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Math Help - please help me solve: cos (x - 20 degrees) = -0.437, for the range 0 to 360 degrees

  1. #1
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    please help me solve: cos (x - 20 degrees) = -0.437, for the range 0 to 360 degrees

    hey i'm doing some revision for a test i have tommorw and this was one of the questions that came up, i'm fine with solving other trigonometric equations but im not too sure where to go with this one eg should i replace the (x-20 degrees) for say just y and come back and then find out x, but i thought i'd ask you intelligent people just to be sure
    NOTE: I HAVEN'T WORKED IN RADIANS BEFORE SO COULD WE KEEP IT IN DEGREES please =)
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  2. #2
    Super Member General's Avatar
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    Do you know the inverse trigonometric functions?
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  3. #3
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    all i know is that
    cos^2(x) + sin^2(x) = 1
    and that
    cos x / sin x = tan x
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  4. #4
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    Quote Originally Posted by rednaxela View Post
    all i know is that
    cos^2(x) + sin^2(x) = 1
    and that
    cos x / sin x = tan x
    that's not what the General asked you.

    are you familiar with the inverse trig function capability of your calculator, which is what you need to solve this problem for x?

    I recommend you visit this website to learn more about it ...

    7. The Inverse Trigonometric Functions


    also, for future reference ... \textcolor{red}{\frac{\sin{x}}{\cos{x}} = \tan{x}}
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  5. #5
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    apologies, im all over the place, yes i am aware of it
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  6. #6
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     <br />
\cos(x-20) = -0.437<br />

    first, realize that if cosine of an angle is negative, then there are two valid angles that will work, one in quad II and one in quad III

    also note that the calculator will only give you back the quad II value

     <br />
\arccos[\cos(x-20)] = \arccos(-0.437)<br />

    for the quad II value of \arccos(-0.437) ...

     <br />
x-20 = \arccos(-0.437)<br />

     <br />
x = \arccos(-0.437) + 20 \approx 136^\circ<br />

    for the quad III value ...

     <br />
x-20 = 360 - \arccos(-0.437)<br />

     <br />
x = 380 - \arccos(-0.437) \approx 264^\circ<br />
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