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Math Help - Trigonometrical equation

  1. #1
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    Trigonometrical equation

    Hi. I have a problem solving this trigonometrical equation:
    6sin(pi+x)+cos(pi/2+x) if cosx=0.6 and 0<x<pi/2
    If you can, please ask questions so that I try to answer them myself and solve the equation step by step. I'm not from an English speaking country, but I hope we have similar terms and you understand me.
    Do I reduce first? I get -7sinx, but I'm not sure how to use then cosx=0.6, if I reduce -7sinx back to cos I get -7cos(3pi/2+x) here also I don't know how to use cosx=0.6....
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Trigonometrical equation

    If \cos(x)=0,6 then x=\arccos(0,6) (1)
    Offcourse there're infinite solutions for x responding to that, but there's given that x\in \]0,\frac{\pi}{2}\[
    Substitute (1) in 6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)

    But you can't call it an equation because 6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right) is not equated to anything ... ?
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  3. #3
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    Re: Trigonometrical equation

    Hm, if it's not an equation is it an expression then?
    Quote Originally Posted by Siron View Post
    If \cos(x)=0,6 then x=\arccos(0,6) (1)
    Wait, if \cos(x)=0,6 then shouldn\t x=\pmarccos0,6+2\pi n, n\in Z? What do I do with it?
    Oh, so yeah, since 0<x<pi/2, then x=arccos0.6, but still, arccos0.6 doesn't have an exact value...
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Trigonometrical equation

    It's indeed not practical to work with, but arccos(0,6)=53,13 or -53,13 because cos(x)=cos(-x)
    Is it the intention to calculate it without a calculator? ...
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  5. #5
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    Re: Trigonometrical equation

    I need an exact value , 53.13 is rounded up.
    So an exact value would be -7sin(arccos0.6)?
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Trigonometrical equation

    Quote Originally Posted by Evaldas View Post
    I need an exact value , 53.13 is rounded up.
    So an exact value would be -7sin(arccos0.6)?
    Almost, 6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)= 5\sin(x).
    So the exact value is 5\sin(\arccos(0,6))

    But this can be simpliefd by usin the fact that \sin[\arccos(x)]=\sqrt{1-x^2}
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  7. #7
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    Re: Trigonometrical equation

    Quote Originally Posted by Siron View Post
    Almost, 6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)= 5\sin(x).
    So the exact value is 5\sin(\arccos(0,6))
    I got -7sinx because I reduced 6sin(pi+x) and because it's the 3rd quarter (pi<x<3pi/2) sin is negative. I then reduced cos(pi/2+x), since it's 90 degrees it changes its name into sin, and it's the 2nd quarter and in the 2nd quarter cos is negative, so -6sinx-sinx=-7sinx (right?)

    Quote Originally Posted by Siron View Post
    But this can be simpliefd by usin the fact that \sin[\arccos(x)]=\sqrt{1-x^2}
    What does [] in this case mean? But I'd have square in this case arccos0.6, no?
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  8. #8
    MHF Contributor Siron's Avatar
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    Re: Trigonometrical equation

    Is it:
    6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)
    or
    -6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)

    ?
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  9. #9
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    Re: Trigonometrical equation

    Quote Originally Posted by Siron View Post
    Is it:
    6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)
    or
    -6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)

    ?
    It's 6\sin(\pi+x)+\cos\left(\frac{\pi}{2}+x\right)
    Here it is calculated by WolframAlpha
    I don't know how to explain but 6sin(pi+x) can be reduced to -6sinx ....
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  10. #10
    MHF Contributor Siron's Avatar
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    Re: Trigonometrical equation

    Ok, so indeed you get 7\sin[\arccos(0.6)]=7\sqrt{1-(0,6)^2}=-5,6
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