Give all six trigonometric function values for each angle θ.&n...

Question

Give all six trigonometric function values for each angle θ. Rationalize denominators when applicable. See Examples 5–7.

sin θ = √2/6 , and cos θ < 0

Accepted Answer

Welcome back. I am so glad you're here. We are asked to enumerate the value of the other five trigonometric functions of M given the following trigonometric function and condition rationalize the denominators. If necessary, we are told that the sine of M is equal to the square root of three divided by six and the cosine of M is less than zero. So let's start off by figuring out which quadrant we're in. If we draw a quick sketch of our four quadrants, we recall the acronym from previous lessons. All students take calculus, which means that in quadrant one, all of our trigonometric functions are positive in quadrant two sign and its reciprocal function K can't are positive in the third quadrant tangent and its reciprocal function cotangent are positive. And in the fourth quadrant, cosine and its reciprocal functions are positive. So here we are given a sine function that's equal to a positive value. That's true in both the first quadrant and the second quadrant, we're also told that our cosine function is negative, it's less than zero cosine is negative in the second and third quadrants. So where are both conditions met? That's in quadrant two. So we're talking about quadrant two. Now, the next thing we're going to do is discern our X Y and our values given those three things, it's pretty easy to figure out the other five trigonometric functions of M. So looking at our sine function, we know that the sine function is Y divided by R. So here are Y is our numerator, Y is equal to the square root of three and R is our denominator R is equal to six. Now, we just need to figure out the X value we recall from previous lessons that we can find X because we know that R squared equals X squared plus Y squared. We know R and Y so we can figure out X R is six. So we have six squared equals X squared plus Y is the square root of three squared. Six squared is 36 equals X squared plus the square root of three squared is three. Subtracting three from both sides on the left, we have 33 equals on the right X squared. Taking the square root of both sides. We have X equals plus or minus the square root of 33. However, we're in quadrant two and in quadrant two, our X values are negative. So we can disregard that positive value and we know that X is only equal to negative square root 33 here. So now that we know Xy and R, we can fill in those other five trigonometric functions. Let's start with cosine the cosine of an angle. In this case, M is equal to X divided by R X is negative square at 33. That's divided by R which is six. And that one is as simplified as it gets. It's nice to have a simple one because pretty soon we're going to be rationalizing quite a bit. All right, our tangents, the tangent of an angle. In this case, M is Y divided by X Y. Here is the square root of three that's divided by X which is negative square root 33. We need to rationalize that. So we're going to multiply both the numerator and the denominator by the square rut of 33. So in the numerator, we've got the square root of, we can bring that negative up from the denominator. So we all have a negative square root of three multiplied by square root of 33. But keeping that all inside of the radical, we can put three multiplied by three, multiplied by 1133 is three multiplied by 11. That's divided by square root of 33 multiplied by itself is just 33. Simplifying in the numerator, those threes can come out of the radical. So we have a negative three square root 11 divided by 33. But that can simplify the three. And the 33 simplifies to be in the numerator, negative square root 11 divided by, in the denominator 11. So the tangent here is negative square root 11 divided by 11. OK. Cotangent of M we know that the cotangent is X divided by Y X is negative square root 33 divided by Y is square root three. Again, to rationalize multiplying both the numerator and the denominator by the square root of three. So very similarly, in the numerator, that's a negative square root of three multiplied by 11 multiplied by three, which is our 33 multiplied by three in the radical divided by the square root of three multiplied by itself is just three. Simplifying in the numerator, we can bring out the three. So we have negative three square root 11 divided by three, those threes cancel out. So this is just negative square root 11. There's our cotangent of M, all right. Next up, we've got the C can't of M C can't is going to be R divided by X R is six in the numerator divided by X is negative square root 33. We do need to rationalize that. So we're going to multiply both the numerator and the denominator by the square rut of 33 in the numerator. That's six square root 33 we can bring that negative up from the denominator that's divided by the square root of 33 multiplied by itself is 33. This can simplify the six and the 33 are both divisible by three. So this would be a negative two square root 33 divided by 11. So the C can of M is negative two square root 33 divided by 11. Now, the last one, the Koi can't, the kant of an angle is R divided by Y. Here R is six divided by Y is square root three. We need to rationalize multiplying both the numerator and the denominator by the square of three in the numerator. That's six square at three divided by the denominator square root three, multiplied by itself is three. And this simplifies six divided by three is two. So we have two square root three and that is our Koi Can of M and there are our other five trigonometric functions of M. Well done. We'll catch you on the next one.

Give all six trigonometric function values for each angle θ. Rationalize denominators when applicable. See Examples 5–7.
sin θ = √2/6 , and cos θ < 0

Key Concepts

Definition of the Six Trigonometric Functions

Using the Pythagorean Identity to Find Missing Values

Sign Determination Based on Quadrants

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