Question

evaluate sin135°cos145°+cos135°sin145°​

Answer 1

$\textbf{Given:}$

$sin\,135^{\circ}\;cos\,145^{\circ}+cos\,135^{\circ}\;sin\,145^{\circ}$

$\textbf{To find:}$

$\text{The value of}$

$sin\,135^{\circ}\;cos\,145^{\circ}+cos\,135^{\circ}\;sin\,145^{\circ}$

$\textbf{Solution:}$

$\text{Consider,}$

$sin\,135^{\circ}\;cos\,145^{\circ}+cos\,135^{\circ}\;sin\,145^{\circ}$

$\text{Using,}$

$\boxed{\bf\,sin(A+B)=sinA\,cosB+cosA\,sinB}$

$=sin(135^{\circ}+145^{\circ})$

$=sin\,280^{\circ}$

$=sin(270^{\circ}+10^{\circ})$

$=-cos\,10^{\circ}$

$\implies\,\bf\,sin\,135^{\circ}\;cos\,145^{\circ}+cos\,135^{\circ}\;sin\,145^{\circ}=-cos\,10^{\circ}$

Answer 2

Answer:

$\sin(135) \cos(145 ) + \cos(135) \sin(145) \\ = \sin(135 + 145) \\ = \sin(280) \\ = \sin(270 + 10) \\ = - \cos(10)$