1. ## 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!

2. ## (sin^2 - cos^2)(1) = sin^2 - cos^2

here is the first one

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

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

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

continue more??

3. 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

4. ## the last 2

$sec x - sin x tan x = cos x$

get all terms in $sinx$ and $cosx$

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

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

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

$cosx = cosx$

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

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

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

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

$cos^{2}x = cos^{2}x$

5. Originally Posted by bigwave
$sec x - sin x tan x = cos x$

get all terms in $sinx$ and $cosx$

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

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

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

$cosx = cosx$
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

6. we are adding 2 fractions not mulitplying
the common denominator is $cosx$

7. 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

8. $sin^{4}x - cos^{4}x = 1-2cos^{2}x$

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

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