Answer:
[tex]\cos(x+y)[/tex] goes with [tex]-\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]
[tex]\sin(x+y)[/tex] goes with [tex]\frac{\sqrt{6}-\sqrt{2}}{4}[/tex]
[tex]\tan(x+y)[/tex] goes with [tex]\sqrt{3}-2[/tex]
Step-by-step explanation:
[tex]\cos(x+y)[/tex]
[tex]\cos(x)\cos(y)-\sin(x)\sin(y)[/tex] by the addition identity for cosine.
We are given:
[tex]\sin(x)=\frac{\sqrt{2}}{2} [/tex] which if we look at the unit circle we should see
[tex]\cos(x)=\frac{\sqrt{2}}{2}[/tex].
We are also given:
[tex]\cos(y)=\frac{-1}{2}[/tex] which if we look the unit circle we should see
[tex]\sin(y)=\frac{\sqrt{3}}{2}[/tex].
Apply both of these given to:
[tex]\cos(x+y)[/tex]
[tex]\cos(x)\cos(y)-\sin(x)\sin(y)[/tex] by the addition identity for cosine.
[tex]\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}[/tex]
[tex]\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}[/tex]
[tex]\frac{-\sqrt{2}-\sqrt{6}}{4}[/tex]
[tex]-\frac{\sqrt{6}+\sqrt{2}}{4}[/tex]
Apply both of the givens to:
[tex]\sin(x+y)[/tex]
[tex]\sin(x)\cos(y)+\sin(y)\cos(x)[/tex] by addition identity for sine.
[tex]\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}[/tex]
[tex]\frac{-\sqrt{2}+\sqrt{6}}{4}[/tex]
[tex]\frac{\sqrt{6}-\sqrt{2}}{4}[/tex]
Now I'm going to apply what 2 things we got previously to:
[tex]\tan(x+y)[/tex]
[tex]\frac{\sin(x+y)}{\cos(x+y)}[/tex] by quotient identity for tangent
[tex]\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}[/tex]
[tex]-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}[/tex]
Multiply top and bottom by bottom's conjugate.
When you multiply conjugates you just have to multiply first and last.
That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.
[tex]-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}[/tex]
[tex]-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}[/tex]
[tex]-\frac{8-2\sqrt{12}}{4}[/tex]
There is a perfect square in 12, 4.
[tex]-\frac{8-2\sqrt{4}\sqrt{3}}{4}[/tex]
[tex]-\frac{8-2(2)\sqrt{3}}{4}[/tex]
[tex]-\frac{8-4\sqrt{3}}{4}[/tex]
Divide top and bottom by 4 to reduce fraction:
[tex]-\frac{2-\sqrt{3}}{1}[/tex]
[tex]-(2-\sqrt{3})[/tex]
Distribute:
[tex]\sqrt{3}-2[/tex]
