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kengeta

  • 5 months ago

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

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  1. myininaya
    • 5 months 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
    • 5 months ago
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    let me know if you still don't know where to go after that

  3. kengeta
    • 5 months ago
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    I'm pretty confused

  4. myininaya
    • 5 months ago
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    on how to factor?

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

  6. myininaya
    • 5 months ago
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    how* (not hot)

  7. kengeta
    • 5 months ago
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    no

  8. myininaya
    • 5 months 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
    • 5 months 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
    • 5 months ago
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    yes

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

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

  13. myininaya
    • 5 months 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
    • 5 months ago
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    yes

  15. myininaya
    • 5 months 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
    • 5 months ago
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    cos^2?

  17. myininaya
    • 5 months ago
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    right so what does that leave us with?

  18. myininaya
    • 5 months 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
    • 5 months ago
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    sin^2(theta)+cos^2(theta)

  20. myininaya
    • 5 months ago
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    And that equals?

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

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

  23. kengeta
    • 5 months ago
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    okay si... 1?

  24. myininaya
    • 5 months ago
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    I will give you a hint: Pythagorean Identity

  25. myininaya
    • 5 months ago
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    yep

  26. kengeta
    • 5 months ago
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    *so

  27. kengeta
    • 5 months ago
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    okay i get it now thanks soooo much

  28. myininaya
    • 5 months 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
    • 5 months ago
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    You don't have to do both ways. Just one way.

  30. myininaya
    • 5 months ago
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    I used a Pythagorean identity in both ways.

  31. myininaya
    • 5 months 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
    • 5 months 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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