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kengeta

  • one year ago

how do i prove : tan^2 θ cos^2 θ + cos^2 θ = 1

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  1. myininaya
    • one year ago
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    well my first attempt to prove this is true is to take the left hand side and try to show the right hand side the left hand side has both terms with the factor cos^2(theta) start by factoring cos^2(theta) out from both terms this will get us closer to writting as 1 term just as the right hand side is (one term)

  2. myininaya
    • one year ago
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    let me know if you still don't know where to go after that

  3. kengeta
    • one year ago
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    I'm pretty confused

  4. myininaya
    • one year ago
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    on how to factor?

  5. myininaya
    • one year ago
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    do you know hot to factor the expression ax+x?

  6. myininaya
    • one year ago
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    how* (not hot)

  7. kengeta
    • one year ago
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    no

  8. myininaya
    • one year ago
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    I will show you how to factor ax+x then you should be able to factor tan^2(theta)*cos^2(theta)+cos^2(theta) so ax+x I see there is a x in both terms |dw:1407522006591:dw| so i will factor that x out like so \[ax+1x=x(a+1)\]

  9. myininaya
    • one year ago
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    you have \[\tan^2(\theta) \cdot \cos^2(\theta)+\cos^2(\theta) \cdot 1 \] \[\text{ do you see that there is a } cos^2(\theta) \text{ in both terms ?}\]

  10. kengeta
    • one year ago
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    yes

  11. myininaya
    • one year ago
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    can you factor that out using what you know about multiplication and division like i did above?

  12. myininaya
    • one year ago
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    also this way i'm asking you to approach it is not the only approach to take

  13. myininaya
    • one year ago
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    ok if you are having trouble factoring the cos^2(theta) out how about we try this another way do you recall tan is sin/cos?

  14. kengeta
    • one year ago
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    yes

  15. myininaya
    • one year ago
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    \[\frac{\sin^2(\theta)}{\cos^2(\theta)} \cos^2(\theta)+\cos^2(\theta)\] in the first term there, do you see anything that cancels?

  16. kengeta
    • one year ago
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    cos^2?

  17. myininaya
    • one year ago
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    right so what does that leave us with?

  18. myininaya
    • one year ago
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    \[\frac{\sin^2(\theta)}{\cancel{\cos^2(\theta)}}\cancel{\cos^2(\theta)}+\cos^2(\theta)\] what does this give us?

  19. kengeta
    • one year ago
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    sin^2(theta)+cos^2(theta)

  20. myininaya
    • one year ago
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    And that equals?

  21. myininaya
    • one year ago
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    you need to recall some trig identities for these trigonometric proofs

  22. myininaya
    • one year ago
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    This one of the most basic ones It actually has a name

  23. kengeta
    • one year ago
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    okay si... 1?

  24. myininaya
    • one year ago
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    I will give you a hint: Pythagorean Identity

  25. myininaya
    • one year ago
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    yep

  26. kengeta
    • one year ago
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    *so

  27. kengeta
    • one year ago
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    okay i get it now thanks soooo much

  28. myininaya
    • one year ago
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    Way 1: \[\tan^2(\theta) \cdot \cos^2(\theta)+\cos^2(\theta) \cdot 1= \\ \cos^2(\theta)(\tan^2(\theta)+1)= \\ \cos^2(\theta) \cdot \sec^2(\theta)= \\ 1\] Way 2: \[\tan^2(\theta) \cos^2(\theta)+\cos^2(\theta)= \\ \frac{\sin^2(\theta)}{\cos^2(\theta)} \cos^2(\theta)+\cos^2(\theta) = \\ \sin^2(\theta) +\cos^2(\theta)= \\ 1\] These are the two ways that I can think of... This does not mean they are the only ways.

  29. myininaya
    • one year ago
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    You don't have to do both ways. Just one way.

  30. myininaya
    • one year ago
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    I used a Pythagorean identity in both ways.

  31. myininaya
    • one year ago
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    way 1 was the way I was trying to get you to go about it the first time around

  32. myininaya
    • one year ago
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    I don't think one way is more harder than the other But you will have to review some algebra especially if you don't remember how to factor. Factoring will come up again.

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