Pre-Calculus - Evaluating Trig Functions

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

$\theta =30^{\circ}$

$\theta =45^{\circ}$

$\theta =120^{\circ}$

$\theta =60^{\circ}$

Explanation

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 $30^{\circ}$ .

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

$\theta =30^{\circ}$

$\theta =45^{\circ}$

$\theta =120^{\circ}$

$\theta =60^{\circ}$

Explanation

Find the decimal value of

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

Explanation

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.

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

Find the decimal value of

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

Explanation

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.

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

Solve for all x on the interval $0^{\circ} \leq x \leq 360^{\circ}$

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

$45^{\circ}$ , $135^{\circ}$

$45^{\circ}$ , $90^{\circ}$

$90^{\circ}$ , $135^{\circ}$

$135^{\circ}$ , $225^{\circ}$

Explanation

Solve for all x on the interval $0^{\circ} \leq x \leq 360^{\circ}$

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

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 $45^{\circ}$ -increment angles. In this case, the angles we are looking for are $45^{\circ}$ and $135^{\circ}$ , because those are the two $45^{\circ}$ -increment angles in the first two quadrants.

Now, you might be saying, "what about $90^{\circ}$ ? That is an increment of 45."

While that is true, $sin(90^{\circ})=1$ , not $\frac{\sqrt{2}}{2}$

So our answer is:

$45^{\circ}$ , $135^{\circ}$

Solve for all x on the interval $0^{\circ} \leq x \leq 360^{\circ}$

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

$45^{\circ}$ , $135^{\circ}$

$45^{\circ}$ , $90^{\circ}$

$90^{\circ}$ , $135^{\circ}$

$135^{\circ}$ , $225^{\circ}$

Explanation

Solve for all x on the interval $0^{\circ} \leq x \leq 360^{\circ}$

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

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 $\frac{\sqrt{2}}{2}$ .

Now, you might be saying, "what about $90^{\circ}$ ? That is an increment of 45."

While that is true, $sin(90^{\circ})=1$ , not $\frac{\sqrt{2}}{2}$

So our answer is:

$45^{\circ}$ , $135^{\circ}$

Solve for $\theta$ :

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

$\frac{5 \pi }{9}$ or $\frac{4 \pi }{9 }$

$5 \pi$ or $4 \pi$

$\frac{5 \pi }{9}$ or $\frac{\pi }{9 }$

$\frac{ \pi }{9}$ or $\frac{2 \pi }{9}$

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

Explanation

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:

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

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

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

$61^\circ$

$42^\circ$

$26^\circ$

$70^\circ$

$57^\circ$

Explanation

One can setup the relationship

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

After taking the arctangent,

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

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

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

Solve for $\theta$ :

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

$\frac{5 \pi }{9}$ or $\frac{4 \pi }{9 }$

$5 \pi$ or $4 \pi$

$\frac{5 \pi }{9}$ or $\frac{\pi }{9 }$

$\frac{ \pi }{9}$ or $\frac{2 \pi }{9}$

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

Explanation

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:

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

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

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

$61^\circ$

$42^\circ$

$26^\circ$

$70^\circ$

$57^\circ$

Explanation

One can setup the relationship

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

After taking the arctangent,

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

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

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

Evaluating Trig Functions

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