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What's the hypotenuse of acosx + bsinx = c

  1. Jul 22, 2012 #1
    1. The problem statement, all variables and given/known data

    What's the hypotenuse of acosx + bsinx = c and by extension sinx - sqrt(2)cosx = 1 ?

    2. Relevant equations
    3. The attempt at a solution

    If cosx = x and sinx = y and a^2 + b^2 = c^2, then going by acosx + bsinx = c the hypotenuse of sinx - sqrt(2)cosx = 1 have to be sqrt(c)= 1? Then how about 1^2 + (- sqrt(2))^2= 3, where c= sqrt(3)?

    Thanks.
     
  2. jcsd
  3. Jul 22, 2012 #2

    Mentallic

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    What do you mean by the hypotenuse? If a right-triangle has shorter sides A and B, then the hypotenuse is given by [itex]\sqrt{A^2+B^2}[/itex], thus the formula satisfying this equality is [itex]A^2+B^2=C^2[/itex] where the hypotenuse is length C.

    So how are you relating this definition to what you've shown us?
     
  4. Jul 22, 2012 #3
    sinx- sqrt(2)cosx= 1 is of the form acosx+ bsinx= c. Comparing the two equations we have c=1. But by a^2+ b^2= c^2 we have c= sqrt(3). If c=1 and a^2+ b^2= c^2, then 1^2+ (-sqrt(2))^2= sqrt(3) ≠ 1.

    cosx= x/r where r=1, so cosx=x and sinx= y. So I thought that must mean acosx=x and bsinx= y, and c= hypotenuse.
     
    Last edited: Jul 22, 2012
  5. Jul 22, 2012 #4

    Mentallic

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    Oh I see, so you're trying to solve for x in the equation [itex]\sin(x)-\sqrt{2}\cos(x)=1[/itex]
    The coefficients (numbers in front of [itex]\sin(x)[/itex] and [itex]\cos(x)[/itex]) don't have anything to do with the hypotenuse of a right-angled triangle.

    You could've also been asked to solve for x in [itex]10\sin(x)-\sqrt{2}\cos(x)=1[/itex] and it also wouldn't satisfy the formula for a right-angled triangle.

    You could also be asked to solve [itex]2x^2+3x=5[/itex] for the coefficients there don't have to satisfy it either, but the quadratic does have its own formula using the coefficient a,b and c, which are completely separate from the formula [itex]a^2+b^2=c^2[/itex] which is also completely separate from the problem you're asked of.

    So to start by solving for x in [itex]a\sin(x)+b\cos(x)=c[/itex], consider letting [itex]c=R\sin(x+\theta)[/itex] for some unknown constant [itex]R, \theta[/itex]. Expand [itex]sin(x+\theta)[/itex] and equate the coefficients of [itex]\sin(x)[/itex] and [itex]\cos(x)[/itex] from the LHS with those from the RHS.
     
  6. Jul 22, 2012 #5

    HallsofIvy

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    What you have written makes no sense. A 'hypotenuse' is a side of a right triangle or the length of that side. 'a cos x+ b sin x' is a number and so might be the length of a side but you haven't told us what 'a', or 'b', or x are. The usual notation is that 'a' and 'b' are legs of a right triangle with hypotenuse 'c' but even assuming that 'x' is an angle in that right triangle, 'a cos x+ b sin x' is no part of that triangle.
     
  7. Jul 22, 2012 #6
    I thought R was the radius and also the hypotenuse.
     
  8. Jul 22, 2012 #7

    HallsofIvy

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    Once again, there is NO 'radius', no 'hypotenuse', and no 'right triangle' in this problem. The use and, indeed, definition, of the trig functions in terms of right triangles is very limited and not generally used in any mathematics beyond basic trigonometry. This is simply a problem about the functions 'sine' and 'cosine' and has nothing to do with right triangles.
     
  9. Jul 22, 2012 #8
    What's the hypotenuse of 52? That's something similar to what your asking.
     
  10. Jul 22, 2012 #9
    Oh, all right. Was just trying to make sense of procedural equations like c= Rsin(x+t) and where they might have originated from in solving these equations.

    Thanks.
     
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