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

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Answer 1 · 2018-03-20T01:44:45

Recall that $sec x = \frac{1}{cos x}$ $secx= 1/cosx$ . Therefore

$1 - (\frac{1}{cos x}) {cos}^{3} x = {sin}^{2} x$ $1 - (1/cosx)cos^3x= sin^2x$

$1 - {cos}^{2} x = {sin}^{2} x$ $1 - cos^2x = sin^2x$

And since ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x =1$ , then ${sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x = 1- cos^2x$ .

${sin}^{2} x = {sin}^{2} x$ $sin^2x= sin^2x$

$L H S = R H S$ $LHS = RHS$

Hopefully this helps!

Answer 2 · 2018-03-20T01:46:08

Pythagorean Identity:
$1 - {cos}^{2} x = {sin}^{2} x$ $1-cos^2x= sin^2x$
And reciprocal identity:
$\frac{1}{sec x} = cos x$ $1/secx=cosx$

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 · 2018-03-20T01:48:24

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!