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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    \[\sin^2\theta+\cos^2\theta=1\]

  2. anonymous
    • 5 years ago
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    use this wherever you see 1 then simplify it till you get your right hand side equation

  3. anonymous
    • 5 years ago
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    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.

  4. anonymous
    • 5 years ago
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    will work it out if you like. it is algebra from there on in.

  5. anonymous
    • 5 years ago
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    oh my, could you work it out?

  6. anonymous
    • 5 years ago
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    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 \[(1-sin^2(\theta))^2=1-2sin^2(\theta)+sin^4(\theta)\]\]

  7. anonymous
    • 5 years ago
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    don't forget that originally you have \[1-cos^4(\theta)\] so now you have \[1-(1-2sin^2(\theta)+sin^4(\theta))=2sin^2(\theta)-sin^4(\theta)\]

  8. anonymous
    • 5 years ago
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    is that enough detail? if not i let me know.

  9. anonymous
    • 5 years ago
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    i don't know why, but i just can't see it

  10. anonymous
    • 5 years ago
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    \[(\sin^2\theta+\cos^2\theta-\cos^2\theta)(\sin^2\theta+\cos^2\theta+\cos^2\theta)=(\sin^2\theta)(sina^2\theta+2\cos^2\theta)=\sin^4\theta+2\sin^2\theta(\cos^2\theta)=\sin^4\theta+2(\sin^2\theta)(1-\sin^2\theta)=sina^4\theta+2\sin^\theta-2\sin^4\theta=2\sin^2\theta-\sin^4\theta\]

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

  12. anonymous
    • 5 years ago
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    so our first job is to multiply this out.

  13. anonymous
    • 5 years ago
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    is it clear that this is the same as multiplying out \[(1-x)(1+x)\]?

  14. anonymous
    • 5 years ago
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    yes, when you write it like that

  15. anonymous
    • 5 years ago
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    \[or rather (x-x^2)(1+x^2)\]

  16. anonymous
    • 5 years ago
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    typo sorry

  17. anonymous
    • 5 years ago
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    ok so lets multiply out \[(1-cos^2(\theta))(1+cos^2(\theta))\]

  18. anonymous
    • 5 years ago
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    \[(1-x^2)(1+x^2)=1-x^4\] so\[((1-cos^2(\theta))(1+cos^2(\theta))=1-cos^4(\theta)\] so far so good?

  19. anonymous
    • 5 years ago
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    yes

  20. anonymous
    • 5 years ago
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    ok now you recall that \[sin^2(\theta)+cos^2(\theta)=1\] so \[cos^2(\theta)=1-sin^2(\theta)\]

  21. anonymous
    • 5 years ago
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    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 \[1-(1-sin^2(\theta))^2\]

  22. anonymous
    • 5 years ago
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    ok?

  23. anonymous
    • 5 years ago
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    oh, ok

  24. anonymous
    • 5 years ago
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    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.

  25. anonymous
    • 5 years ago
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    ooohhh

  26. anonymous
    • 5 years ago
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    clear or no?

  27. anonymous
    • 5 years ago
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    it is when you replace the sin/cos with the 'x'. easier to see

  28. anonymous
    • 5 years ago
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    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.

  29. anonymous
    • 5 years ago
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    thanks....

  30. anonymous
    • 5 years ago
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    wanna do the next one?

  31. anonymous
    • 5 years ago
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    i'm not sure where to start

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