Question

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

Answer 1

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!

Answer 2

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`

Answer 3

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!