Answer:
Step-by-step explanation:
Our interval is 180 to 270 so that's quadrant III. In quadrant 3, both x and y are negative. If we set up our right triangle in this quadrant and label it accordingly with the values given for tangent theta, we find that the missing length is the hypotenuse. Solve for that using Pythagorean's Theorem. Doing that gives us that the hypotenuse is [tex]\sqrt{61}[/tex]. Now we can solve for your other identities. First,
[tex]cos(2\theta)=2cos^2\theta-1[/tex] (There are 3 identities for cos(2θ) and regardless of which one you pick, the answer will be the same every time...promise!)
Filling in now:
[tex]cos(2\theta)=2(\frac{-5}{\sqrt{61} })^2-1[/tex] and simplified a bit:
[tex]cos(2\theta)=2(\frac{25}{61})-1[/tex] and a bit more:
[tex]cos(2\theta)=\frac{50}{61}-\frac{61}{61}[/tex] giving us, finally:
[tex]cos(2\theta)=-\frac{11}{61}[/tex]
Now for b. We have an identity for the half angle of sin:
[tex]sin(\frac{\theta}{2})=[/tex] ±[tex]\sqrt{\frac{1-cos\theta}{2} }[/tex] and filling in:
[tex]sin(\frac{\theta}{2})=[/tex] ±[tex]\sqrt{\frac{1-(-\frac{5}{\sqrt{61} } )}{2} }[/tex] which simplifies a bit to:
[tex]sin(\frac{\theta}{2})=[/tex] ±[tex]\sqrt{\frac{1+\frac{5}{\sqrt{61} } }{2} }[/tex] and a bit more to:
[tex]sin(\frac{\theta}{2})=[/tex] ±[tex]\sqrt{\frac{\sqrt{61}+5 }{2\sqrt{61} } }[/tex] I'm not sure if this is the form you need it in, but this is the answer, as convoluted and crazy as it may look.
As for the cosine half-angle, the identity is the same, except there's a + sign in the numerator of the identity instead of a -, giving us the solution:
[tex]cos(\frac{\theta}{2})=[/tex] ±[tex]\sqrt{\frac{\sqrt{61}-5 }{2\sqrt{61} } }[/tex]
