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anonymous

  • one year ago

Verify the identity. Show your work. (1 + tan2u)(1 - sin2u) = 1 my answer > cos^2x+sin^2x=1

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  1. anonymous
    • one year ago
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    but my teacher said 6. You can re-write each factor as another trig function squared. 1+tan^2u=(standard identity)? and 1-sin^2u=(standard identity)? Once you do that, the next steps are clear. Please refer to section 5.1 of your text for examples and sample problems.

  2. anonymous
    • one year ago
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    @phi @Vocaloid

  3. freckles
    • one year ago
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    what does 6 mean?

  4. freckles
    • one year ago
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    also do you mean tan^2(u) and sin^2(u) because I see tan(2u) and sin(2u)

  5. freckles
    • one year ago
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    Also are you asking us to refer to section 5.1 of your book?

  6. freckles
    • one year ago
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    I don't see how that is possible.

  7. Loser66
    • one year ago
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    I am with @freckles It must be \((1+tan^2u)(1-sin^2u) =1\)

  8. anonymous
    • one year ago
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    idk yes my book 6 the number of the problem im on and idk what she means '

  9. Nnesha
    • one year ago
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    well you are right cos^2+ sin^2 =1 right ! BUT you need to prove( 1+tan^2 ) (1-sin^2i) equal to 1

  10. anonymous
    • one year ago
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    sohow do i do that

  11. Nnesha
    • one year ago
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    \[\huge\rm (1+\tan^2u)(1-\sin^2u)=1\] show that R.H.S=L.H.S are u familiar with the trig identities you have to apply that :=)

  12. freckles
    • one year ago
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    \[\cos^2(x)+\sin^2(x)=1 \\ \text{ subtract } \sin^2(x) \text{ on both sides } \\ \cos^2(x)=1-\sin^2(x) \\ \text{ hint: now divide both sides of } \cos^2(x)+\sin^2(x)=1 \\ \text{ by } \cos^2(x)\]

  13. freckles
    • one year ago
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    other than that I can't refer to any thing in your book because I don't think I have your book

  14. freckles
    • one year ago
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    but i can help you prove the identity above

  15. anonymous
    • one year ago
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    lol i just need help with it i seriouly dont know what else to do

  16. freckles
    • one year ago
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    try to see if you can follow what I said above

  17. freckles
    • one year ago
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    divide both sides by cos^2(x) of the identity I spoke about just now and tell me what you have

  18. anonymous
    • one year ago
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    i have sin= 1 right idk im horrible at math

  19. freckles
    • one year ago
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    Did you try to divide both sides of cos^2(x)+sin^2(x)=1 by cos^2(x)?

  20. anonymous
    • one year ago
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    yes

  21. freckles
    • one year ago
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    so do you know what cos^2(x)/cos^2(x)=? or what sin^2(x)/cos^2(x)=? or what 1/cos^2(x)=?

  22. anonymous
    • one year ago
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    how did you gwt all that

  23. freckles
    • one year ago
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    Well I asked you to divide both sides of cos^2(x)+sin^2(x)=1 by cos^2(x)

  24. freckles
    • one year ago
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    \[\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=\frac{1}{cos^2(x)}\]

  25. anonymous
    • one year ago
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    ohhh i did it way wrong

  26. freckles
    • one year ago
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    \[\frac{\cos^2(x)}{\cos^2(x)}+\frac{\sin^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}\]

  27. freckles
    • one year ago
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    cos^2(x)/cos^2(x)=1 sin^2(x)/cos^2(x)=tan^2(x) 1/cos^2(x)=sec^2(x)`

  28. freckles
    • one year ago
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    though you could just leave it as 1+tan^2(x)=1/cos^2(x) if you want

  29. freckles
    • one year ago
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    now plug in both of those results we got from the Pythagorean identity

  30. Loser66
    • one year ago
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    \(1+tan^2u = sec^2 u =\dfrac{1}{cos^2u}\\1-sin^2 u = cos^2u\\hence,(1+tan^2u)(1-sin^2u)= \dfrac{1}{\cancel{cos^2u}}*\cancel{cos^2u}=1\)

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