Simplifying trigonometric expression

(1-4cos^2(t)*sin^2(t))/((cos(t)+sin(t))^…
Please show all the workings

(1-4cos^2 t *sin^2 t)/(cos t + sin t)^2
=[1-(2cos t sin t)^2]/(cos^2 t + sin^2 t + 2sin t cos t)
=[1-sin^2 2t]/(1 + sin 2t)
=[(1+sin 2t) (1-sin 2t)]/(1+sin 2t)
=(1-sin 2t)

Hi

Your expression is not complete, so i will not be able to solve your problem completely, but i can help you with part of your problem concerning ' (1-4cos^2(t)*sin^2(t)) '.

Please note the trigonometry formula which stares:

>>>> sin( 4t ) = sin ( 2*(2t) ) = 2*sin(2t)cos(2t)

So, your first part of the expression can be written as:

>>>> (1 - 4cos^2(t)*sin^2(t)) = (1 - [ 2cos(2t)sin(2t) ]^2 ]

......................................… (1 - [ sin( 4t ) ]^2 )

Now please note that: >> [ cos( 4t ) ]^2 = 1 - [ sin( 4t ) ]^2 { Formula }

So: >> (1 - [ sin( 4t ) ]^2 ) = [ cos( 4t ) ]^2

Now i will assume that the other expression of your equation be ' [ cos(t) + sin(t) ] ^2 ' which can be solved like this:

First use remarkable identity formula: ( A + B )^2 = A^2 + 2*A*B + B^2

So: [ cos(t) + sin(t) ] ^2 = [ cos(t) ]^2 + 2*sin(t)*cos(t) + [ sin(t) ] ^2

Please remember that as formulas you have:

1 ) [ cos(t) ]^2 + [ sin(t) ] ^2 = 1

2 ) sin(2t) = 2*sin(t)*cos(t)

So: [ cos(t) + sin(t) ] ^2 = [ cos(t) ]^2 + 2*sin(t)*cos(t) + [ sin(t) ] ^2

...................................= 1 + sin(2t)

Hope that helps , Good Luck

Please do not forget to rate my answer.

(1-4cos^2(t)*sin^2(t))/((cos(t)+sin(t))^… ---> what after ... ??

incomplete question

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