Question

, , and angle x and angle y are both in the first quadrant.

Answer 1

`\tan(x+y) = 3.73`

Step-by-step explanation:

The missing part of the question are:

`\sin(x) = (1)/(2)`

`\cos(y) = (\sqrt 2)/(2)`

Required

`\tan(x + y)`

First, we calculate
`\sin(y)` and
`\cos(x)`

We have:

`\sin^2(x) + \cos^2(x) = 1`

So:

`(1/2)^2 + \cos^2(x) = 1`

Collect like terms

`\cos^2(x) = 1 - (1/2)^2`

`\cos^2(x) = 1 - (1)/(4)`

Take LCM

`\cos^2(x) = (4-1)/(4)`

`\cos^2(x) = (3)/(4)`

Square roots of both sides

`\cos(x) = (\sqrt 3)/(2)`

Similarly,

`\sin^2(y) + \cos^2(y) = 1`

So:

`\sin^2(y)+(\sqrt 2/2)^2 = 1`

`\sin^2(y)+ (2/4) = 1`

`\sin^2(y)+1/2 = 1`

Collect like terms

`\sin^2(y) = 1 - 1/2`

Take LCM

`\sin^2(y) = (2 -1)/(2)`

`\sin^2(y) = (1)/(2)`

Square roots of both sides

`\sin(y) = (1)/(\sqrt2)`

Rationalize

`\sin(y) = (\sqrt2)/(2)`

So, we have:

`\sin(x) = (1)/(2)`
`\cos(x) = (\sqrt 3)/(2)`

`\cos(y) = (\sqrt 2)/(2)`
`\sin(y) = (\sqrt2)/(2)`

`\tan(x) = \sin(x) / \cos(x)`

`\tan(x) = (1)/(2) / (\sqrt 3)/(2)`

Rewrite as:

`\tan(x) = (1)/(2) * (2)/(\sqrt 3)`

`\tan(x) = (1)/(\sqrt 3)`

Rationalize

`\tan(x) = (\sqrt 3)/(3)`

Similarly

`\tan(y) = \sin(y) / \cos(y)`

`\tan(y) = (\sqrt 2)/(2) / (\sqrt 2)/(2)`

`\tan(y) = 1`

Lastly,

`\tan(x + y)= (\tan(x) + \tan(y))/(1 - \tan(x) \cdot \tan(y))`

`\tan(x + y)= ((\sqrt3)/(3) + 1)/(1 - (\sqrt3)/(3) \cdot 1)`

`\tan(x + y)= ((\sqrt3)/(3) + 1)/(1 - (\sqrt3)/(3))`

Combine fractions

`\tan(x + y)= ((\sqrt3+3)/(3))/((3 - \sqrt3)/(3))`

Cancel out 3

`\tan(x + y)= (\sqrt3+3)/(3 - \sqrt3)`

Using a calculator

`\tan(x+y) = 3.73`