An equation of the terminal side of an angle θ in standard positi...

Question

An equation of the terminal side of an angle θ in standard position is given with a restriction on x. Sketch the least positive such angle θ , and find the values of the six trigonometric functions of θ .  See Example 3.   
                                                                                                                                                                                                                                                                                                                                                                                   
  2x + y = 0 , x ≥ 0

Accepted Answer

Draw a sketch of the least positive angle beta whose equation of the terminal side in center position is given below. Also find the values of sine beta cosine beta and tangent beta where we have three X plus two Y is equals to zero and X is greater than and equals to zero. Russell given a coordinate plane to graph our equation on to solve this. Let's first find our graph. We have three X plus two Y equals zero. Let's go ahead and solve for A Y to find points supply in our graph, we get two Y equals negative three X and we get Y equals negative three halves X. Now we notice that X has to be gradient or equals to zero, which means we can plot a line here on our graph that starts at equal zero. If we were to plug in zero, we get the 0.00. It would keep on going down. We can go down 3/2 and make a new point there. We now have our vector. We also have a point at X equals two Y equals negative three. This can form a triangle for us. Now we can use this triangle to find our sine cosine and tangent. No beta will be ry divided by R. There is R which will be negative three divided by our partners R R. We can solve by taking the square root of two squared plus negative three squared square root four plus nine get gives the square root of 13. So our sine beta is negative three divided by the square root of 13. We can rationalize this by multiplying by the squared of 13 divided by the squared of to get negative three squares of 13 divided by 13. You can now find coon beta by taking X divided by R which will be two divided by the square root of 13 which by rationalizing, we get two squares of 13 divided by 13. Finally, tangent beta will be Y divided by X which we have negative three divided by two. That's just negative three halves. We now have everything we need for our answer. Our angle with start at zero. You can go around plan or clockwise until we get to our vector. We now have our graph and all of our measures being sine beta with negative three squares of 13 divided by 13. I assign beta being two squares of 13 divided by 13 and tangent beta being negative three divided by two. OK. I hope to help you solve the problem. Thank you for watching. Goodbye.

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An equation of the terminal side of an angle θ in standard position is given with a restriction on x. Sketch the least positive such angle θ , and find the values of the six trigonometric functions of θ . See Example 3. 2x + y = 0 , x ≥ 0