If sin ⁡ θ = 17 17 \sin\theta=\frac{\sqrt{17}}{17} sin θ = 17 17 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , find the values of the five other trigonometric functions. Rationalize the denominators if necessary.

Welcome back, everyone. So let's give this problem a try. Here, we are asked if the sine of theta is equal to the square root of 17 over 17. Find the values of the 5 other trigonometric functions and rationalize the denominators if necessary. Okay. So solve this problem, what we first need to do is figure out the 5 trigonometric functions that we're looking for. And I can see that we have the sine of theta, and so what we don't have is the cosine of theta, the tangent of theta, the cosecant of theta, the secant of theta, and then the cotangent of theta. These are all the trigonometric functions we've learned about thus far. And so to find these trigonometric functions, we're going to use the sine of theta to do this, because the sine of theta is given to us. And what I'm going to do is draw a right triangle over here, and this right triangle is going to give us a visual that we can follow to figure out these missing sides. We're gonna say this is our angle theta, and then this is a right triangle. And what I can do is use the memory tool of SOHCAHTOA to figure out what the sides of this triangle would be. Because recall that SOHCAHTOA shows us how we can relate these trigonometric functions sine, cosine, and tangent, to the sides of the right triangle. Now we're given the sine of theta, and so tells us that sine is equal to opposite over hypotenuse. So we're gonna have opposite over hypotenuse, and I can see here that the opposite side of this right triangle is going to be the square root of 17. So if I go opposite to this angle right there, we're going to get square root 17, and then I can see the hypotenuse is 17. And the hypotenuse is always the longest side so that's going to be 17. Now what I need to do from here is figure out what this last missing side is. See that the square root of 17 is a, I'm going to call this other missing side b, and then the hypotenuse is always going to be equal to c. Now what I can do is use the pythagorean theorem a squared plus b squared equals c squared, and this is going to allow me to find this missing side b. Now a squared is going to be the square root of 17 squared. B squared is not a known value, and then c squared is going to be 17 squared. I can simplify this because the square root and the square are gonna cancel each other giving me just 17. And then this is gonna be plus b squared, which we don't know. And then 17 squared is the same thing as 289, Yes. I can move this 17 to the other side of the equation by subtracting 17 on both sides, that'll get the 17s to cancel on the left, leaving me with just b squared. And then we have 289 minus 17, which is 272. Now from here, we'll take the square root on both sides of this equation, canceling the square on the b. And so we'll have b is equal to the square root of 272. And if you simplify the square root, this is actually the same thing as 4 times the square root of 17. So this right here is the simplified answer to the square root, and that's also going to be the missing side of this right triangle. So we'll have 4 times the square root of 17. So notice at this point, we've gone ahead and calculated all sides of this right triangle. Now what I'm gonna do is start with the cosine. And for the cosine beta, we have from SOHCAHTOA that cosine is adjacent over hypotenuse. So we're gonna have adjacent over hypotenuse. Now the adjacent side of this right triangle we can see here is 4 times the square root of 17, And the hypotenuse is just 17. So that's gonna be the cosine. Now as for the tangent, we have from SOHCAHTOA, the tangent is opposite over adjacent. Now I can see the opposite side of this right triangle is the square root of 17, and, and I also gonna write this out here. So we have that tangent is opposite over adjacent, and the opposite side is square root 17, and the adjacent side is gonna be 4 times the square root of 17. And notice here that the square root of 17 will cancel, meaning all we'll be left with is 1 over 4. So that's the tangent. Next, we need to find the cosecant. And what I'm gonna do to find the cosecant is use the reciprocal identity, because cosecant is the same thing as 1 over sine. Now what I can do, since I can see we have one over sine theta, and sine theta is a value that was given to us originally, I can take the square root of 17 over 17, and just flip it. So this is the same thing as 17 over the square root of 17 for the cosecant. And from here, notice that we have a square root in the denominator, and we're told to rationalize the denominator if necessary. So what I'm actually gonna do is multiply the top and bottom of this fraction by the square root of 17, because because that is going to allow me to cancel the square root in the denominator, leaving me with 17 times the square root of 17 over 17. But notice here that these seventeens are going to cancel, leaving me with just the square root of 17, and that's the cosecant. Now Now next, we need to find the secant, which according to the reciprocal identities is the same thing as 1 over cosine theta. Now you can see that we calculated cosine theta to be 4 square root 17 over 17. So what I'm going to do is flip this, because we have one over cosine, to be 17 over 4 times the square root of 17. And what I can do is rationalize this denominator by multiplying the top and bottom of the fraction by the square root of 17. This will get the square roots to cancel, giving us 17 square root 17 over 4 times 17. And here the 17s are going to cancel, leaving us with the square root of 17 over 4, and that's the secant. Now lastly, we need to find the cotangent. And the cotangent of theta is the same thing as one over the tangent theta according to the reciprocal identities. Now we can see here that tangent theta is 1 fourth, so that means that the cotangent of theta is just gonna be 1 fourth flipped, which is just 4 over 1, or simply just 4. So notice how using the SOHCAHTOA memory tool we were able to find the cosine, I'm gonna do this in green here so we're able to find the cosine, we were able to find the tangent, and then using the reciprocal identities that we had, we were able to find the cosecant, the secant, and then the cotangent theta, all from just being given the sine of theta in the initial problem. So that is how you can use these identities and memory tools to solve for all the other trigonometric functions. So hope you found this video helpful. Thanks for watching.