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anonymous

  • one year ago

If sin θ = 12 over 37, use the Pythagorean Identity to find cos θ.

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  1. Nnesha
    • one year ago
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    Identity is \[\huge\rm sin^2 \theta +\cos^2 \theta =1\] substitute sin for 12/37 solve for cos

  2. Nnesha
    • one year ago
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    |dw:1440269530637:dw|

  3. Nnesha
    • one year ago
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    or you can draw a right triangle ^^^\[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

  4. anonymous
    • one year ago
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    |dw:1440269641978:dw|

  5. anonymous
    • one year ago
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    alright thx. wasn't sure i was solving it right.

  6. Nnesha
    • one year ago
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    np :)

  7. anonymous
    • one year ago
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    Refer to the attachment.

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  8. anonymous
    • one year ago
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    cos θ = ± 35 over 37 cos θ = ± 23 over 37 cos θ = ± 35 over 12 cos θ = ± 20 over 35 are my answer choices... little confused now.

  9. Nnesha
    • one year ago
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    you need cos theta for right triangle you hve to apply Pythagorean theorem \[\huge\rm a^2+b^2=c^2\]

  10. Nnesha
    • one year ago
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    cos = adjacent over hypotenuse

  11. Nnesha
    • one year ago
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    \[\huge\rm \sin \theta =\frac{ opposite }{ hyp } =\frac{ 12 }{ 37 }\]|dw:1440270658205:dw| find 3rd side of right triangle

  12. anonymous
    • one year ago
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    thats where i get lost. do i just substitute each of those in for the variables

  13. Nnesha
    • one year ago
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    remember c is the longest side of right triangle (hypotenuse )

  14. anonymous
    • one year ago
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    a^2+12^2=37^2... right?

  15. Nnesha
    • one year ago
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    yes right!!

  16. anonymous
    • one year ago
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    and if i rearrange it to solve for a i get 35^2?

  17. Nnesha
    • one year ago
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    hmm not square

  18. Nnesha
    • one year ago
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    you would subtract 12^2 both sides or you can take square of 12 first and the subtract it doesn't matter \[c^2= \sqrt{37^2-12^2}\]

  19. anonymous
    • one year ago
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    35

  20. Nnesha
    • one year ago
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    yes right

  21. anonymous
    • one year ago
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    so the answer is +-35/37?

  22. Nnesha
    • one year ago
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    yes right :=)

  23. anonymous
    • one year ago
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    thx much

  24. anonymous
    • one year ago
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    can you help with one more?

  25. Nnesha
    • one year ago
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    i'll try

  26. anonymous
    • one year ago
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    What are the sine, cosine, and tangent of 5 pi over 4 radians?

  27. anonymous
    • one year ago
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    i can't find the tan

  28. Nnesha
    • one year ago
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    \[\huge\rm tan \theta =\frac{ \sin \theta }{ \cos \theta }\] :=)

  29. Nnesha
    • one year ago
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    on the unit circle there are cos and sin (x,y)(cos,sin) x-coordinate represent cos y-coordinate represents sin so tan is sin over cos

  30. anonymous
    • one year ago
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    so then its -1?

  31. anonymous
    • one year ago
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    sin Θ = negative square root 2 over 2; cos Θ = negative square root 2 over 2; tan Θ = -1?

  32. Nnesha
    • one year ago
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    yes right :=)

  33. anonymous
    • one year ago
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    thank you, would give 2 medals.. but one at a time :]

  34. Nnesha
    • one year ago
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    ahha that fine one medal is enough :) & my pleasure

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