Answer:
[tex]\tan(x+y) = 3.73[/tex]
Explanation:
The missing part of the question are:
[tex]\sin(x) = \frac{1}{2}[/tex]
[tex]\cos(y) = \frac{\sqrt 2}{2}[/tex]
Required
[tex]\tan(x + y)[/tex]
First, we calculate [tex]\sin(y)[/tex] and [tex]\cos(x)[/tex]
We have:
[tex]\sin^2(x) + \cos^2(x) = 1[/tex]
So:
[tex](1/2)^2 + \cos^2(x) = 1[/tex]
Collect like terms
[tex]\cos^2(x) = 1 - (1/2)^2[/tex]
[tex]\cos^2(x) = 1 - \frac{1}{4}[/tex]
Take LCM
[tex]\cos^2(x) = \frac{4-1}{4}[/tex]
[tex]\cos^2(x) = \frac{3}{4}[/tex]
Square roots of both sides
[tex]\cos(x) = \frac{\sqrt 3}{2}[/tex]
Similarly,
[tex]\sin^2(y) + \cos^2(y) = 1[/tex]
So:
[tex]\sin^2(y)+(\sqrt 2/2)^2 = 1[/tex]
[tex]\sin^2(y)+ (2/4) = 1[/tex]
[tex]\sin^2(y)+1/2 = 1[/tex]
Collect like terms
[tex]\sin^2(y) = 1 - 1/2[/tex]
Take LCM
[tex]\sin^2(y) = \frac{2 -1}{2}[/tex]
[tex]\sin^2(y) = \frac{1}{2}[/tex]
Square roots of both sides
[tex]\sin(y) = \frac{1}{\sqrt2}[/tex]
Rationalize
[tex]\sin(y) = \frac{\sqrt2}{2}[/tex]
So, we have:
[tex]\sin(x) = \frac{1}{2}[/tex] [tex]\cos(x) = \frac{\sqrt 3}{2}[/tex]
[tex]\cos(y) = \frac{\sqrt 2}{2}[/tex] [tex]\sin(y) = \frac{\sqrt2}{2}[/tex]
[tex]\tan(x) = \sin(x) \div \cos(x)[/tex]
[tex]\tan(x) = \frac{1}{2} \div \frac{\sqrt 3}{2}[/tex]
Rewrite as:
[tex]\tan(x) = \frac{1}{2} * \frac{2}{\sqrt 3}[/tex]
[tex]\tan(x) = \frac{1}{\sqrt 3}[/tex]
Rationalize
[tex]\tan(x) = \frac{\sqrt 3}{3}[/tex]
Similarly
[tex]\tan(y) = \sin(y) \div \cos(y)[/tex]
[tex]\tan(y) = \frac{\sqrt 2}{2} \div \frac{\sqrt 2}{2}[/tex]
[tex]\tan(y) = 1[/tex]
Lastly,
[tex]\tan(x + y)= \frac{\tan(x) + \tan(y)}{1 - \tan(x) \cdot \tan(y)}[/tex]
[tex]\tan(x + y)= \frac{\frac{\sqrt3}{3} + 1}{1 - \frac{\sqrt3}{3} \cdot 1}[/tex]
[tex]\tan(x + y)= \frac{\frac{\sqrt3}{3} + 1}{1 - \frac{\sqrt3}{3}}[/tex]
Combine fractions
[tex]\tan(x + y)= \frac{\frac{\sqrt3+3}{3}}{\frac{3 - \sqrt3}{3}}[/tex]
Cancel out 3
[tex]\tan(x + y)= \frac{\sqrt3+3}{3 - \sqrt3}[/tex]
Using a calculator
[tex]\tan(x+y) = 3.73[/tex]
