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Math Help - Angle measures in radians & degrees

  1. #1
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    Angle measures in radians & degrees

    just a couple problems i need some help with.

    1. Find the measure of the angle in radians and degrees.
    sin^-1 (0.6)

    2. Find the six trigonometric values of theta= cos^-1 (3/7). Give exact answers. I remember doing this during the school year but it's been a long summer and i've competely lost touch with this topic unfortunately

    3. Solve for x: e^-0.2x = 4
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  2. #2
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    Quote Originally Posted by fezz349 View Post
    just a couple problems i need some help with.

    1. Find the measure of the angle in radians and degrees.
    sin^-1 (0.6)

    2. Find the six trigonometric values of theta= cos^-1 (3/7). Give exact answers. I remember doing this during the school year but it's been a long summer and i've competely lost touch with this topic unfortunately

    3. Solve for x: e^-0.2x = 4
    One problem per thread please and show some effort as well. I'll help you with #3.

    Take the natural log of both sides.

    \ln(e^{-0.2x})=\ln(4)

    Also remember the rule that \ln(a^b)=b\ln(a)
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  3. #3
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    right, so i took what u said and got 0.2xlne = ln4

    i feel like this is wrong but i divide ln4 by ln of e in my calculator. it shows as ln(e^( . so i'm assuming the is being raised to the first power right? I got 1.386294361. Then divide by 0.2 on each side and my final answer was x=6.931471806. Just doesn't sound right to me for some reason.
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  4. #4
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    \ln(e)=1

    That's a very important identity. I wouldn't a decimal answer. Just keep the natural logs as is.
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  5. #5
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    Gotcha, it's starting to come back to me now from precalc last year. Just another quick question, from the six trig functions questions, i thoguht about it for a bit and cos^-1 = (3/7) is equivalent to secant=3/7 right? Meaning cosine would be 7/3. I draw my triangle, and solve for the last side by using pythagorean theorem... i think
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  6. #6
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    Quote Originally Posted by fezz349 View Post
    Gotcha, it's starting to come back to me now from precalc last year. Just another quick question, from the six trig functions questions, i thoguht about it for a bit and cos^-1 = (3/7) is equivalent to secant=3/7 right? Meaning cosine would be 7/3. I draw my triangle, and solve for the last side by using pythagorean theorem... i think
    No, that's not true. It's a common mistake. cos^(-1) means inverse cosine, not 1 over cosine. When you take the cosine of an angle you get a ratio. When you are given a ratio you use the inverse function to find the angle.
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