Answer to Question #139925 in Trigonometry for Michael

Question

Determine the values of sin(2a), cos(2a) and tan(2a) , given cos(a) = -4/5 and Pi over 2 ≤ 0 ≤ Pi

Answer 1

$cos(a)=-\frac{4}{5}$

lets find sin(a):

$sin^2(a)+cos^2(a)=1\implies sin(a)=\sqrt{1-cos^2(a)}$

$\sqrt{1-(-\frac{4}{5})^2}=\sqrt{\frac{9}{25}}=\pm\frac{3}{5}$

$\frac{\pi}{2}\leq a \leq \pi$ then $\sin(a)=\frac{3}{5}$

tan(a):

$tan(a)=\frac{sin(a)}{cos(a)}=\frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4}$

using double angle formulas lets find sin(2a), cos(2a), tan(2a):

$sin(2a)=2sin(a)cos(a)=2\cdot\frac{3}{5}(-\frac{4}{5})=-\frac{24}{25}$

$cos(2a)=cos^2(a)-sin^2(a)=(-\frac{4}{5})^2-(\frac{3}{5})^2=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$

$tan(2a)=\frac{sin(2a)}{cos(2a)}=\frac{-\frac{24}{25}}{\frac{7}{25}}=-\frac{24}{7}$