Question

Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first quadrant and angle y is in the second quadrant. Information provided in the picture. PLEASE HELP

Answer 1

`\cos(x+y)` goes with
`-(√(6)+√(2))/(4)`

`\sin(x+y)` goes with
`(√(6)-√(2))/(4)`

`\tan(x+y)` goes with
`√(3)-2`

Explanation:

`\cos(x+y)`

`\cos(x)\cos(y)-\sin(x)\sin(y)` by the addition identity for cosine.

We are given:

`\sin(x)=(√(2))/(2)` which if we look at the unit circle we should see

`\cos(x)=(√(2))/(2)`.

We are also given:

`\cos(y)=(-1)/(2)` which if we look the unit circle we should see

`\sin(y)=(√(3))/(2)`.

Apply both of these given to:

`\cos(x+y)`

`\cos(x)\cos(y)-\sin(x)\sin(y)` by the addition identity for cosine.

`(√(2))/(2)(-1)/(2)-(√(2))/(2)(√(3))/(2)`

`(-√(2))/(4)-(√(6))/(4)`

`(-√(2)-√(6))/(4)`

`-(√(6)+√(2))/(4)`

Apply both of the givens to:

`\sin(x+y)`

`\sin(x)\cos(y)+\sin(y)\cos(x)` by addition identity for sine.

`(√(2))/(2)(-1)/(2)+(√(3))/(2)(√(2))/(2)`

`(-√(2)+√(6))/(4)`

`(√(6)-√(2))/(4)`

Now I'm going to apply what 2 things we got previously to:

`\tan(x+y)`

`(\sin(x+y))/(\cos(x+y))` by quotient identity for tangent

`(√(6)-√(2))/(-(√(6)+√(2)))`

`-(√(6)-√(2))/(√(6)+√(2))`

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.

`-(√(6)-√(2))/(√(6)+√(2)) \cdot (√(6)-√(2))/(√(6)-√(2))`

`-(6-√(2)√(6)-√(2)√(6)+2)/(6-2)`

`-(8-2√(12))/(4)`

There is a perfect square in 12, 4.

`-(8-2√(4)√(3))/(4)`

`-(8-2(2)√(3))/(4)`

`-(8-4√(3))/(4)`

Divide top and bottom by 4 to reduce fraction:

`-(2-√(3))/(1)`

`-(2-√(3))`

Distribute:

`√(3)-2`