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Thread: Solving sin (2x) cos x = cos (2x) using sum or difference identities

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    Solving sin (2x) cos x = cos (2x) using sum or difference identities

    Can anybody answer this question:

    Use the sum or difference identities to solve the equation: sin (2x) cos x = cos (2x)

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    Which identities did you have in mind? Looks like sin(2x) and cos(2x) might be helpful. Let's see your efforts.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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    Quote Originally Posted by tkhunny View Post
    Which identities did you have in mind? Looks like sin(2x) and cos(2x) might be helpful. Let's see your efforts.
    I got to 2 sinx cosx (cos x)= cos^2x-sin^2x but don't know where to go from there. I tried moving everything to one side and setting it equal to zero, then factoring, but that didn't seem to work either. Suggestions?

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    Wink

    Start from scratch remember these type of quest chechks how well u knew ur definition
    use fundamental law of trigonometry
    cos(a+b)=????
    cos(a-b)=???? Add and subtract both to get ans

    sin(a+b)=????????
    sin(a-b)=?????? And and subtract both to get ans

    compare your ques with the ans of above equation to know what you need

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    Quote Originally Posted by mgabor View Post
    I got to 2 sinx cosx (cos x)= cos^2x-sin^2x but don't know where to go from there. I tried moving everything to one side and setting it equal to zero, then factoring, but that didn't seem to work either. Suggestions?
    It can be a good idea to try to reduce the equation to one with only one trig function. The left side has cos^2(x), and so does the right; can you use the Pythagorean identity to rewrite the equation using only the sine? Then you can solve what amounts to a polynomial equation.

    When you write back, be sure to show your actual work, not just a summary of the kinds of things you did. Then we can see if you made a mistake we can correct.

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    Here's what I got, but I'm still stuck:

    2 sin x(1-sin^2x)=1-sin^2x-sin^2x
    2 sin x - 2 sin^3 x=1 - 2 sin^2 x
    2 sin^3 x - 2 sin^2 x - 2 sin x +1=0

    I can't factor by grouping - I've tried every combination and it doesn't work.

    I also thought I might be able to move the 1 to the other side and then factor, but that didn't work either.

    2 sin x(sin^2 x - sin x - 1)=-1 but what is left inside the parentheses is not factorable.

    Help!

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    Cool

    Quote Originally Posted by mgabor View Post
    Here's what I got, but I'm still stuck:

    2 sin x(1-sin^2x)=1-sin^2x-sin^2x
    2 sin x - 2 sin^3 x=1 - 2 sin^2 x
    2 sin^3 x - 2 sin^2 x - 2 sin x +1=0

    I can't factor by grouping - I've tried every combination and it doesn't work.

    I also thought I might be able to move the 1 to the other side and then factor, but that didn't work either.

    2 sin x(sin^2 x - sin x - 1)=-1 but what is left inside the parentheses is not factorable.

    Help!
    The advice you've been given is good, and is standard. I'm wondering, though, if anybody else has tried following this through to the end. There are solutions (here), but I'm having a devil of a time arriving at them independently.

    The equation can be rearranged thusly:

    . . . . .[tex]2\, \sin(x)\, \cos^2(x)\, =\, \cos^2(x)\, -\, \sin^2(x)[/tex]

    . . . . .[tex]\sin^2(x)\, +\, 2\, \sin(x)\, \cos^2(x)\, -\, \cos^2(x)\, =\, 0[/tex]

    This is a quadratic in sine, so we can apply the Quadratic Formula:

    . . . . .[tex]\sin(x)\, =\, \dfrac{-(2\, \cos^2(x))\, \pm\, \sqrt{(2\, \cos^2(x))^2\, -\, 4(1)(-\cos^2(x))\,}}{2(1)}[/tex]

    . . . . .[tex]\sin(x)\, =\, \dfrac{-2\, \cos^2(x)\, \pm\, \sqrt{4\, \cos^4(x)\, +\, 4\, \cos^2(x)\,}}{2}[/tex]

    ...which eventually ends up in the vicinity of:

    . . . . .[tex]\sin(x)\, =\, -\cos(x)\, \pm\, \cos(x)\,\sin(x)[/tex]

    This technique is sometimes helpful, but it doesn't appear to be, in this case. I'm open to corrections and further hints, but I'm grinding to a halt, too.

    I don't think it's you (your skills or your imagination); this equation is in fact a bit of a bear....

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    Quote Originally Posted by mgabor View Post
    Here's what I got, but I'm still stuck:

    2 sin x(1-sin^2x)=1-sin^2x-sin^2x
    2 sin x - 2 sin^3 x=1 - 2 sin^2 x
    2 sin^3 x - 2 sin^2 x - 2 sin x +1=0

    I can't factor by grouping - I've tried every combination and it doesn't work.

    I also thought I might be able to move the 1 to the other side and then factor, but that didn't work either.

    2 sin x(sin^2 x - sin x - 1)=-1 but what is left inside the parentheses is not factorable.
    What you have here is a cubic equation in the sine, 2s^3 - 2s^2 - 2s + 1 = 0. I got the same. It has no rational roots, so it can't be factored; a graph shows that it has three real roots, two of them in the range of the sine, but they are irrational and likely not expressible in exact form without great effort. In general, cubics are hard.

    What I would do next is to make sure you copied the problem correctly!

    Perhaps you can tell us the context.

    It is also odd that the problem as you quoted it says "using sum or difference identities", considering that we used double-angle identities (which are derived from sum identities) instead. Did you mix together parts of two problems?

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