Cosine - ACT Math

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Question

The value of a cosine is positive in which quadrants?

Answer

The cosine is positive in the 1st and 4th quadrants and negative in 2nd and 3rd

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Question

Which of the following is equal to $cos\ x * csc\ x$ ?

Answer

Here, we use the SOHCAHTOA ratios and the fact that csc x = 1 / sin x.

Cosine x = adjacent side length / hypotenuse length

Cosecant x = 1 / sin x = hypotenuse / opposite

(Adjacent / hypotenuse) * (hypotenuse / opposite) = Adjacent / opposite = Cotangent x.

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Question

and is between and $\pi$ . What is the value of ?

Answer

For to $\pi$ , we know that $cos(\frac{\pi}{3}) = 0.5$ . So, the question asks, what is the value of , where $x=\frac{\pi}{3}$ . Therefore, it is asking what the value of $cos(\frac{\pi}{6})$ is, which is $\frac{\sqrt{3}}{2}$ .

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Question

To the nearest , what is the cosine formed from the origin to ? Assume counterclockwise rotation.

Answer

If the point to be reached is , then we may envision a right triangle with sides and , and hypotenuse . The Pythagorean Theorem tells us that , so we plug in and find that:

Thus, $c = \sqrt{61}$

Now, SOHCAHTOA tells us that $\textup{cosine} = \frac{\textup{adjacent}}{\textup{hypotenuse}}$ , so we know that:

$\textup{cos }x^{\circ} = \frac{5}{\sqrt{61}} \approx 0.6402$

Thus, our cosine is approximately .

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Question

Two drivers race to a finish line. Driver A drives north blocks, and east blocks and crosses the goal. Driver B drives the shortest direct route between the two points. Relative to east, what is the cosine of the angle at which Driver B raced? Round to the nearest .

Answer

If the point to be reached is blocks north and blocks east, then we may envision a right triangle with sides and , and hypotenuse . The Pythagorean Theorem tells us that , so we plug in and find that:

Thus, $c = \sqrt{436} = 2\sqrt{109}$

Now, SOHCAHTOA tells us that $\textup{cosine} = \frac{\textup{adjacent}}{\textup{hypotenuse}}$ , so we know that:

$cos x^{\circ} = \frac{20}{2\sqrt{109}} = \frac{10}{\sqrt{109}} \approx .9578$

Thus, our cosine is approximately .

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Question

Simplify (cosΘ – sinΘ)2

Answer

Multiply out the quadratic equation to get cosΘ2 – 2cosΘsinΘ + sinΘ2

Then use the following trig identities to simplify the expression:

sin2Θ = 2sinΘcosΘ

sinΘ2 + cosΘ2 = 1

1 – sin2Θ is the correct answer for (cosΘ – sinΘ)2

1 + sin2Θ is the correct answer for (cosΘ + sinΘ)2

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Question

Which of the following represents a cosine function with a range of to ?

Answer

The range of a cosine wave is altered by the coefficient placed in front of the base equation. So, if you have , this means that the highest point on the wave will be at and the lowest at ; however, if you then begin to shift the equation vertically by adding values, as in, , then you need to account for said shift. This would make the minimum value to be and the maximum value to be .

For our question, the range of values covers . This range is accomplished by having either or as your coefficient. ( merely flips the equation over the -axis. The range "spread" remains the same.) We need to make the upper value to be instead of . To do this, you will need to add , or , to . This requires an upward shift of . An example of performing a shift like this is:

Among the possible answers, the one that works is:

The parameter does not matter, as it only alters the frequency of the function.

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Question

Which of the following represents a cosine function with a range of to ?

Answer

The range of a cosine wave is altered by the coefficient placed in front of the base equation. So, if you have , this means that the highest point on the wave will be at and the lowest at ; however, if you then begin to shift the equation vertically by adding values, as in, , then you need to account for said shift. This would make the minimum value to be and the maximum value to be .

For our question, the range of values covers . This range is accomplished by having either or as your coefficient. ( merely flips the equation over the -axis. The range "spread" remains the same.) We need to make the upper value to be instead of . To do this, you will need to subtract , or , from . This requires an downward shift of . An example of performing a shift like this is:

Among the possible answers, the one that works is:

The parameter does not matter, as it only alters the frequency of the function.

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Question

Which of the following represents a cosine function with a range from to ?

Answer

The range of a cosine wave is altered by the coefficient placed in front of the base equation. So, if you have , this means that the highest point on the wave will be at and the lowest at ; however, if you then begin to shift the equation vertically by adding values, as in, , then you need to account for said shift. This would make the minimum value to be and the maximum value to be .

For our question, the range of values covers . This range is accomplished by having either or as your coefficient. ( merely flips the equation over the -axis. The range "spread" remains the same.) We need to make the upper value to be instead of . This requires an upward shift of . An example of performing a shift like this is:

Among the possible answers, the one that works is:

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Question

What is the range of the trigonometric function defined by $S(t) = -3\sin (1.5t)$ ?

Answer

The range of a sine or cosine function spans from the negative amplitude to the positive amplitutde. The amplutide is given by $\left|a\right|$ in the equation $S(t)=a\sin(bt)$ . Thus the range for our function is

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Question

What is the range of the given trigonometric equation:

$f(t) = \cos(-3t)$

Answer

For the sine and cosine funcitons, the range is equal to the negative amplitude to the positive amplitude.

The amplitude is found by taking $\left|a\fight|$ from the general equation:

$f(t) = a\cos(bt)$ .

We see in our equation that $a = 1 \rightarrow \left|a\right| = 1$

(when no coefficient is written, it is a 1).

Thus we get that the amplitude is

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Question

What is the range of the function ?

Answer

The range of the function represents the spread of possible answers you can get for , given all values of . In this case, the ordinary range for a cosine function is , since the largest value that cosine can solve to is (for a cosine of $0/2\pi/360^{\circ}$ or a multiple of one of those values), and the smallest value cosine can solve to is (for a cosine of $\pi/180^{\circ}$ or a multiple of one of those values).

However, in this case our final answer is increased by after the cosine is applied to . This results in a final range of to (or, in other words, to , plus ).

So, our final range is .

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Question

What is the range of the function ?

Answer

The range of the function represents the spread of possible answers you can get for , given all values of . In this case, the ordinary range for a cosine function is , since the largest value that cosine can solve to is (for a cosine of $0/2\pi/360^{\circ}$ or a multiple of one of those values), and the smallest value cosine can solve to is (for a cosine of $\pi/180^{\circ}$ or a multiple of one of those values).

However, in this case our final answer is multiplied by -3 after the cosine is applied to . This results in a final range of to (or, in other words, to , multiplied by ).

So, our final range is .

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Question

What is the range of the function ?

Answer

The range of the function represents the spread of possible answers you can get for , given all values of . In this case, the ordinary range for a cosine function is , since the largest value that cosine can solve to is (for a cosine of $0/2\pi/360^{\circ}$ or a multiple of one of those values), and the smallest value cosine can solve to is (for a cosine of $\pi/180^{\circ}$ or a multiple of one of those values).

However, in this case our final answer is first multiplied by , then decreased by after the cosine is applied to . Multiplying the initial range by results in a new range of . Next, subtracting from this range gives us a new range of .

Note that the does not change our range. This is because, irrespective of other multipliers, a cosine operation can only return values between and . To think of this a different way, will give us the same returns as , only we will move around the unit circle five times as much before finding our answer.

Thus, our final range is .

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Question

Which of the following functions has a range of ?

Answer

The range of the function represents the spread of possible answers you can get for , given all values of . In this case, the ordinary range for a cosine function is , since the largest value that cosine can solve to is (for a cosine of $0/2\pi/360^{\circ}$ or a multiple of one of those values), and the smallest value cosine can solve to is (for a cosine of $\pi/180^{\circ}$ or a multiple of one of those values).

One fast way to match a range to a function is to look for the function which has a vertical shift equal to the mean of the range values. In other words, for the standard trigonometric function $f (x) = \alpha cos(bx-c) +d$ , where represents the vertical shift, $d = \frac{range_{max}+range_{min}}{2}$ .

In this case, since our range is , we expect our to be $\frac{8-22}{2} = -7$ .

Of the answer choices, only has , so we know this is our correct choice.

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Question

Which of the following functions has a range of ?

Answer

The range of the function represents the spread of possible answers you can get for , given all values of . In this case, the ordinary range for a cosine function is , since the largest value that cosine can solve to is (for a cosine of $0/2\pi/360^{\circ}$ or a multiple of one of those values), and the smallest value cosine can solve to is (for a cosine of $\pi/180^{\circ}$ or a multiple of one of those values).

One fast way to match a range to a function is to look for the function which has a vertical shift equal to the mean of the range values. In other words, for the standard trigonometric function $f (x) = \alpha cos(bx-c) +d$ , where represents the vertical shift, $d = \frac{range_{max}+range_{min}}{2}$ .

In this case, since our range is , we expect our to be $\frac{7+9}{2} = 8$ .

Of the answer choices, only has , so we know this is our correct choice.

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Question

Which of the following functions has a range of ?

Answer

The range of the function represents the spread of possible answers you can get for , given all values of . In this case, the ordinary range for a cosine function is , since the largest value that cosine can solve to is (for a cosine of $0/2\pi/360^{\circ}$ or a multiple of one of those values), and the smallest value cosine can solve to is (for a cosine of $\pi/180^{\circ}$ or a multiple of one of those values).

One fast way to match a range to a function is to look for the function which has a vertical shift equal to the mean of the range values. In other words, for the standard trigonometric function $f (x) = \alpha cos(bx-c) +d$ , where represents the vertical shift, $d = \frac{range_{max}+range_{min}}{2}$ .

In this case, since our range is , we expect our to be $\frac{-8-12}{2} = -10$ .

Of the answer choices, only has , so we know this is our correct choice.

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Question

If $\dpi{100} \small \tan (x)=\frac{5}{12}$ , then what is $\dpi{100} \small \cos (x)=$ ?

Answer

If $\dpi{100} \small \tan (x)=\frac{5}{12}$ ,

and being that $\dpi{100} \small \tan (x)=\frac{opposite}{adjacent}$ ,

$\dpi{100} \small \frac{5}{12}=\frac{opposite}{adjacent}$ .

From this, you can construct a right triangle where 5 is the opposite side to the reference angle and 12 is the adjacent side to the reference angle.

After constructing this triangle, you can then proceed to apply the Pythagorean Theorem.

$\dpi{100} \small C^{2}=A^{2}+B^{2}$

As 5 and 12 are the legs of the constructed right triangle, 5 and 12 are $\dpi{100} \small a$ and $\dpi{100} \small b$ in no particular order. Hence:

$\dpi{100} \small A=5,\ B=12$

$\dpi{100} \small C^{2}=A^{2}+B^{2}$

$\dpi{100} \small C^{2}=5^{2}+12^{2}$

$\dpi{100} \small C^{2}=25+144$

$\dpi{100} \small C^{2}=169$

$\dpi{100} \small \sqrt{C^{2}}=\sqrt{169}$

$\dpi{100} \small C=13$

We have calculated the hypotenuse of our right triangle. With this knowledge, we can now assign a value to $\dpi{100} \small \cos (x)$ . As $\dpi{100} \small \cos (x)=\frac{adjacent}{hypotenuse}$ , our values of adjacent (12) and our hypotenuse (13) will now be very useful.

The $\dpi{100} \small \cos (x)=\frac{12}{13}$ .

A helpful pneumonic to remember trig functions: $\dpi{100} \small SOH-CAH-TOA$ . (Sound it out as phonetically as you can)

$\dpi{100} \small SOH$ : $\dpi{100} \small \sin (x)= \frac{opposite}{hyotenuse}$

$\dpi{100} \small CAH$ : $\dpi{100} \small cos(x)=\frac{adjacent}{hypotenuse}$

$\dpi{100} \small TOA$ : $\dpi{100} \small tan(x) = \frac{opposite}{adjacent}$

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Question

Find the length of in the right triangle below to the nearest tenth.

Answer

Finding a missing side in a right triangle generally utliizes one of two concepts: the Pythagorean Theorem or trigonometry. When we have two sides given and need the third side, we use the Pythagorean Theorem. When we only have one side and need a second but do have the help of an angle other than the right angle, we can use trigonometry. That is the case our problem falls into.

The three sides of every right triangle can be identified in terms of one of the non-right angles. The hypotenuse is always the side opposite the right angle (and the longest side). But if we take one of the non-right angles as our starting point, one of the legs is opposite the angle, and another is adjacent to the angle. For example, in our triangle, from the starting point of the the 18-degree angle, our hypotenuse is the side of length 17, the adjacent leg is the side labeled , and the unlabeled side is the opposite side.

Given these distinctions, right triangle trigonometry hinges on three ratios: sine, cosine, and tangent. Sine is the ratio of the opposite leg to the hypotenuse. Cosine is the ratio of the adjacent leg to the hypotenuse. Tangent is the ratio of the opposite leg to the adjacent leg.

Our goal is to find , the length of the adjacent leg. We know the length of the hypotenuse is 17. Therefore, our ratio of choice is cosine. Cosine of our angle (18 degrees) is the ratio of to 17. We can write this as an equation.

$cos(18^{\circ})=\frac{x}{17}$

Fortunately for us living in our technological age, most calculators have buttons for these three ratios. But before you can use your calculator, you have to be sure it is in the proper mode. The calculator needs to be in degree mode, but the way to check and/or change modes depend on the calculator, so if you don't know how to do this, check the manual or consult a knowledgeable person.

But with a calculator in degree mode we can continue. For most calculators, simply hit the cosine button (abbreviated "cos"), followed by the angle measure (18), and then hit the equal button.

You should get 0.9510565. That means we get the following equation.

$0.95105652=\frac{x}{17}$

From there we have an easy equation to solve. We simply multiply both sides of the equation by 17.

$x=17\cdot0.95105652=16.167961$

Rounded to the nearest tenth, our answer is 16.2.

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Question

The following figure is not drawn to scale.

Using cosine, find the measure of the missing side labeled as .

Answer

Trig functions are only possible to use when the triangle is a right triangle. Although the figure may provide ample information and look like it is a right triangle, the latter cannot be assumed because there is no indication of a right angle. It lacks the notation that may be observed in the following:

Therefore the correct answer to this problem is that it is not possible using a Trigonometric function.

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