plane trigonometry

Submitted by james perez on

trigo2.jpg

can you help me solve ths? thank you

Jhun Vert Thu, 10/08/2015 - 09:00

Prove the identity
$\dfrac{2\sin x - 2\cos x}{\tan x - \cot x} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2(\sin x - \cos x)}{\dfrac{\sin x}{\cos x} - \dfrac{\cos x}{\sin x}} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2(\sin x - \cos x)}{\dfrac{\sin^2 x- \cos^2 x}{\sin x \, \cos x}} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2(\sin x - \cos x)\sin x \, \cos x}{\sin^2 x- \cos^2 x} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2(\sin x - \cos x)\sin x \, \cos x}{(\sin x - \cos x)(\sin x + \cos x)} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{2\sin x \, \cos x}{\sin x + \cos x} = \dfrac{\sin 2x}{\sin x + \cos x}$

$\dfrac{\sin 2x}{\sin x + \cos x} = \dfrac{\sin 2x}{\sin x + \cos x}$

Orion1213 Fri, 10/09/2015 - 00:00

since the given triangle is a right triangle, that is, C=90 degrees... the sum of A + B = C... denotes it... hypotenuse can be solve, one of the legs, either side "a" (opposite to angle A) or side "b" (opposite to angle B) must be known... proceed solving hypotenuse "h" (opposite to angle C=90 deg)... and use either cosine or sine function.