How do I prove a trig identity??

Use sin2(x)+cos2(x)=1\displaystyle \sin^2(x)+\cos^2(x)=1.
 
(sin(x)+cos(x))(sin(x)+cos(x))(sin(x)+cos(x))=(sin2(x)+2sin(x)cos(x)+cos2(x))(sin(x)+cos(x))=(1+2sin(x)cos(x))(sin(x)+cos(x))\displaystyle \begin{array}{l} \left( {\sin (x) + \cos (x)} \right)\frac{{\left( {\sin (x) + \cos (x)} \right)}}{{\left( {\sin (x) + \cos (x)} \right)}} \\ = \frac{{\left( {\sin ^2 (x) + 2\sin (x)\cos (x) + \cos ^2 (x)} \right)}}{{\left( {\sin (x) + \cos (x)} \right)}} \\ = \frac{{\left( {1 + 2\sin (x)\cos (x)} \right)}}{{\left( {\sin (x) + \cos (x)} \right)}} \\ \end{array}
 
Thanks I never actually saw that. Our teacher always says to start on the most complex side, and so I never even considered the right side of the expression.
 
1+2sinxcosxsinx+cosx=sin2x+cos2x+2sinxcosxsinx+cosx=\displaystyle \frac{1+2\sin x\cos x}{\sin x+\cos x} = \frac{\sin^2x + \cos^2x +2 \sin x\cos x}{\sin x+\cos x}=

(sinx+cosx)(sinx+cosx)sinx+cosx=sinx+cosx\displaystyle \frac{(\sin x + \cos x)(\sin x + \cos x)}{\sin x + \cos x} = \sin x + \cos x
 
Top