystyle y-intercept of \displaystyle 1
  • \displaystyle f(\pi) = -1
Possible Answers:

\displaystyle f(x) = sin(x + \frac{\pi}{2})

\displaystyle f(x) = sin(x)

\displaystyle f(x) = sin(\pi x + \frac{\pi}{2})

\displaystyle f(x) = sin(x - \frac{\pi}{2})

\displaystyle f(x) = sin(\pi x)

Correct answer:

\displaystyle f(x) = sin(x + \frac{\pi}{2})

Explanation:

Examining the equation of this form:

\displaystyle f(x) = asin(bx + c) + d

We can find \displaystyle a, \displaystyle b\displaystyle c and \displaystyle d using the clues given:

  • \displaystyle a = 1 because the range of \displaystyle [-1, 1] indicates an amplitude of \displaystyle 1.
  • \displaystyle d = 0 because \displaystyle 0 is the midpoint between \displaystyle -1 and \displaystyle 1.
  • \displaystyle b = 1 because the period is not changed from the standard \displaystyle 2\pi.
  • \displaystyle c = \frac{\pi}{2}. The combined fact that \displaystyle f(0) = 1 and \displaystyle f(\pi) (halfway through the period) is \displaystyle -1 indicates a cosine function would work - only the answer must be given as a sine function. Fortunately, though, we can use a shift-related property of \displaystyle cos(x) = sin(x + \frac{\pi}{2}).

All told, once we realize these, then we can see that
\displaystyle f(x) = sin(x + \frac{\pi}{2}) fits our criteria.

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