Try this solution:
Part A: note, that tanx=1/ contanx. Using this property and formula 1+contan²x=1/ sin²x, it is possible to find sinx:
[tex] sinx= \sqrt { \frac{1}{1+ctg^2x}}=\sqrt{\frac{1}{1+\frac{144}{25}}}= \frac{5}{13} [/tex]
Part B: if tanx=5/12, then using the formula 1+tan²x=1/cos²x it is possible to find cosx:
[tex] cosx=\sqrt{\frac{1}{1+tan^2x}}=\sqrt{\frac{1}{1+\frac{25}{144}}}=\frac{12}{13}. [/tex]
Part C: note, that sin(x+2pi/3)=sinx*cos2pi/3+sin2pi/3*cosx=-0.5sinx+√3/2cosx, where sinx=5/13 and cosx=12/13.
[tex] sin(x+\frac{2 \pi}{3})=\frac{\sqrt{3}}{2} *\frac{12}{13}-\frac{1}{2} *\frac{5}{13}=\frac{12\sqrt{3}-5}{26} [/tex]
