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Math Help - Inverse functions and substitution

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    Inverse functions and substitution

    Hi, I am having trouble finding resources on how to apply the properties of inverse functions arcsin, arccos and arctan, as well as how to use substitution to solve equations such as:

    5cos(x) + 12cos(x) = 3~

    ~0\leq x \leq 2\pi

    Everything I come across seems to go over the properties of the inverse functions and graph them etc. which I already understand; however, I am still confused as to how I would apply the properties to solve trig equations.

    How should I go about solving the above problem and can anyone direct me to some more examples?

    Thanks.
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    Re: Inverse functions and substitution

    Quote Originally Posted by terrorsquid View Post
    Hi, I am having trouble finding resources on how to apply the properties of inverse functions arcsin, arccos and arctan, as well as how to use substitution to solve equations such as:

    5cos(x) + 12cos(x) = 3~

    ~0\leq x \leq 2\pi

    Everything I come across seems to go over the properties of the inverse functions and graph them etc. which I already understand; however, I am still confused as to how I would apply the properties to solve trig equations.

    How should I go about solving the above and can anyone direct me to some more examples?

    Thanks.
    Solve for cos(x), then solve for x!
    Since 5 + 12 = 17, we have
    17cos(x) = 3
    cos(x) = 3/17
    x = arccos(3/17) and 2pi - arccos(3/17)

    Now if the problem had been
    5cos(x) + 12sin(x) = 3, then we would take a different (and more interesting) approach...
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    MHF Contributor Siron's Avatar
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    Re: Inverse functions and substitution

    Are you also searching for other types of goniometric equations? ...
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    Re: Inverse functions and substitution

    Quote Originally Posted by TheChaz View Post
    Now if the problem had been
    5cos(x) + 12sin(x) = 3, then we would take a different (and more interesting) approach...
    This is what I meant to write and is what I have been trying to find examples of but I can only seem to find definitions of inverse functions or trig substitution used to integrate and not for solving trig equations.

    The only thing I seem to spot is that the squares of the coefficients on the LHS add up to a clean 169 so if you divided both sides by 13, you would you have the squares of the coefficients = 1.

    How would you apply substitution or the properties of inverse functions to the eqation:

    5cos(x) + 12sin(x) = 3

    Never thought I'de like trig, but this has got me interested again
    Last edited by terrorsquid; August 17th 2011 at 09:19 PM.
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    Re: Inverse functions and substitution

    Quote Originally Posted by Siron View Post
    Are you also searching for other types of goniometric equations? ...
    If 5cos(x) + 12sin(x) = 3 is a goniometric equation then yes
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    MHF Contributor Siron's Avatar
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    Re: Inverse functions and substitution

    I think you can try something with the t-formulas, let t=\tan\left(\frac{x}{2}\right).
    Are you familiar with this? ...
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    Super Member TheChaz's Avatar
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    Re: Inverse functions and substitution

    Quote Originally Posted by terrorsquid View Post
    This is what I meant to write and is what I have been trying to find examples of but I can only seem to find definitions of inverse functions or trig substitution used to integrate and not for solving trig equations.

    The only thing I seem to spot is that the squares of the coefficients on the LHS add up to a clean 169 so if you divided both sides by 13, you would you have the squares of the coefficients = 1.

    How would you apply substitution or the properties of inverse functions to the eqation:

    5cos(x) + 12sin(x) = 3

    Never thought I'de like trig, but this has got me interested again
    You are on the right track! And while (5, 12, 13) is a pythagorean triple, you can perform this with any values of A and B for

     A\cos(x) + B\sin(x) = D . Just divide by \sqrt{(A^2 + B^2)}

    So here we go.
    Dividing by 13, we have

    \frac{5}{13} \cos(x) + \frac{12}{13} \sin(x) = \frac{3}{13}

    Let t be the angle with sine 5/13 (and hence cosine of 12/13).

    Then our equation becomes

    \sin(t) \cos(x) + \cos(t) \sin(x) = \frac{3}{13}

    The left hand side is the sine angle sum...

     \sin(t + x) = \frac{3}{13}

    arcsine
    find the value of t
    etc.
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    Re: Inverse functions and substitution

    Ah, ok. I got to the point where I could see that with the new coefficients, the equation was obviously on the unit circle; however, when I drew in the coordinates 5/13 and 12/13, I didn't know which was cos(x) and which was sin(x) (x,y) on the circle. I tried once just guessing and put +5/13 on the x axis and +12/13 on the y axis. Therefore, I had:

    cos(t) = \frac{5}{13}

    sin(t) = \frac{12}{13}

    cos(t)cos(x) + sin(t)sin(x) = \frac{3}{13}

    LHS = cos angle difference:

    cos(t - x) = \frac{3}{13}

    arccos\left(\frac{3}{13}\right) = t - x

    x = t - arccos\left(\frac{3}{13}\right) = arccos\left(\frac{5}{13}\right) - arccos\left(\frac{3}{13}\right)

    Is it that the same answer? how do I know which coefficient to use on which axis?

    Thanks.
    Last edited by terrorsquid; August 18th 2011 at 05:54 AM.
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    Re: Inverse functions and substitution

    Quote Originally Posted by Siron View Post
    I think you can try something with the t-formulas, let t=\tan\left(\frac{x}{2}\right).
    Are you familiar with this? ...
    I have something written down:

    sin(2\theta) = \frac{2t}{1 + t^2}

    cos(2\theta) = \frac{1 - t^2}{1 + t^2}

    I don't know where they came from though and don't understand their use; the only identity I know for sin(2\theta) and cos(2\theta) would be the double angle identities.
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    MHF Contributor Siron's Avatar
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    Re: Inverse functions and substitution

    That's ok, now you can wright the equation as:
    5\cdot \left(\frac{1-t^2}{1+t^2}\right)+12\cdot \left(\frac{2t}{1+t^2}\right)-3=0

    \Leftrightarrow \frac{5(1-t^2)+24t-3(1+t^2)}{1+t^2}=0

    Solve this equation for t and afterwards do the back-substitution.
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  11. #11
    Super Member TheChaz's Avatar
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    Re: Inverse functions and substitution

    Quote Originally Posted by terrorsquid View Post

    cos(t) = \frac{5}{13}

    sin(t) = \frac{12}{13}

    cos(t)cos(x) + sin(t)sin(x) = \frac{3}{13}

    LHS = cos angle difference:

    cos(t - x) = \frac{3}{13}

    arccos\left(\frac{3}{13}\right) = t - x

    x = t - arccos\left(\frac{3}{13}\right) = arccos\left(\frac{5}{13}\right) - arccos\left(\frac{3}{13}\right)

    Is it that the same answer? how do I know which coefficient to use on which axis?

    Thanks.
    I prefer sine, so I chose it as such.
    They are the same, but don't forget that there are infinite (though coterminal) solutions - and that arccos(5/13) = arcsin(12/13) and vice versa.
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    Re: Inverse functions and substitution

    Ok, cool. Would the function still have period \pi? and therefore an infinite amount of answers at \pi k where k \in \mathbb{Z} (I'm a little confused on how to determine the solution set).

    Also, with the other method I get:

    t = \frac{\pm\sqrt{10} + 3}{2}

    So, I subbed the two values in for cos(x) and solved for sin(x) and vice versa to get 4 different answers. approx ~ 9,-9, 35 and 144. Again, I don't know how to define the infinite set of answers or, if there was a given restriction, how to discern which are valid answers.
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  13. #13
    MHF Contributor Siron's Avatar
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    Re: Inverse functions and substitution

    Is this the t you become by using the t-formulas?
    The only thing you've to do now is the back-substitution:
    t=\tan\left(\frac{x}{2}\right) \Leftrightarrow \frac{x}{2}=\arctan(t)+k\cdot \pi \Leftrightarrow x=2\cdot \arctan(t)+2k\pi

    Can you continue now? (as The Chaz noticed there are infinite solutions)
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    Re: Inverse functions and substitution

    Quote Originally Posted by Siron View Post
    Is this the t you become by using the t-formulas?
    The only thing you've to do now is the back-substitution:
    t=\tan\left(\frac{x}{2}\right) \Leftrightarrow \frac{x}{2}=\arctan(t)+k\cdot \pi \Leftrightarrow x=2\cdot \arctan(t)+2k\pi

    Can you continue now? (as The Chaz noticed there are infinite solutions)
    I'm ok solving the equations now, I am just having trouble visualising which quadrants the function appears in because there is a sin and a cos in it. If it was just cos(x) = 1/2, I could just think cos(x) is positive therefore it is in the 1st and 4th quadrants and then think it is either 60 or 360-60 (and 2pi k multiples of each). Or if it was sin(x) = 0 I can imagine the sin curve and know that it will be 0 at 0 and any integer multiple of pi. But how do you interpret a heterogeneous function like this one and know where it will reoccur?
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    Re: Inverse functions and substitution

    Quote Originally Posted by terrorsquid View Post
    cos(t) = \frac{5}{13}

    sin(t) = \frac{12}{13}

    cos(t)cos(x) + sin(t)sin(x) = \frac{3}{13}

    LHS = cos angle difference:

    cos(t - x) = \frac{3}{13}

    arccos\left(\frac{3}{13}\right) = t - x

    x = t - arccos\left(\frac{3}{13}\right) = arccos\left(\frac{5}{13}\right) - arccos\left(\frac{3}{13}\right)
    Say, for example, the above was between -\pi and \pi what would you be thinking when figuring out the solution set?
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