sin-12-13-and-is-in-the-2nd-quadrent-prove-cos-5-13-

Question Number 71146 by sadimuhmud 136 last updated on 12/Oct/19

\sin α = \frac{12}{13} and α is in the 2 nd quadrent .

prove \cos α = - \frac{5}{13}

Commented by Rio Michael last updated on 12/Oct/19

s i n α = \frac{12}{13}

f r o m {s i n}^{2} α + {c o s}^{2} α = 1

\Rightarrow {c o s}^{2} α = 1 - {s i n}^{2} α

{c o s}^{2} α = 1 - {(\frac{12}{13})}^{2}

c o s α = \pm \sqrt{\frac{25}{169}}

c o s α = - \frac{5}{13}

r e a s o n b e i n g t h a t t h e c o s i n e r a t i o i s n e g a t i v e i n t h e s e c o n d q u a d r a n t .

Answered by Kunal12588 last updated on 12/Oct/19

s i n α = \frac{12}{13}

u s i n g i d e n t i t y : - \sin^{2} θ + \cos^{2} θ = 1

\Rightarrow c o s α = - \frac{\sqrt{13^{2} - 12^{2}}}{13} [∵ \frac{π}{2} < α ⩽ π]

\Rightarrow c o s α = - \frac{\sqrt{169 - 144}}{13} = - \frac{\sqrt{25}}{13} = - \frac{5}{13}