Find the remaining five trigonometric functions of θ.
co...

Question

Find the remaining five trigonometric functions of θ.

cos θ = -1/4, sin θ > 0

Accepted Answer

Hey, everyone in this problem, we're asked to determine the exact values of the other five circular functions of the. We're told that cosine of the is equal to negative four divided by 15 and that sine of the is less than zero. So let's start by writing what we've been told, we have cosine of theta and recall that cosine of theta can be written as X divided by R. OK. When we write it in terms of those components or the coordinates and we know that this is equal to negative four divided by 15. No, if cosine is negative, we know that our R value is always positive. OK? Remember that it measures the distance it, that radius. So R is always positive. So this means that the X value must be negative. OK. So we must have that X can be written as negative or an R can be written as 15. All right. Now we have our X and our R value, we're gonna put a red box around those because we're gonna need those to write of these other functions. And when the question says the circular functions circular is another word for trigonometric OK. So we're thinking about those trigonometric ratios sine of the tangent of theta C can't co C can't cotangent, right. All right. So we have our X value, we have our R value. The only other thing we need in order to write all of these ratios out is the Y value. OK. Well, good news. Xr and Y are related to the Pythagorean theorem. They recall that they make up a nice right triangle. So we can write X squared plus Y squared is equal to R squared. Substituting in our values negative four squared plus Y squared is equal to 15 squared. Simplifying 16 plus Y squared is equal to 225 subtracting by 16 on both sides, Y squared is equal to 209. And finally, we take the square root. And one thing that's really, really important when we're dealing with these trigonometric problems to remember is that when we take the square root, we get both the positive and negative root. So we have plus or minus the square root of 209. And that's the result mathematically. And then we have to infer from the problem which of those roots makes sense. Now we're told that sin of the, OK, which is equal to Y divided by R is less than zero. Now, just like we talked about with cosine R is always positive. So if sine is less than zero, that must mean that Y is less than zero. OK. So Y must be less than zero. And so what we're gonna do is we're gonna take why as the negative root 209. OK. All right. So we have our X value, we have our Y value, we have our R value. That's everything we need to write out all of the trigonometric ratios. Now, let's start with the first one sign of data and sign of theta, we've just written this is why divided by R we found our Y value to be negative the square root of 209 divided by the R value which is 50. OK. So s of the negative 209 divided by 15 cosine of data we were given in the problem. So we don't need to worry about that one tangent of the I recall is going to be the Y value divided by the X value. Now, alternatively, we could also write tangent of data as sine the divided by cosine theta and use those sine and cosine values that we've already found to calculate tangent. The, it's gonna work out the exact same way. OK. So using the Y and X values instead, we get negative the square root of 209 divided by negative four, the two negatives are gonna divide out. We're just gonna have the square root of 209 divided by four. All right. So we've got sine cosine tangent Now we're gonna move to Koi Can of theta. OK? Are reciprocal identity and Kosi Can of theta recall is one divided by sine of theta. Well, one divided by sin of theta is going to be negative 15 divided by the square root of 209. Now we wanna rationalize this, we don't want this square root in the denominator. So we're gonna multiply numerator and denominator by the square root of 209. We get that KC can of the is negative 15, multiplied by the square root of 209 all divided by the square root of 209. OK. Next is CC can't C can of theta is going to be one divided by cosine of theta. OK? Cosine of theta is what we were given in the problem negative four divided by 15. And so C can of theta is going to be negative 15 divided by four. And finally, the very last trigonometric ratio we need to deal with is co tangent and cotangent of theta. Recall is the U reciprocal of tangent of theta. So this is going to be equal to four divided by the square root of 209. And just like we did for Kosi Camp, we're gonna rationalize this denominator. We're gonna multiply numerator and denominator by the square root of 209. And we get that code tangent is four, multiplied by the square root of 209 divided by 209. Ok. So those are the other five trigonometric ratios we were given cosine and we've now found sine tangent, cosecant Zant and Cotangent. Thanks everyone for watching. I hope this video helped see you in the next one.

Find the remaining five trigonometric functions of θ.
cos θ = -1/4, sin θ > 0

Key Concepts

Pythagorean Identity

Signs of Trigonometric Functions in Quadrants

Definitions of the Six Trigonometric Functions

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