How do I proof this equation is an identity? 1-sec(x)cos³(x)=sin²(x)

3 Answers
Mar 20, 2018

Recall that #secx= 1/cosx#. Therefore

#1 - (1/cosx)cos^3x= sin^2x#

#1 - cos^2x = sin^2x#

And since #sin^2x + cos^2x =1#, then #sin^2x = 1- cos^2x#.

#sin^2x= sin^2x#

#LHS = RHS#

Hopefully this helps!

Mar 20, 2018

Pythagorean Identity:
#1-cos^2x= sin^2x#
And reciprocal identity:
#1/secx=cosx#

Applied to:
#1-sec(x)cos³(x)=sin²(x)#
#1-1/cancel(cosx)*coscancel(³)(x)=sin²(x)#
#1-cos^2x= sin^2x#
#sin^2x= sin^2x#

Mar 20, 2018

Use these three identities (the third one is pretty much the same as the second, but switched around a little):

#sec(x)=1/cos(x)#

#cos^2(x)+sin^2(x)=1#

#=>cos^2(x)=1-sin^2(x)#

To complete the proof, I'll start with the left-hand side of the equation and manipulate it until it equals the right-hand side of the equation:

#LHS=1-sec(x)cos^3(x)#

#color(white)(LHS)=1-sec(x)*cos^3(x)#

#color(white)(LHS)=1-1/cos(x)*cos^3(x)#

#color(white)(LHS)=1-cos^3(x)/cos(x)#

#color(white)(LHS)=1-color(red)cancelcolor(black)(cos^3(x))^(cos^2(x))/color(red)cancelcolor(black)cos(x)#

#color(white)(LHS)=1-cos^2(x)#

#color(white)(LHS)=1-(1-sin^2(x))#

#color(white)(LHS)=1-1+sin^2(x)#

#color(white)(LHS)=color(red)cancelcolor(black)(1-1)+sin^2(x)#

#color(white)(LHS)=sin^2(x)#

#color(white)(LHS)=RHS#

That is the proof. Hope this helped!