Find the exact values of the six trigonometric functions of θ

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Solve the following:  

Correct answer:

Find the exact values of the six trigonometric functions of θ

Explanation:

Rewrite  in terms of sine and cosine functions.

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.

Find the exact values of the six trigonometric functions of θ

Find the value of .

Correct answer:

Explanation:

Using trigonometric relationships, one can set up the equation

.

Solving for ,

Thus, the answer is found to be 29.

Find the exact values of the six trigonometric functions of θ

Find the value of .

Correct answer:

Explanation:

Using trigonometric relationships, one can set up the equation

.

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

.

Solving for ,

.

Thus, the answer is found to be 106.

Find all of the angles that satistfy the following equation:

Correct answer:

OR 

Explanation:

The values of  that fit this equation would be:

 and  

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

.

This is why the answer 

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

Thus the answer would be each  multiple of  and  , which would provide the following equations:

  OR    

Evaluate: 

Correct answer:

Explanation:

To evaluate , break up each term into 3 parts and evaluate each term individually.

Simplify by combining the three terms.

 

What is the value of  ?

Correct answer:

Explanation:

Convert  in terms of sine and cosine.

Since theta is  radians, the value of  is the y-value of the point on the unit circle at  radians, and the value of  corresponds to the x-value at that angle.

The point on the unit circle at  radians is .  

Therefore,  and .  Substitute these values and solve.

Solve:  

Correct answer:

Explanation:

First, solve the value of .  

On the unit circle, the coordinate at  radians is .  The sine value is the y-value, which is .  Substitute this value back into the original problem.

Rationalize the denominator.

Find the exact answer for:  

Correct answer:

Explanation:

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 .

 refers to the y-value of the coordinate at 30 degrees.  The y-value of this special angle is .

 refers to the x-value of the coordinate at 30 degrees.  The x-value is .

Combine the terms to solve .

Find the value of 

.

Correct answer:

Explanation:

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 .

Simplify the following expression:

Correct answer:

Explanation:

Simplify the following expression:

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 

So, we know that 

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

So therefore, our answer must be 1

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What are the six trigonometric functions of θ?

Thus, for each θ, the six ratios are uniquely determined and hence are functions of θ. They are called the trigonometric functions and are designated as the sine, cosine, tangent, cotangent, secant, and cosecant functions, abbreviated sin, cos, tan, cot, sec, and csc, respectively.

How do you find the exact values of the six trigonometric ratios of the angle θ in the triangle?

We have to find the exact values of the six trigonometric ratios of the angle θ in the triangle. Therefore, the exact values of six trigonometric ratios are sin θ = 0.9756, cos θ = 0.2195, tan θ = 4.4444, cosec θ = 1.025, sec θ = 4.5556 and cot θ = 0.2250.

How do you find the value of θ?

Just remember the cosine of an angle is the side adjacent to the angle divided by the hypotenuse of the triangle. In the diagram, the adjacent side is a and the hypotenuse is c , so cosθ=ac . To find θ , you use the arccos function, which has the same relationship to cosine as arcsin has to sine.