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anonymous

  • one year ago

If sin(theta)=X, then express sec(theta) in terms of x

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  1. anonymous
    • one year ago
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    The answer I got was \[\sqrt{1-(1/x)^2}\] Would this be correct?

  2. freckles
    • one year ago
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    |dw:1438813890812:dw| use pythagorean theorem to find the other side

  3. freckles
    • one year ago
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    and recall for secant use hyp/adj

  4. anonymous
    • one year ago
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    I'm sorry, I'm a bit confused where to go from there

  5. freckles
    • one year ago
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    can you find the adjacent side by use of Pythagorean theorem ?

  6. freckles
    • one year ago
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    \[(\text{ adjacent })^2+x^2=1^2 \\ \text{ solve for adjacent }\]

  7. freckles
    • one year ago
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    let me know if you don't know how maybe I can give you a step

  8. anonymous
    • one year ago
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    adjacent= -x+1?

  9. freckles
    • one year ago
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    \[( \text{ adjacent } )^2+x^2=1 \\ \text{ subtract} x^2 \text{ on both sides } \\ ( \text{ adjacent })^2=1-x^2 \\ \text{ take square root of both sides } \\ \text{ adjacent } = \pm \sqrt{1-x^2}\] you cannot simplify this ... if that is what you tried to do

  10. freckles
    • one year ago
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    |dw:1438814521770:dw| now find sec(theta) using the right triangle

  11. anonymous
    • one year ago
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    The whole negative thing kinda threw me off. Thanks

  12. anonymous
    • one year ago
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    What I tried to do initially was rewrite the identity sin^2+Cos^2=1 and go from there

  13. freckles
    • one year ago
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    you can use that too

  14. freckles
    • one year ago
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    \[\sin^2(\theta)+\cos^2(\theta)=1 \\ \text{ we are given } \sin(\theta)=x \text{ so } \sin^2(\theta)=x^2 \\ x^2+\cos^2(\theta)=1 \\ \\ \text{ subtract} x^2 \text{ on both sides } \\ \cos^2(\theta)=1-x^2 \\ \text{ take square root of both sides } \\ \cos(\theta)=\pm \sqrt{1-x^2}\] then just flip both sides to find sec(theta)

  15. freckles
    • one year ago
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    its the same thing really

  16. anonymous
    • one year ago
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    Ooh, I tried to use the recirpocals first, hence why I had (1/x)^2 Thanks a bunch mate (:

  17. freckles
    • one year ago
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    np

  18. freckles
    • one year ago
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    well csc^2(x) would actually be (1/x)^2

  19. anonymous
    • one year ago
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    Yeah, I figured if the identity works with sin and cos, it'd work with their reciprocals. I see the error

  20. anonymous
    • one year ago
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    Do you think you might be able to help me with one more question?

  21. freckles
    • one year ago
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    i can try

  22. anonymous
    • one year ago
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    Evaluate cos1+cos2+cos3+...+cos357+cos358+cos359 (all measurements are in degrees) The only thing I can imagine is that it repeats at a certain point or something

  23. freckles
    • one year ago
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    \[\cos(x)+\cos(180-x) \\=\cos(x)+\cos(180)\cos(x)-\sin(180)\sin(x) \\ =\cos(x)-1\cos(x)-0\sin(x)\\ =0\] So I think maybe we can use cos(x)+cos(180-x)

  24. freckles
    • one year ago
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    So I think maybe we can use cos(x)+cos(180-x)=0*

  25. freckles
    • one year ago
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    \[\cos(1)+\cos(179)=0 \\ \cos(2)+\cos(178)=0 \\ \cos(3)+\cos(177)=0 \\ \cos(4)+\cos(176) =0 \\ \cdots \\ \] and I have to come back to finish this i'm being called away unfortunately

  26. anonymous
    • one year ago
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    Alright

  27. freckles
    • one year ago
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    notice above I put a minus but it doesn't matter that one part is still 0 \[\cos(x)+\cos(180-x) \\=\cos(x)+\cos(180)\cos(x)+\sin(180)\sin(x) \\ =\cos(x)-1\cos(x)+0\sin(x)\\ =0\] hmm... \[\cos(180)=-1 \\ \text{ hmmm... we need to think now for the rest of the finite series }\] like we already know cos(1)+cos(2)+...+cos(180)=-1 we need to figure out would to do with cos(181)+cos(182)+cos(183)+...+cos(359) we should be able to think of something similar

  28. freckles
    • one year ago
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    how about for that other sign try to see if something like cos(180+x)+cos(360-x)=0 works.. cos(180)cos(x)-sin(180)sin(x)+cos(360)cos(x)+sin(360)sin(x) =-cos(x)+cos(x)=0

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