# Trig proofs help!

• Dec 13th 2009, 10:04 PM
steelersgirl18
Trig proofs help!
Hi! I needed help verifying the following proofs! If anyone wants to contribute by solving either of them, I will greatly appreciate it!

Sin^(4) x - cos^(4) x = 1-2cos^(2) x

and:

sec x - sin x • tan x = cos x

and:

{(cos^(2) x)/(1+ sin x)} -1= -sin x

thank you!
• Dec 13th 2009, 10:15 PM
bigwave
(sin^2 - cos^2)(1) = sin^2 - cos^2
here is the first one

$\displaystyle Sin^{4}x - cos^{4} x = 1-2cos^{2} x$

$\displaystyle (sin^2 - cos^2)(sin^2 + cos^2) = sin^2 + cos^2 - 2cos^2$

$\displaystyle (sin^2 - cos^2)(1) = sin^2 - cos^2$

continue more??
• Dec 13th 2009, 10:25 PM
steelersgirl18
Wow! Thank you so much! Yes, can you please show me the other ones?
Also, can you try verifying them by only working on the left side of the equation? I have to do it that way sorry
• Dec 13th 2009, 10:40 PM
bigwave
the last 2
$\displaystyle sec x - sin x tan x = cos x$

get all terms in $\displaystyle sinx$ and $\displaystyle cosx$

$\displaystyle \frac{1}{cosx} - sinx \frac{sinx}{cosx} = cosx$

$\displaystyle \frac{1 - sin^{2}x}{cosx} = cosx$

$\displaystyle \frac{cos^{2}x}{cosx} = cosx$

$\displaystyle cosx = cosx$

--------------------------------------------------------

$\displaystyle \frac{cos^{2}x}{1 + sinx} - 1 = -sinx$

$\displaystyle cos^{2}x = (-sin+1)(1+ sinx)$

or $\displaystyle cos^{2}x = (1-sinx)(1+ sinx) = 1 - sin^{2}x$

$\displaystyle cos^{2}x = cos^{2}x$
• Dec 13th 2009, 10:52 PM
steelersgirl18
Quote:

Originally Posted by bigwave
$\displaystyle sec x - sin x tan x = cos x$

get all terms in $\displaystyle sinx$ and $\displaystyle cosx$

$\displaystyle \frac{1}{cosx} - sinx \frac{sinx}{cosx} = cosx$

$\displaystyle \frac{1 - sin^{2}x}{cosx} = cosx$

$\displaystyle \frac{cos^{2}x}{cosx} = cosx$

$\displaystyle cosx = cosx$

wait a sec.... When you combined the left side to form (1-sin^2x)/(cosx) , aren't you also supposed to multiple the denominators? So would the correct combination be : (1-sin^2x)/(cos^2x) ?

EDIT: nevermind, you're right whoops
• Dec 13th 2009, 10:57 PM
bigwave
we are adding 2 fractions not mulitplying
the common denominator is $\displaystyle cosx$
• Dec 13th 2009, 11:10 PM
steelersgirl18
One laaaast thing ;)
can you complete the first one but this time only work on the left side? Thank you so much, you're very helpful
• Dec 13th 2009, 11:34 PM
bigwave
$\displaystyle sin^{4}x - cos^{4}x = 1-2cos^{2}x$

$\displaystyle (sin^{2}x - cos^{2}x)(sin^{2}x + cos^{2}x) = 1-2cos^{2}x$

$\displaystyle ((1-cos^{2}x) - cos^{2}x)(1) = 1-2cos^{2}x$