Find exact values of the six trigonometric functions of each a...

Question

Find exact values of the six trigonometric functions of each angle. Rationalize denominators when applicable. See Examples 2, 3, and 5. -1860°

Accepted Answer

Welcome back everyone. In this problem, we want to list the six trigonometric function values of the anger negative 2010 degrees. We want to provide exact values and when necessary, rationalize the denominator. Now, if we're trying to find the six trigonometric values for the anger negative 2010 degrees, it's going to prove a bit of a challenge to find the exact value. Instead, what we'll have to do is to try and find a core terminal value for negative 2010 degrees between 0 to degrees for which we can easily find that exact value. What we're really seeing here is that if I were to draw the unit circle on the right here, and let's say that this angle is negative 2010 degrees, we want to find some anger that has the same terminal side as negative 2010 degrees. The best way to do that is to divide negative 2010 by a revolution 360 degrees. And then see how we can get it to the least positive core terminal value. So let's do that. No negative 2010 degrees divided by 360 degrees. When you divide those, you will get approximately negative 5.58. So to find the least positive core terminal angle, we can add six revolutions to negative 2010 degrees. So the least positive co terminal a of negative 2010, 10 degrees. Let me write that properly would be equal to negative 2010 plus six multiplied by 360. Now six multiplied by 360 is negative 2010 plus 2000, 160 equals 150 degrees. So this would be the least positive core terminal angle. In other words, if I adjust my diagram here, oops, I still need my circle. But that's telling us that negative 2010 degrees would correspond or share the same terminal side as 150 degrees in the second quadrant. So they'll have the same exact values. Now, why, why does this help us? Well, again, recall that on the unit circle, OK, on the unit circle, each coordinate is written in the form of the cosine of theater and the sign of theater where the cosine of theater is the X value and the sign of theta is the Y value. So if we can find the cosine and the sign of 150 degrees, it will be the same as the cosine and sine of negative degrees. Now we know those values because recall that the co sign of 150 degrees equals the negative square root of three divided by two and the sign of 150 degrees equals a half. So that tells us, then that would tell us that the core sign of negative 2010 degrees would also be equal to the negative skit of three divided by two. And the sign of negative 2010 degrees would also be equal to a half. So we have the first two of the six trigonometric functions. And we can use these two to figure out the rest of the trigonometric functions. Let's see how now what do you know about the tangent recall that the tangent of theater equals the sign of theater divided by the cosine of theater. So that means then that the tangent of negative 2010 degrees would be equal to the sign of two negative 2010 degrees, which is a half divided by the core sign of that value, which is the negative square root of three divided by two. No, we can go ahead and simplify for that value of the tangent. So we could rewrite this as a half multiplied by a negative two divided by the square of 32. And the negatives two share a common factor of 22 goes into 21 time two goes into negative two negative one times and one multiplied by negative one equals negative 11 multiplied by the square of three equals the square of three. We can simplify even further by rationalizing. So we can multiply both the numerator and the denominator by the square root of three negative one multiplied by the square of three equals negative square of three and squared of three multiplied by the square of three equals three. Therefore, this tells us that the tangent of negative 2010 degrees equals a negative square of three divided by three. Now that we have the values for the cosine, the sign and the tangent, we can find the reciprocal values. Recall, recall that the sequence of an angle is the reciprocal of its sign. So that tells us then that a core S count of negative 2010 degrees would be the reciprocal of sine. Since sine is a half, then the core sequent would have been equal to two. Next, let me draw a line here to separate these. Recall that the second is the reciprocal of the cosine. And since the cosine of negative 2010 is a negative spirit of three divided by two, that means that the sequence of negative 2010 degrees equals two divided by the negative square root of three or negative two divided by the square of three. We can simplify this by rationalizing. That means multiplying the numerator and the denominator by the square of three negative two multiplied by the square root of three equals negative two multiplied by the skirt of three. Well, obviously, and the square of three divide multiplied by the square of three equals three. So that would be the second count of negative 2010 degrees. Finally, we can find the core tangent because it's the reciprocal of the tangent call that the core tangent is the reciprocal of the tangent. So no, that means the core tangent of negative degrees is the reciprocal of negative skirt of three divided by three. So that would be negative three divided by the skirt of three. And again, we can rationalize to simplify negative three, multiplied by the square root of three will be our numerator. And our denominator will be the square of three multiplied by the square of three which equals three. Notice that the numerator and denominator have a common factor of three. So if we factor 03, we will be left with the core tangent of negative 2010 degrees being equal to the negative square root of three. No, we have all our values. Notice that the core tangent of negative 10 degrees equals negative square of three. It's C equals negative two multiplied by the square of three divided by three. Its cos is two, its tangent is negative square root of three divided by three. Its sign is a, a half and its core sign is negative squared of three divided by two. Thanks a lot for watching. Everyone. I hope this video helped.

Find exact values of the six trigonometric functions of each angle. Rationalize denominators when applicable. See Examples 2, 3, and 5. -1860°

Key Concepts

Angle Coterminality and Reduction

Definition of the Six Trigonometric Functions

Rationalizing Denominators

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