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Show that 8sinx + 15cosx <= 17

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that 8sinx + 15cosx ≤ 17


    2. Relevant equations
    8^2 + 15^2 = 17^2, or (8,15,17) is a Pythagorean triplet
    Identities like sin^2x + cos^2x = 1


    3. The attempt at a solution
    8sinx + 15cosx <= y
    square both sides, clean up
    64sin^2x + 240sinxcosx + 225cos^2x ≤y^2
    substitute sin^2x for 1-cos^2x and likewise for cos^2x, derived from Pythagorean Identity.
    289 - 64cos^2x -225sin^2x +240sinxcosx ≤ y^2

    I need to show that the trigonometric functions on the left side equal 0, so I can take the square root of both sides and find 17≤ y, but I don't know how to do that. It is also possible that I'm approaching this from the wrong viewpoint.
     
  2. jcsd
  3. Mar 9, 2012 #2
    Do you know derivatives yet? Or are you working strictly with trig identities?

    If you take the derivative of the function and find the zeroes you will have the maximum/minimum of the original function.
     
  4. Mar 9, 2012 #3
    I am working strictly with trig identities.
     
  5. Mar 9, 2012 #4

    SammyS

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    Hello Bearded Man. Welcome to PF !

    You demonstrated that a triangle with sides of length 8, 15, and 17 is a right triangle.

    Let θ be the larger of the two acute angles of such a triangle.

    Then cos(θ) = 8/17 and sin(θ) = 15/17

    Get the left hand side of your inequality to have a factor of [itex]\cos(\theta)\sin(x)+\sin(\theta)\cos(x)[/itex].
     
  6. Mar 9, 2012 #5

    Mark44

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    This can be done strictly with trig.
     
  7. Mar 10, 2012 #6
    Now that I have cos(x), sin(x), cos(θ), and sin(θ) defined, can't I just do this:
    8sin(x) + 15cos(x) ≤ y
    64/17 + 225/17 ≤ y
    289/17 ≤ y
    17 ≤ y
     
  8. Mar 10, 2012 #7

    LCKurtz

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    You aren't trying to prove ##17\le y##. You are trying to prove ##8\sin x + 15\cos x \le 17##. Try writing$$
    8\sin x + 15\cos x = 17\left( \frac 8 {17}\cos x + \frac{15}{17}\sin x\right )$$and see if you can use the previous hints. Think about an addition formula.
     
  9. Mar 10, 2012 #8

    HallsofIvy

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    First, I don't see any reason to introduce "y". Second, it is not true that "8 sin(x)= 64/17" and "15 cos(x)= 225/17" which is what you are implying.

    Better assert that [itex]sin(x+ y)= sin(x)cos(y)+ cos(x)sin(y)[/itex] and compare that to [itex]17((8/17) sin(x)+ (15/17)cos(x))[/itex]. Can you find y so that cos(y)= 8/17 and [itex]sin(y)= 15/17[/itex]? Is so, what does that tell you about 8sin(x)+ 15cos(x)?
     
  10. Mar 10, 2012 #9
    Of course, how silly of me.

    Thank you everyone for the help.
     
  11. Mar 11, 2012 #10

    Curious3141

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    The simplest approach in these cases is to express [itex]a\sin x + b\cos x[/itex] as [itex]R\sin(x + \theta)[/itex], where [itex]R = \sqrt{a^2 + b^2}[/itex] and [itex]tan \theta = \frac{b}{a}[/itex].

    You can then make the observation that [itex]\sin(x+\theta) \leq 1[/itex], meaning that the expression has a maximal value of [itex]R[/itex].
     
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