Evaluating Trig Functions - Pre-Calculus

Card 0 of 52

Find the value of to the nearest tenth if and .

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Rewrite in terms of sine and cosine.

$cot(30)+tan(30)=\frac{cos(30)}{sin(30)}$+\frac{sin(30)}{cos(30)}$$

Substitute the known values and evaluate.

$$\frac{0.866}{0.5}$+\frac{0.5}{0.866}$=1.732+0.577367 =2.309367$

The answer to the nearest tenth is .

Find the value of

$\sin(270^\circ)+\cos(60^\circ)$ .

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Before beginning this problem on a calculator, though this is not necessary since these are special angles, ensure that the mode of the calculator is in degrees.

Input the values of the expression and solve.

Find the decimal value of

$2\cos\left($\frac{\pi}{6}\right)$

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To determine the decimal value of the following trig function, $2cos($\frac{\pi}{6}$)$ , make sure that the calculator is in radian mode.

Compute the expression.

$2cos($\frac{\pi}{6}$)=2(0.866)=1.732$

Determine the value of $cot($\frac{\pi}{3}$)$ in decimal form.

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Ensure the calculator is in radian mode since the expression shows the angle in terms of $\pi$ . Also convert cotangent to tangent.

$cot($\frac{\pi}{3}$)=\frac{1}{tan(\frac{\pi}${3})}= $\frac{1}{\sqrt3}$=0.577$

Determine the correct value of

.

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The question $sin($\frac{1}{2}$)$ asks for the y-coordinate on the unit circle when the degree angle is $\frac{1}{2}$ .

Be careful not to confuse finding the value of the angle when the y-value of the coordinate of the unit circle is $\frac{1}{2}$ .

Ensure that the calculator is in degree mode.

$sin($\frac{1}{2}$)=0.00872=0.009$

What is the value of $\theta$ (in degrees)?

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One can setup the relationship

$\tan(\theta) = $\frac{opposite}{adjacent}$\ tan(\theta)=\frac{9}{5}$$ .

After taking the arctangent,

the arctangent cancels out the tangent and we are left with the value of $\theta$ .

$\ $\theta=tan^{-1}$\left($\frac{9}{5}$ \right ) \ \theta=60.945^\circ \ \theta=61^\circ$

The above triangle is a right triangle. Find the value of $\theta$ (in degrees).

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One can setup the relationship

$cos(\theta)=\frac{adjacent}{hypotenuse}$$

$cos(\theta)=\frac{6}{11}$$ .

After taking the arccosine,

the arccosine cancels out the cosine leaving just the value of $\theta$ .

$\ $\theta=cos^{-1}$\left($\frac{6}{11}$ \right ) \ \theta= 56.944^\circ \ \theta= 57^\circ$

Solve for all x on the interval

$sin(x)=\frac{\sqrt{2}$}{2}$

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Solve for all x on the interval

$sin(x)=\frac{\sqrt{2}$}{2}$

We can begin by recalling which two quadrants have a positive sine. Because sine corresponds to the y-value, we know that sine is positive in quadrants I and II.

Next, recall where we get $$\frac{\sqrt{2}$}{2}$ .

$$\frac{\sqrt{2}$}{2}$ always corresponds to our -increment angles. In this case, the angles we are looking for are and , because those are the two -increment angles in the first two quadrants.

Now, you might be saying, "what about ? That is an increment of 45."

While that is true, , not $$\frac{\sqrt{2}$}{2}$

So our answer is:

,

Solve for all x on the interval

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Solve for all x on the interval

Remember Soh, Cah, Toa?

For this problem it helps to recall that $Tangent=\frac{opposite}{adjacent}$$

Since our tangent is equal to 1 in this problem, we know that our opposite and adjacent sides must be the same (otherwise we wouldn't get "1" when we divided them)

Can you think of any angles in the first quadrant which yield equal x and y values?

If you guessed you guessed right! Remember that your angle in the unit circle will give you a triangle, which will have equal height and base.

Given the equation $sec(\Theta)=-2$ , what is one possible value of $\Theta$ ?

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Find 1 possible value of $\Theta$ Given the following:

$sec(\Theta)=-2$

Recall that

$sec(\Theta)=\frac{1}{cos(\Theta)}$$

So if $sec(\Theta)=-2$ , then $cos(\Theta)=\frac{-1}{2}$$

Thinking back to our unit circle, recall that cosine corresponds to the x-value. Therefore, we must be in quadrants II or III.

So, which angles correspond to an x-value of -0.5? Well, they must be the angles closest to the y-axis, which are our increment angles.

This means our angle must be either

or

It must be , because 240 is not an option.

Note that there are technically infinte solutions, because we are not given a specific interval. However, we only need to worry about one.

Solve for $\theta$ :

$\sin(3\theta) = -$\frac{\sqrt{3}$}{2}$

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If the sine of an angle, in this case $3 \theta$ is $-$\frac{ \sqrt 3 }{2}$$ , the angle must be $\frac{4 \pi }{3}$ or $\frac{5 \pi }{3 }$ .

Then we need to solve for theta by dividing by 3:

$\frac{4 \pi }{3}$ \div 3 = $\frac{4 \pi }{3}$ \cdot $\frac{1}{3}$ = $\frac{4 \pi }{9 }$

$\frac{5 \pi }{3}$ \div 3 = $\frac{5 \pi }{3}$ \cdot $\frac{1}{3}$ = $\frac{5 \pi }{9 }$

Which of the following could be a value of $\theta$ ?

$tan(\theta)=-1$

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Which of the following could be a value of $\theta$ ?

$tan(\theta)=-1$

To begin, it will be helpful to recall the following property of tangent:

$tan=\frac{sin}{cos}$=\frac{y}{x}$$

This means that if $tan(\theta)=-1$ our sine and cosine must have equal absolute values, but with opposite signs.

The only place where we will have equal values for sine and cosine will be at the locations halfway between our quadrantal angles (axes). In other words, our answer will align with one of the angles.

Additionally, because our sine and cosine must have opposite signs (one negative and one positive), we need to be in either quadrant 2 or quadrant 4. There is only answer from either of those two, so our answer must be .

Find $\theta$ if $\sin\theta =\frac{1}{2 }$$ and it is located in Quadrant I.

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Since we know the value of the trigonometric function and the triangle is located in Quadrant I, we can draw the triangle and get a sense of it. If the opposite side is 1 and the hypotenuse is 2, we know that we're dealing with a 30-60-90 special triangle. And since the opposite side of the angle is 1, we know that the angle is .

Find the value of to the nearest tenth if and .

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Rewrite in terms of sine and cosine.

$cot(30)+tan(30)=\frac{cos(30)}{sin(30)}$+\frac{sin(30)}{cos(30)}$$

Substitute the known values and evaluate.

$$\frac{0.866}{0.5}$+\frac{0.5}{0.866}$=1.732+0.577367 =2.309367$

The answer to the nearest tenth is .

Find the value of

$\sin(270^\circ)+\cos(60^\circ)$ .

Tap to see back →

Before beginning this problem on a calculator, though this is not necessary since these are special angles, ensure that the mode of the calculator is in degrees.

Input the values of the expression and solve.

Find the decimal value of

$2\cos\left($\frac{\pi}{6}\right)$

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To determine the decimal value of the following trig function, $2cos($\frac{\pi}{6}$)$ , make sure that the calculator is in radian mode.

Compute the expression.

$2cos($\frac{\pi}{6}$)=2(0.866)=1.732$

Determine the value of $cot($\frac{\pi}{3}$)$ in decimal form.

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Ensure the calculator is in radian mode since the expression shows the angle in terms of $\pi$ . Also convert cotangent to tangent.

$cot($\frac{\pi}{3}$)=\frac{1}{tan(\frac{\pi}${3})}= $\frac{1}{\sqrt3}$=0.577$

Determine the correct value of

.

Tap to see back →

The question $sin($\frac{1}{2}$)$ asks for the y-coordinate on the unit circle when the degree angle is $\frac{1}{2}$ .

Be careful not to confuse finding the value of the angle when the y-value of the coordinate of the unit circle is $\frac{1}{2}$ .

Ensure that the calculator is in degree mode.

$sin($\frac{1}{2}$)=0.00872=0.009$

What is the value of $\theta$ (in degrees)?

Tap to see back →

One can setup the relationship

$\tan(\theta) = $\frac{opposite}{adjacent}$\ tan(\theta)=\frac{9}{5}$$ .

After taking the arctangent,

the arctangent cancels out the tangent and we are left with the value of $\theta$ .

$\ $\theta=tan^{-1}$\left($\frac{9}{5}$ \right ) \ \theta=60.945^\circ \ \theta=61^\circ$

The above triangle is a right triangle. Find the value of $\theta$ (in degrees).

Tap to see back →

One can setup the relationship

$cos(\theta)=\frac{adjacent}{hypotenuse}$$

$cos(\theta)=\frac{6}{11}$$ .

After taking the arccosine,

the arccosine cancels out the cosine leaving just the value of $\theta$ .

$\ $\theta=cos^{-1}$\left($\frac{6}{11}$ \right ) \ \theta= 56.944^\circ \ \theta= 57^\circ$