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Trigonometric problem

  1. Nov 16, 2015 #1

    diredragon

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    1. The problem statement, all variables and given/known data
    cos(x - 3p/2) = - 4/5
    p <x< p/2
    sin(x/2)*cos(5p/4)= ?
    2. Relevant equations

    3. The attempt at a solution
    I made it as far as to determine that sinx= 4/5 and cosx = - 3/5 but cant seems to progress any further. Im looking for an easier way to find the solution without having to deal with halfangle roots and all of that. Any hints?
     
  2. jcsd
  3. Nov 16, 2015 #2

    Mentallic

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    For the first problem, what would your solution for y be to the question [itex]\cos(y)=-4/5[/itex]?

    For the second, are you trying to simplify [itex]\sin(x/2)*\cos(5\pi/4)[/itex] into something in terms of sin(x) and cos(x)? Sometimes you are forced to deal with half angle formulas, especially when you're given a half angle.
     
  4. Nov 16, 2015 #3

    diredragon

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    Actually i made an error, the question reads sin(x/3)*cos(5x/4)= ?
    And for what you said y would be y= arccos(-4/5) right?
     
  5. Nov 16, 2015 #4

    Mentallic

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    Ok, but that's just an expression. It's like saying, [itex]x^2+2x+1=?[/itex] there's nothing to do with it from here, unless you're expected to turn it into another equal form, such as turning that example into [itex](x+1)^2[/itex]

    Yes, right. Now let [itex]y=x-3\pi/2[/itex].
     
  6. Nov 16, 2015 #5

    diredragon

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    Ok but how do i get sinx/2 ( its 2 not over 3) and cos(5x/4)? What does y = x - 3p/2 give me. I dont know y or x just siny and sinx ( through cos(A-B) being 4/5)
     
  7. Nov 16, 2015 #6

    Mentallic

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    What I'm trying to get you to understand is that if
    [tex]\cos(A)=B[/tex]
    where A and B are anything at all (avoiding impossibilities such as B>1 for example and others that I'll ignore), then it's always going to be the case that
    [tex]A=\arccos(B)[/tex]

    so if you instead have that [itex]\cos(x-3\pi/2)=-4/5[/itex] then using the exact rules from above, where [itex]A=x-3\pi/2[/itex] and [itex]B=-4/5[/itex], what do you get? But we aren't done by that point, because we need to solve x on its own, which just requires one more simple step.

    With the other problem, you're going to have to give me the exact question that you're trying to solve, word for word. At the moment, it's not a question but just an expression. I also don't know what you mean when you say "how do I get sin(x/2) and cos(5x/4)" because it makes no sense in this context. You showed me the expression [itex]\sin(x/3)*\cos(5x/4)[/itex]. It equals exactly that at the moment. Plug in x=0 and you'll have a value, plug in x=1 and you'll get another value. There's nothing to be done here.
     
  8. Nov 16, 2015 #7

    diredragon

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    It says evaluate the x from above expression and substitute into expression sin(x/2)*(cos5x/2) to get some number. This is correct form as i made an error above. So now i see that x - 3p/2 = arccos(-4/5) but without a calculator how would i get what x is?
     
  9. Nov 16, 2015 #8

    Mentallic

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    That was definitely an important detail that you excluded until now. The procedure for this situation would be to get the exact value of x (make x the subject), then plug it into the second expression as it is exactly, simplify a bit if possible, then use your calculator to give a numerical approximation.

    The one part you need to include though is that [itex]\pi < x < \pi/2[/itex], but this doesn't make sense because x can't be greater than [itex]\pi[/itex] and simultaneously less than [itex]\pi/2[/itex] so I'll assume you meant [itex]\pi/2 < x < \pi[/itex]. If you're unsure. Draw a graph of y=cos(x-3pi/2) and find a few x values where its y value equals -4/5. Choose the one that lands within the desired domain of x.
     
  10. Nov 16, 2015 #9

    diredragon

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    I can solve a problem using a calculator but i posted it so i can solve it by means of some simplifications and identityes, its from a recent test in which no calculators were used.
     
  11. Nov 17, 2015 #10

    Mentallic

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    Then you've likely forgotten the exact question - which is also evident in the fact that you changed the question three times throughout the thread - or that you weren't expected to give a numerical answer. What you have at the moment isn't going to simplify easily at all, and this is generally the case in trigonometry.

    What you may be interested in is trying to figure out what a more simple form of [itex]\cos(x-3\pi/2)[/itex] is equivalent to. When sines and cosines are shifted by a factor of [itex]\pi/2[/itex], they alternate between each other.
     
  12. Nov 17, 2015 #11

    SammyS

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    I suggest you give a complete statement of the complete problem, with complete sentences and using as clear mathematics as you possibly can.
     
  13. Nov 17, 2015 #12

    diredragon

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    Knowing that cos(x - 3p/2) = -4/5 and (p/2)<x<p the value of the expression sin(x/2)cos(5x/2) equals what?
    The solution i know must be is 82/125 but how to get there i dont know
    Sorry for the mistakes before i was careless. This is complete problem read.
     
  14. Nov 17, 2015 #13

    Samy_A

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    Hint:

    ##\sin(2x)=\sin(\frac{5}{2}x-\frac{1}{2}x)##
    ##\sin(3x)=\sin(\frac{5}{2}x+\frac{1}{2}x)##
     
  15. Nov 17, 2015 #14

    Ray Vickson

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    Is ##p## some other variable?
     
  16. Nov 17, 2015 #15

    SammyS

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    Good question.

    I'm pretty sure that Mentallic, Samy_A, and I (SammyS) have been assuming that p is indeed pi, a.k.a. π .
     
  17. Nov 17, 2015 #16

    diredragon

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    No :) its pie
     
  18. Nov 17, 2015 #17

    Ray Vickson

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    I suppose you mean 'pi', not 'pie'. Just writing pi instead of p would make all the difference!
     
  19. Nov 17, 2015 #18

    diredragon

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    I have used the property of cos(A - B) = cosAcosB + sinAsinB to get sinx = 4/5 and then by sinx^2 + cosx^2 = 1 i get cosx= -3/5.
    Now for the sin(x/2)*cos(5x/2) i can substitute sin(x/2) with ((1-cosx)/2)^(1/2) and how to simplify cos((5x/2)) but thats just clearly the hearder way. Somehow this must simplify into 82/125
     
  20. Nov 18, 2015 #19

    Samy_A

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    Did you try this?
    Using
    ##\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)##
    ##\sin(a-b)=\sin(a)\cos(b)-\sin(b)\cos(a)##
    computing the two expressions in my hint and substracting them will give you ##\sin(\frac{1}{2}x)\cos(\frac{5}{2}x)## in terms of ##\sin(2x)## and ##\sin(3x)##, and these two can be computed in terms of ##\sin(x)=4/5## and ##\cos(x)=-3/5##.
     
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