## establishing identities with ^3 and ^4 in them

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
bikeman
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### establishing identities with ^3 and ^4 in them

i am asked two similar problems. my question is how it was done as the example in the book i am reading is useless in explaining.
the first question is:
(sin^3t+cos^3t)/(sint+cost)=1-sint cost
the example plays out to what i assume is factoring sint+cost out of it to look like (sint+cost)(sin^2t-sint cost+cos^2)/(sint+cost)

my question is, where did the "sin^2t-sint cost+cos^2" come from? when i factor it it doesnt look like that.

the next problem is (cos^4t-sin^4t)/(cos[2t]) how do i get rid of the ^4?

thanks for any help provided.

Martingale
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### Re: establishing identities with ^3 and ^4 in them

bikeman wrote:i am asked two similar problems. my question is how it was done as the example in the book i am reading is useless in explaining.
the first question is:
(sin^3t+cos^3t)/(sint+cost)=1-sint cost
the example plays out to what i assume is factoring sint+cost out of it to look like (sint+cost)(sin^2t-sint cost+cos^2)/(sint+cost)

my question is, where did the "sin^2t-sint cost+cos^2" come from? when i factor it it doesnt look like that.

the next problem is (cos^4t-sin^4t)/(cos[2t]) how do i get rid of the ^4?

thanks for any help provided.

$a^3+b^3=(a+b)(a^2-ab+b^2)$

$a^4-b^4=(a+b)(a-b)(a^2+b^2)$

stapel_eliz
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bikeman wrote: where did the "sin^2t-sint cost+cos^2" come from? when i factor it it doesnt look like that.

How did you factor? What did you get?

(To review how to factor differences of squares and sums and differences of cubes, try here).

bikeman
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Joined: Wed Jun 09, 2010 11:48 pm
Location: west virginia university
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### Re: establishing identities with ^3 and ^4 in them

I factored wrong.
i found that (cos^4t-sin^4t) is the same as saying (cos^2t+sin^2t)(cos^2t-sin^2t) one of which is a Pythagorean identity. the lower part of the problem, Cos(2t) is part of a series of identities which which cancels one half of the upper part of the equation. meaning you are left is 1 X1 which equals 1