anonymous
  • anonymous
(1-cos²θ)(1+cos²θ)=2sin²θ-sin⁴θ. Need to show this is an identity.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\sin^2\theta+\cos^2\theta=1\]
anonymous
  • anonymous
use this wherever you see 1 then simplify it till you get your right hand side equation
anonymous
  • anonymous
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
will work it out if you like. it is algebra from there on in.
anonymous
  • anonymous
oh my, could you work it out?
anonymous
  • anonymous
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 \[(1-sin^2(\theta))^2=1-2sin^2(\theta)+sin^4(\theta)\]\]
anonymous
  • anonymous
don't forget that originally you have \[1-cos^4(\theta)\] so now you have \[1-(1-2sin^2(\theta)+sin^4(\theta))=2sin^2(\theta)-sin^4(\theta)\]
anonymous
  • anonymous
is that enough detail? if not i let me know.
anonymous
  • anonymous
i don't know why, but i just can't see it
anonymous
  • anonymous
\[(\sin^2\theta+\cos^2\theta-\cos^2\theta)(\sin^2\theta+\cos^2\theta+\cos^2\theta)=(\sin^2\theta)(sina^2\theta+2\cos^2\theta)=\sin^4\theta+2\sin^2\theta(\cos^2\theta)=\sin^4\theta+2(\sin^2\theta)(1-\sin^2\theta)=sina^4\theta+2\sin^\theta-2\sin^4\theta=2\sin^2\theta-\sin^4\theta\]
anonymous
  • anonymous
ok lets to slowly. first of all, the left hand side of the equation is \[(1-cos^2(\theta))(1-cos^2(\theta))\]
anonymous
  • anonymous
so our first job is to multiply this out.
anonymous
  • anonymous
is it clear that this is the same as multiplying out \[(1-x)(1+x)\]?
anonymous
  • anonymous
yes, when you write it like that
anonymous
  • anonymous
\[or rather (x-x^2)(1+x^2)\]
anonymous
  • anonymous
typo sorry
anonymous
  • anonymous
ok so lets multiply out \[(1-cos^2(\theta))(1+cos^2(\theta))\]
anonymous
  • anonymous
\[(1-x^2)(1+x^2)=1-x^4\] so\[((1-cos^2(\theta))(1+cos^2(\theta))=1-cos^4(\theta)\] so far so good?
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok now you recall that \[sin^2(\theta)+cos^2(\theta)=1\] so \[cos^2(\theta)=1-sin^2(\theta)\]
anonymous
  • anonymous
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 \[1-(1-sin^2(\theta))^2\]
anonymous
  • anonymous
ok?
anonymous
  • anonymous
oh, ok
anonymous
  • anonymous
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.
anonymous
  • anonymous
ooohhh
anonymous
  • anonymous
clear or no?
anonymous
  • anonymous
it is when you replace the sin/cos with the 'x'. easier to see
anonymous
  • anonymous
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.
anonymous
  • anonymous
thanks....
anonymous
  • anonymous
wanna do the next one?
anonymous
  • anonymous
i'm not sure where to start

Looking for something else?

Not the answer you are looking for? Search for more explanations.