FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative
Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached
Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13
cos α = side adjacent to the angle / hypotenuse
cos α = -5/13
Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°
[tex]cos \beta = -\frac{1}{2} \sqrt{3} [/tex]
[tex]tan \beta= \frac{1}{3} \sqrt{3} [/tex]
SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
[tex]sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} ) [/tex]
[tex]sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )[/tex]
[tex]sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})[/tex]
Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
[tex]cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )[/tex]
[tex]cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )[/tex]
[tex]cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )[/tex]
Find tan (α - β)
[tex]tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }[/tex]
[tex]tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}[/tex]
Simplify the denominator
[tex]tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}[/tex]
[tex]tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }[/tex]
Simplify the numerator
[tex]tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }[/tex]
[tex]tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }[/tex]
Simplify the fraction
[tex]tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})[/tex]
[tex]tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}[/tex]
