Question

(1-cos²θ)(1+cos²θ)=2sin²θ-sin⁴θ. Need to show this is an identity.

Answer 1

\[\sin^2\theta+\cos^2\theta=1\]

Answer 2

use this wherever you see 1 then simplify it till you get your right hand side equation

Answer 3

best bet may be to write first as \[1-cos^4(\theta)\] then replace \[cos^4(\theta)\] by \[(1-sin^2(\theta))^2\]
then multiply out and i believe this gives it to you.

Answer 4

will work it out if you like. it is algebra from there on in.

Answer 5

oh my, could you work it out?

Answer 6

sure
i will multiply out \[(1-sin^2(\theta))^2\] but it is just the same as \[(1-x^2)^2\] with x replaced by sine.
\[(1-x^2)^2=1-2x^2+x^4\]
so
\[(1-sin^2(\theta))^2=1-2sin^2(\theta)+sin^4(\theta)\]\]

Answer 7

don't forget that originally you have
\[1-cos^4(\theta)\]
so now you have \[1-(1-2sin^2(\theta)+sin^4(\theta))=2sin^2(\theta)-sin^4(\theta)\]

Answer 8

is that enough detail? if not i let me know.

Answer 9

i don't know why, but i just can't see it

Answer 10

\[(\sin^2\theta+\cos^2\theta-\cos^2\theta)(\sin^2\theta+\cos^2\theta+\cos^2\theta)=(\sin^2\theta)(sina^2\theta+2\cos^2\theta)=\sin^4\theta+2\sin^2\theta(\cos^2\theta)=\sin^4\theta+2(\sin^2\theta)(1-\sin^2\theta)=sina^4\theta+2\sin^\theta-2\sin^4\theta=2\sin^2\theta-\sin^4\theta\]

Answer 11

ok lets to slowly.
first of all, the left hand side of the equation is
\[(1-cos^2(\theta))(1-cos^2(\theta))\]

Answer 12

so our first job is to multiply this out.

Answer 13

is it clear that this is the same as multiplying out
\[(1-x)(1+x)\]?

Answer 14

yes, when you write it like that

Answer 15

\[or rather (x-x^2)(1+x^2)\]

Answer 16

typo sorry

Answer 17

ok so lets multiply out \[(1-cos^2(\theta))(1+cos^2(\theta))\]

Answer 18

\[(1-x^2)(1+x^2)=1-x^4\]
so\[((1-cos^2(\theta))(1+cos^2(\theta))=1-cos^4(\theta)\]
so far so good?

Answer 19

yes

Answer 20

ok now you recall that
\[sin^2(\theta)+cos^2(\theta)=1\]
so \[cos^2(\theta)=1-sin^2(\theta)\]

Answer 21

and of course \[cos^4(\theta)=(cos^2(\theta))^2\] so replace
\[cos^2(\theta)\]by \[(1-sin^2(\theta))\] in this expression to get
\[1-(1-sin^2(\theta))^2\]

Answer 22

ok?

Answer 23

oh, ok

Answer 24

now multiply out and you get your answer exactly. this is like
\[1-(1-x^2)^2=1-(1-2x^2+x^4)=2x^2-x^4\]
with x replaced by sine.

Answer 25

ooohhh

Answer 26

clear or no?

Answer 27

it is when you replace the sin/cos with the 'x'. easier to see

Answer 28

yes of course. the only little bit of trig here was replacing \[cos^2(\theta)\]by
\[(1-sin^2(\theta)\]
every other step was algebra. multiply, collect like terms etc.

Answer 29

thanks....

Answer 30

wanna do the next one?

Answer 31

i'm not sure where to start