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

Question

Find exact values of the six trigonometric functions of each angle. Rationalize denominators when applicable. See Examples 2, 3, and 5.1305°

Accepted Answer

Hello, today we're going to be identifying the six trigonometric functions for the given angle. We're going to be providing the exact values of these functions. And when necessary, we're going to be rationalizing the denominator. So the angle that is given to us is positive degrees. One thing to note about this angle is that it lies outside the first rotation of the unit circle. Normally, when we are working with problems like this, it'll be easier to work with an angle that lies within the first rotation of the unit circle, which will be between the region of zero and 360 degrees. So if we want to rewrite or find an equivalent angle to 1290 what we can do is we can take our current angle and subtract 360 degrees from this value. So 1290 minus 360 will give us the value of 930 degrees. This is still outside or beyond 360 degrees. So we're going to take 930 degrees and subtract another 360 degrees from this value doing so will give us the value of 570 degrees. Again, we're going to take 570 degrees and subtract degrees. Doing so will give us the value of 210 degrees. So what this means is that the angle, 1290 degrees is equivalent to the angle of 210 degrees on the unit circle. So let's first start by identifying the location of 210 degrees on the unit circle. Now, 210 degrees can be found in the third quadrant of the unit circle. And the corresponding terminal point at this angle is given to us as the negative square root three divided by two comma negative one divided by two. So now that we have the location of the angle of 1290 degrees and we have the corresponding terminal point, we can use this information to find the six trigonometric functions. Let's go ahead and start by identifying sine cosine and tangent. Now recall that any terminal point on the unit circle is identified as cosine comma S sign. This is where cosine is the X values of the terminal points and sine is the Y value of the terminal points. So that means if we want to start by identifying sine of 1290 we just need to identify the Y value of the terminal point. Now again, the Y value of the terminal point is negative one divided by two. So what this means is that sign of 1290 is equal to negative one divided by two. This is going to be the value of our first trigonometric function. Now let's go ahead and identify cosine of 1290 degrees. Now again, cosine is the X values of the terminal points. The current X value of the terminal point is negative square root three divided by two. So what this means is that cosine of 1290 degrees is equal to negative square root three divided by two. This will be the value of the second trigonometric function. Now let's go ahead and identify the value for tangent. Now recall that the terminal points are identified using cosine and sine. They don't use tangent. So how can we find the value of tangent of 1290 degrees will recall that the identity for tangent of data is equal to the Y value divided by the X value. So in order to get tangent of degrees, we're going to be taking the Y value of the terminal point and dividing it by the X value of the terminal point. Now again, the Y value is negative one divided by two. We're going to take this value and divide it by the X value which is negative square root three divided by two. So what this leaves us is a complex fraction. In order to simplify the complex fraction, we're going to multiply the denominator by its reciprocal which will be negative two divided by the square root of three. And since we are multiplying the denominator by this value, we must also multiply the numerator by this value. So negative square root three divided by two multiplied by negative two, divided by square root of three will reduce the denominator to just one. In the numerator. We can use cross reduction to simplify both of the twos to just one. This will leave us with negative one multiplied by negative one divided by the square root of three which leaves off us leave us with positive one divided by the square root of three. So currently tangent of degrees is equal to one divided by the square root of three. Now we have a square root in the denominator. We're going to need to rationalize this value. So we're going to multiply the numerator and denominator by the square root of three. And this will allow us to rewrite the value of tangent as the squared of three divided by three. This is going to be the value for tangent of degrees. So now that we have the values of sine cosine and tangent. Let's go ahead and identify the values for co sequent sequent and cotangent. Now, in order to get the value of sequent recall, that the co sequent of data is equal to the inverse of sine theta. So in order to identify co sequent of degrees, we're going to have to take the inverse of the current sine value. Now, earlier, we identified that sine is equal to negative one divided by two. If we take the inverse of that value, that would mean that sequence of 1290 degrees is equal to negative two divided by one. And this value can be rewritten as just negative two. So this is going to be the value of our fourth trigonometric function. Now let's go ahead and identify the value for second. Now recall that seeking of data is equal to the inverse of cosine theta. So if we want second of 1290 degrees, we're going to have to take the inverse of the current cosine value. The cosine value was identified to be negative square with three divided by two. And the inverse of this value will be negative two divided by the square root of three. We're going to have to rationalize this value since we have a square root in the denominator. So we're going to multiply the numerator and denominator by the square root of three. This will allow us to rewrite second of degrees as negative two multiplied by the square root of three, that quantity divided by three. This will be the value of our fifth trigonometric function. And finally, we're going to identify our cotangent value. Now, in order to get our cotangent value, recall that cotangent of data is equal to the inverse of tangent data. So in order to get co tangent, we're going to take the inverse of the current tangent value. Now recall that the current tangent value is equal to the square root three divided by three. So that means that the cotangent value will equal to the inverse, which will be three divided by the square root of three. We're going to have to rationalize this quantity. So we're going to multiply the numerator and denominator by the square root of three. This will allow us to rewrite the code tangent value as three multiplied by the square of three divided by three. And since we have a three in both the numerator and denominator, we can reduce these values to just one. And that will leave us with cotangent of degrees as just the square root of three. And this will be the final trigonometric value that we are looking for. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video

Find exact values of the six trigonometric functions of each angle. Rationalize denominators when applicable. See Examples 2, 3, and 5.1305°

Key Concepts

Unit Circle

Reference Angles

Rationalizing Denominators

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