Answer:
Part 1) [tex]sin(2x)=8\frac{\sqrt{5}}{81}[/tex]
Part 2) [tex]cos(2x)=\frac{79}{81}[/tex]
Part 3) [tex]tan(2x)=8\frac{\sqrt{5}}{79}[/tex]
Step-by-step explanation:
Part 1) Find sin(2x)
we know that
[tex]sin(2x)=2sin(x)cos(x)[/tex]
we have
[tex]sin(x)=\frac{1}{9}[/tex]
Find cos(x)
Remember that
[tex]sin^{2}(x)+cos^{2}(x)=1[/tex]
substitute
[tex](\frac{1}{9})^{2}+cos^{2}(x)=1[/tex]
[tex]cos^{2}(x)=1-(\frac{1}{9})^{2}[/tex]
[tex]cos^{2}(x)=1-(\frac{1}{81})[/tex]
[tex]cos^{2}(x)=\frac{80}{81}[/tex]
[tex]cos(x)=\frac{\sqrt{80}}{9}[/tex]
[tex]cos(x)=4\frac{\sqrt{5}}{9}[/tex] -----> is positive because angle x belong to the I quadrant
Find sin(2x)
we have
[tex]sin(x)=\frac{1}{9}[/tex]
[tex]cos(x)=4\frac{\sqrt{5}}{9}[/tex]
so
[tex]sin(2x)=2sin(x)cos(x)[/tex]
[tex]sin(2x)=2(\frac{1}{9})(4\frac{\sqrt{5}}{9})[/tex]
[tex]sin(2x)=8\frac{\sqrt{5}}{81}[/tex]
Part 2) Find cos(2x)
we know that
[tex]cos(2x)=cos^{2}(x)-sin^{2}(x)[/tex]
we have
[tex]sin(x)=\frac{1}{9}[/tex]
[tex]cos(x)=4\frac{\sqrt{5}}{9}[/tex]
so
[tex]cos(2x)=(4\frac{\sqrt{5}}{9})^{2}-(\frac{1}{9})^{2}[/tex]
[tex]cos(2x)=\frac{80}{81}-\frac{1}{81}[/tex]
[tex]cos(2x)=\frac{79}{81}[/tex]
Part 3) Find tan(2x)
we know that
[tex]tan(2x)=\frac{sin(2x)}{cos(2x)}[/tex]
we have
[tex]sin(2x)=8\frac{\sqrt{5}}{81}[/tex]
[tex]cos(2x)=\frac{79}{81}[/tex]
so
[tex]tan(2x)=\frac{8\frac{\sqrt{5}}{81}}{\frac{79}{81}}[/tex]
[tex]tan(2x)=8\frac{\sqrt{5}}{79}[/tex]
