Find the value of any of the six trigonometric functions - Pre-Calculus

Card 0 of 84

Question

Determine the value of: $\cot(30^{\circ})$

Answer

To determine the value of , simplify cotangent into sine and cosine.

$cot(30)=\frac{cos(30)}{sin(30)}= \frac{\frac{\sqrt3}{2}}{\frac{1}{2}}= {\frac{\sqrt3}{2}}\cdot{\frac{2}{1}}=\sqrt3$

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Question

In a 3-4-5 right triangle, which of the following is a possible value for $sin(\theta)$ ?

Answer

Sine theta is defined as the leg opposite to the angle over the hypothenuse. Write the definition of sine. The hypotenuse is the longest side of the right triangle, which indicates that 5 should be in the denominator.

$sin(\theta)= \frac{\textup{Opposite leg}}{\textup{Hypothenuse}}$

The ratio of the legs opposite to theta over the hypothenuse can either be $\frac{3}{5}$ or $\frac{4}{5}$ .

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Question

What is the value of $sin\left(\frac{\pi}{4}\right)$ ?

Answer

The sine of an angle corresponds to the y-component of the triangle in the unit circle. The angle $\frac{\pi}{4}$ is a special angle. In the unit circle, the hypotenuse is the radius of the unit circle, which is 1. Since the angle is $\frac{\pi}{4}$ , the triangle is an isosceles right triangle, or a 45-45-90.

Use the Pythagorean Theorem to solve for the leg. Both legs will be equal to each other.

$a=\sqrt\frac{1}{2}$

Rationalize the denominator.

$a=\frac{1}{\sqrt2}\cdot \frac{\sqrt2}{\sqrt2} = \frac{\sqrt2}{2}$

Therefore, $sin(\frac{\pi}{4})= \frac{\sqrt2}{2}$ .

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Question

Evaluate: $sin\left(\frac{\pi}{2}\right)-sin\left(\frac{\pi}{3}\right)+sin\left(\frac{\pi}{4}\right)$

Answer

To evaluate $sin(\frac{\pi}{2})-sin(\frac{\pi}{3})+sin(\frac{\pi}{4})$ , break up each term into 3 parts and evaluate each term individually.

$sin(\frac{\pi}{2})=1$

$sin(\frac{\pi}{3})=\frac{\sqrt3}{2}$

$sin(\frac{\pi}{4})=\frac{\sqrt2}{2}$

Simplify by combining the three terms.

$1-\frac{\sqrt3}{2}+\frac{\sqrt2}{2}= \frac{2-\sqrt3+\sqrt2}{2}$

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Question

What is the value of $tan(\pi)+sin(\pi)$ ?

Answer

Convert $tan(\pi)+sin(\pi)$ in terms of sine and cosine.

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

Since theta is $\pi$ radians, the value of $sin(\pi)$ is the y-value of the point on the unit circle at $\pi$ radians, and the value of $cos(\pi)$ corresponds to the x-value at that angle.

The point on the unit circle at $\pi$ radians is .

Therefore, $sin(\pi)=0$ and $cos(\pi)=-1$ . Substitute these values and solve.

$\frac{sin(\pi)}{cos(\pi)}+sin(\pi)=\frac{0}{-1}+0=0$

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Question

Solve: $\frac{1}{sin(\frac{\pi}{4})}$

Answer

First, solve the value of $sin\left(\frac{\pi}{4}\right)$ .

On the unit circle, the coordinate at $\frac{\pi}{4}$ radians is $\left(\frac{\sqrt2}{2}, \frac{\sqrt2}{2}\right)$ . The sine value is the y-value, which is $\frac{\sqrt2}{2}$ . Substitute this value back into the original problem.

$\frac{1}{sin(\frac{\pi}{4})}= \frac{1}{\frac{\sqrt2}{2}}= \frac{2}{\sqrt2}$

Rationalize the denominator.

$\frac{2}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}= \frac{2\sqrt2}{2}= \sqrt2$

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Question

Simplify the following expression:

$sin(-270^{\circ})$

Answer

Simplify the following expression:

$sin(-270^{\circ})$

Begin by locating the angle on the unit circle. -270 should lie on the same location as 90. We get there by starting at 0 and rotating clockwise $270^{\circ}$

So, we know that

$sin(-270^{\circ})=sin(90^{\circ})$

And since we know that sin refers to y-values, we know that

$sin(90^{\circ})=1$

So therefore, our answer must be 1

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Question

Solve the following:

$cot\left(\frac{\pi}{2}\right)+tan(\pi)-sec\left(\frac{\pi}{3}\right)$

Answer

Rewrite $cot\left(\frac{\pi}{2}\right)+tan(\pi)-sec\left(\frac{\pi}{3}\right)$ in terms of sine and cosine functions.

$\frac{cos(\frac{\pi}{2})}{sin(\frac{\pi}{2})}+\frac{sin(\pi)}{cos(\pi)}-\frac{1}{cos(\frac{\pi}{3})}$

Since these angles are special angles from the unit circle, the values of each term can be determined from the x and y coordinate points at the specified angle.

Solve each term and simplify the expression.

$\frac{0}{1}+\frac{0}{-1}-\frac{1}{\frac{1}{2}}= 0+0-2=-2$

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Question

Find the value of

$\sin(300)$ .

Answer

The value of refers to the y-value of the coordinate that is located in the fourth quadrant.

This angle is also from the origin.

Therefore, we are evaluating .

$sin(-60)=-sin(60)=-\frac{\sqrt3}{2}$

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Question

Find the exact answer for:

Answer

To evaluate , solve each term individually.

refers to the x-value of the coordinate at 60 degrees from the origin. The x-value of this special angle is $\frac{1}{2}$ .

refers to the y-value of the coordinate at 30 degrees. The y-value of this special angle is $\frac{1}{2}$ .

refers to the x-value of the coordinate at 30 degrees. The x-value is $\frac{\sqrt3}{2}$ .

Combine the terms to solve .

$\frac{1}{2}-\frac{1}{2}+\frac{\sqrt3}{2}=\frac{\sqrt3}{2}$

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Question

Find the value of $\csc\theta$ .

$\theta=\frac{11\pi}{6}$

Answer

Since

$\csc\theta=\frac{1}{\sin\theta}$

we begin by finding the value of $\sin\theta$ .

$\sin\frac{11\pi}{6}=-\frac{1}{2}$ .

Then,

$\csc\frac{11\pi}{6}=\frac{1}{-1/2}=-2$

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Question

Compute $\sec(300^\circ)$ , if possible.

Answer

Rewrite the expression in terms of cosine.

$sec(300)=\frac{1}{cos(300)}$

Evaluate the value of , which is in the fourth quadrant.

$cos(300)= \frac{1}{2}$

Substitute it back to the simplified expression of .

$sec(300)=\frac{1}{cos(300)}=\frac{1}{\frac{1}{2}}=2$

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Question

Find the value of .

Answer

Using trigonometric relationships, one can set up the equation

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

$cos(30^\circ)=\frac{25}{x}$ .

Solving for ,

$\x\cdot cos(30^\circ)=25\ \x=\frac{25}{cos(30^\circ)}\ \x=28.8675... \ \x=29$

Thus, the answer is found to be 29.

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Question

Find the value of .

Answer

Using trigonometric relationships, one can set up the equation

$sin(\theta)=\frac{opposite}{hypotenuse}$ .

Plugging in the values given in the picture we get the equation,

$sin(25) = \frac{45}{x}$ .

Solving for ,

$\x\cdot sin(25)=45\ \x=\frac{45}{sin(25)}\ \x=106.479...\ \x=106$ .

Thus, the answer is found to be 106.

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Question

Find the value of $sec(\pi)+cos(-\pi)$ , if possible.

Answer

In order to solve $sec(\pi)+cos(-\pi)$ , split up the expression into 2 parts.

$sec(\pi)= \frac{1}{cos(\pi)}=\frac{1}{-1}=-1$

$cos(-\pi)= -1$

$sec(\pi)+cos(-\pi)= -1-1=-2$

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Question

Choose the answer which is equivalent to the following trig expression:

$Cos(\frac{3\pi}{2})$

Answer

Choose the answer which is equivalent to the following trig expression:

$Cos(\frac{3\pi}{2})$

Begin by finding the location of our given angle. If we start at 0 on the unit circle and go clockwise, every quadrant covers $90^{\circ}/\frac{\pi}{2}$

Therefore, our given angle will correspond to $270^{\circ}$ , which is the bottom half of the y-axis.

Now, because cosine corresponds to the x-value, we know that this expression must be equivalent to 0. If we are on the y-axis, we have no x-value, and therefore, cosine must equal 0

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Question

Which of the following is equivalent to the given expression?

$cot(\frac{3\pi}{2})$

Answer

Which of the following is equivalent to the given expression?

$cot(\frac{3\pi}{2})$

To simplify cotangent expressions, we can think of the expression as tangent and then simply take the reciprocal. So:

$tan(\frac{3\pi}{2})=\frac{1}{0}$ , which is undefined.

So,

$cot(\frac{3\pi}{2})=\frac{0}{1}=0$

Our answer is

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Question

Simplify the following expression:

$\sec(-2\pi)$

Answer

Simplify the following expression:

$sec(-2\pi)$

I would begin here by recalling that secant is the reciprocal of cosine. Therefore, we can take the cosine of the given angle and then find its reciprocal.

So,

$cos(-2\pi)=1$

(Because cosine refers to x-values and $-2\pi$ lies on the x-axis)

Therefore,

$sec(-2\pi)=1$

Because $\frac{1}1{}=1$ .

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Question

Find all of the angles that satistfy the following equation:

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

Answer

The values of $\theta$ that fit this equation would be:

$\frac{\pi}{3}$ and $\frac{2\pi}{3}$

because these angles are in QI and QII where sin is positive and where

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

This is why the answer

$\theta= \frac{\pi k}{3}$

is incorrect, because it includes inputs that provide negative values such as:

$sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt{3}}{2}$

Thus the answer would be each $2\pi$ multiple of $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ , which would provide the following equations:

$\theta= \frac{\pi}{3}+2\pi k$ OR $\theta= \frac{2\pi}{3}+2\pi k$

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Question

Calculate the value of the following trig function:

$sin270^{\circ}$

Answer

Calculate the value of the following trig function:

$sin270^{\circ}$

This is a problem which can be instantly solved by a calculator, provided it isn't in radian mode.

However, we are going to run through how to figure this out without a calculator, because knowing how to do it is far more powerful.

Begin by placing the angle in the unit circle. $270^{\circ}$ is a multiple of $90^{\circ}$ , more specifically, it is three times ninety.

Because we know our angle is three times ninety, we know the angle we are dealing with is one of our four "quadrantal" angles. These are the four that make up our x-y grid.

So, $270^{\circ}$ is the angle directly opposite of $90^{\circ}$ . This means our x-value must be zero, and our y-value must be .

Now, since we know that sine is basically our y-value, the value of $sin270^{\circ}$ must be equal to .

$sin270^{\circ}=-1$

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