Setting Up Trigonometric Equations - Trigonometry
Card 0 of 13
A triangle has sides 
, 
, 
of lengths 
, 
, 
respectively. The angle opposite each side is called 
, 
, 
, respectively. The sine of which angle and the cosine and which different angle will both yield 
In the answer, list the sine first and the cosine second.
A triangle has sides
In the answer, list the sine first and the cosine second.
This is the figure being described in the problem, and as sine is opposite over hypotenuse and cosine adjacent over hypotenuse, the sine of 
and the cosine of 
will yield the correct answer.
This is the figure being described in the problem, and as sine is opposite over hypotenuse and cosine adjacent over hypotenuse, the sine of
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Solve the equation over the domain 
(answer in degrees).
Solve the equation over the domain
Rearrange algebraically so that,
.
Over the interval 0 to 360 degrees, sinx = 1 when x equals 90 degrees.
Rearrange algebraically so that,
Over the interval 0 to 360 degrees, sinx = 1 when x equals 90 degrees.
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Solve each equation over the domain 
(answer in degrees).
Solve each equation over the domain
First, think of the angle values for which 
. (This is the equivalent of taking the arctan.)
.
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.
First, think of the angle values for which
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.
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Solve each equation over the domain 
(answer in degrees).
Solve each equation over the domain
Rearrange the equation so that,
.
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.
Rearrange the equation so that,
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.
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Solve each equation over the interval 
Solve each equation over the interval
Rearrange the equation so that,
.
Take the square of both sides and then recall the angle measures for which,
.
These measures over the interval 
.
Rearrange the equation so that,
Take the square of both sides and then recall the angle measures for which,
These measures over the interval
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Solve each quation over the interval 
Solve each quation over the interval
Rearrange the equation so that,
.
Take the square of both sides and find the angles for which
.
These two angles are 
and 
.
Rearrange the equation so that,
Take the square of both sides and find the angles for which
These two angles are
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Solve for 
using trigonometric ratios.
Solve for
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:
The sine of 
is 
, so we can substitue that in:
cross multiplying gives us 
.
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:
The sine of
cross multiplying gives us
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Solve for 
using trigonometric ratios.
Solve for
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:
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 
is 
, and the cosine is 
. Find the tangent by dividing:
Now we can substitute that value into the original equation we set up:
multiply both sides by 3
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:
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
Now we can substitute that value into the original equation we set up:
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Solve for 
using trigonometric ratios.
Solve for
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:
The cosine of 
is 
, so this makes our equation now:
. x must equal 2.
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:
The cosine of
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Which of these sine functions fulfills the following criteria?
- Range of
- Period of
-intercept of 
Which of these sine functions fulfills the following criteria?
- Range of
- Period of
Examining the equation of this form:
We can find 
, 
, 
and 
using the clues given:
because the range of 
indicates an amplitude of 
.
because 
is the midpoint between 
and 
.
because the period is not changed from the standard 
.
. The combined fact that 
and 
(halfway through the period) is 
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 
.
All told, once we realize these, then we can see that
fits our criteria.
Examining the equation of this form:
We can find
All told, once we realize these, then we can see that
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Which of these functions fulfills these criteria?
- Amplitude of
- Period of
-intercept of 
- No phase shift
- Minimum of
Which of these functions fulfills these criteria?
- Amplitude of
- Period of
- No phase shift
- Minimum of
Combining a good deal of our information - an amplitude of 
, a 
-intercept of 
and a minimum of 
with no phase shift - means we are looking for a cosine function. In other words, we can start right at...
We also know that 
because there is a period of 
(since 
when 
. In other words, we can conclude that the function we are looking for is...
Combining a good deal of our information - an amplitude of
We also know that
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Which of these functions includes the following points?
Which of these functions includes the following points?
Judging from those five given points, we can draw the following clues:
- Amplitude of
- Period of
Also, note that none of the answer choices have a phase shift, meaning that you can instantly start looking for a sine function because of the zero-rise-zero-fall-zero period that is marked by those five points. Furthermore, we can also realize that we are looking for...
...because we need a period of 4 - i.e,
when 
.
Judging from those five given points, we can draw the following clues:
- Amplitude of
- Period of
Also, note that none of the answer choices have a phase shift, meaning that you can instantly start looking for a sine function because of the zero-rise-zero-fall-zero period that is marked by those five points. Furthermore, we can also realize that we are looking for...
...because we need a period of 4 - i.e,
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A sine function 
where 
is time measured in seconds has the following properties:
- Amplitude of
- Minimum of
- No phase shift
- Frequency of
Hz (cycles per second)
is which of these functions?
A sine function
- Amplitude of
- Minimum of
- No phase shift
- Frequency of
One important thing to realize is that the frequency is the reciprocal of the period. So if the function has a frequency of 
Hertz (or 
cycles per second), the period has to be 
or 
seconds. Because 
when 
, we know we are looking for a equation that includes 
.
Also, because we have an amplitude of 
but a minimum of 
, there must be a shift upwards by 
units. Only one function fulfills those two criteria and the period criteria:
One important thing to realize is that the frequency is the reciprocal of the period. So if the function has a frequency of
Also, because we have an amplitude of
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