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

Question

Find the remaining five trigonometric functions of θ.

cos θ = 1/5, θ in quadrant I

Accepted Answer

Hey, everyone in this problem, we're asked to determine the exact values of the other five circular functions of theta. We're told that cosine of theta is equal to negative three divided by seven and that theta is in quadrant two. Now, when we see circular functions, that's just another word for trigonometric functions. OK. So we're looking for those other trigonometric functions sine of data, tangent of data co C can of data, C can of data and cotangent of data. And that's what we're looking for here. Now, let's think about what we were given. So we were given cosine of the and recall that cosine of theta is equal to X divided by R when we're talking about our coordinates. And so we have that cosine of theta is negative three divided by seven. Now R is always positive, which tells us that this X value must be negative three. OK. Another way we can think about this is to think about the quadrant one. We're in quadrant two. So we draw our axis, OK. Quadrant two is in the top left. And if we think about labeling points in that quadrant, they're gonna have negative X values because they're to the left of the Y AX. OK. So that's another way we can think about it, which tells us that that X value is going to be negative. And so what we have is that X can be written as negative three, then R can be written as positive set. OK? That gives us that correct cosine ratio. Now, if we know the X value and we know the R value, the only other thing we need in order to write all of the values for all of the functions is the Y value. And recall that X squared plus Y squared is equal to R squared. And we can use the Pythagorean theorem to relate all three of these values. We know X, we know R we can go ahead and solve for Y. So let's substitute in our values negative three squared plus Y squared is going to be equal to seven squared, negative three squared is nine. So nine plus Y squared is going to be equal to 49. We're gonna subtract nine from both sides. We get that Y squared is equal to 40. And finally, we can take the square root yet that Y is going to be equal to plus or minus the square root of 40. Now, this is really, really, really important when we take a square root, we get both the positive and negative roots. OK? We don't want to forget about either of them, but we want to think about the context of our problem and figure out which one makes sense here. So if we go back to the little diagram we drew, we're in quadrant two, Q two is the top left. We are above the X axis. So the Y values are positive. OK. So we are in quadrant two which tells us that Y is positive. So we are gonna take the positive route there. We get, why is this square root of 40? And we can simplify this a little bit 40 can be written as 10 multiplied by four. We know that the square root of four is two. So we can write this as two multiplied by the square root of 10. And now we have our X value ry value and our R value, we can go ahead and write down all of those trigonometric functions. So let's get ourselves some more room and I am just gonna rewrite X and R. We have the Y value here. So X recall was negative three. R was said, OK. So let's go ahead now and write these functions. Now sine of the recall is going to be Y divided by R. So this is gonna be two multiplied by the square root of 10 divided by seven. And there's no simplifying that we can do there. So we have sign of the cosign of data we were given in the problem. So no need to do that one. Tangent of theta. It's going to be Y divided by X or alternatively, tangent of theta is sine theta divided by cosine theta. So we could work at it like that as well. Kry value is two multiplied by the square root of 10. We're dividing by the X value of negative three. And so tangent of the is going to be two root 10 divided by negative three. And now we're into our reciprocal trig ratio. So Koc cant of the I recall this is gonna be one divided by sin of data. Sign of data is two or 10 divided by seven. And so Koi can is going to be seven divided by two route 10. We're gonna multiply numerator and denominator by the square root of 10. And we don't want that square root in the denominator. And so we can write this as seven multiplied by the square root of 10 divided by 20. That's KC cant of the, we can move to C cant of data ac can of theta is one divided by cosine of theta, cosine of theta was negative three divided by seven. So cosine of theta is going to be negative seven divided by three and finally cotangent of data, the reciprocal of tangent of the. So this is going to be negative three divided by 210. And just like we did for Koc can the, we're gonna multiply the numerator and denominator by the square root of 10. We don't wanna leave that in the denominator. So we get that negative three multiplied by the square root of 10 divided by 20. Once we rationalize that. OK. All right. So those are the values of the other five trigonometric ratios. We were given cosine, we found that sine of theta is two multiplied by route 10 divided by seven tangent of theta is two multiplied by route 10 divided by negative three. Koc can of theta is seven multiplied by the square root of 10 divided by 20 C Can of theta is negative seven divided by three. And code tangent of theta is negative three multiplied by the square root of 10 divided by 20. 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 = 1/5, θ in quadrant I

Key Concepts

Pythagorean Identity

Signs of Trigonometric Functions in Quadrants

Definitions of Trigonometric Functions

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