Find the six trigonometric function values for each angle. Ration...

Question

Find the six trigonometric function values for each angle. Rationalize denominators when applicable.

Accepted Answer

Determine the values of the six trigonometric functions for the angle illustrated when necessary rationalize the denominator. And we notice here we have a vector pointing in the direction of the point A the 44. Now to solve this, let's create a triangle, our triangle will be Xynr. Now we have our X which is negative four, we have the Y which is also four. Now we just need to find art R will be the square root of X squared plus Y squared. This gives us part equals the square root of four squared plus negative four square which gives us the square root of 32. If we were to simplify, we get four square roots of two. No, we can go ahead and find our trick functions. We know that sign data as equals two Y divided by R cosine data is equals to X divided by R and tangent data is equals to Y divided by X. For our side, we can then say four divided by four squares of two, which will give us one divided by the square of two was by rationalizing by multiplying by the squared of two divided by the squared of two, we get the squared of two divided by two. As our answer cosine, we would have negative four divided by four squares to two, which we can then see is negative squares to two divided by two. Next, let's go ahead and find our tangent. We know Y is four and our X is negative four which gives us negative one for tangent. Now we have our first three trick functions. We can go ahead and find our reciprocal functions which are S data sin theta and code theta. Whereas sequent is one divided by cosine which that would be one divided by negative squared to two divided by square just two by itself. That gives us two divided by the square of two which is negative which that would just give us negative squares of two. When simplified for co sequent, this is one divided by the side. This is one divided by the square of two divided by two which gives gives us the square of two. Finally code and data will be one divided by 10 data which is one divided by negative one or just negative one. We now have all six trig functions sine which is squared of two divided by two cosine which is negative squared of two divided by two tangent which is negative one CK which is negative squared to two sequent to squared of two and code tangent at negative one. OK. I hope to help you solve the problem. Thank you for watching. Good bye.

Find the six trigonometric function values for each angle. Rationalize denominators when applicable.

Key Concepts

Trigonometric Functions

Unit Circle

Rationalizing Denominators

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