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Homework Help: Trig principle values problem

  1. May 14, 2006 #1
    I have a trig problem that i need help with. I have done a) & b) but not c) because i don't get it.

    Solve the following for:

    a) principle values of x
    b) 0°≤x≥180°
    c) all values of x

    1. 2sin²x-1=0

    my work:

    1a. principle values of x

    sinx(2sin)-1 = 0




    1b. 0°≤x≥180°


    sin½= 30°,90°

    1c. all values of x
    Last edited: May 14, 2006
  2. jcsd
  3. May 14, 2006 #2
    Your initial working should be the same for all 3 parts, it's just the difference in the range of x for each part that results in you having different answers.

    In your working, you wrote:
    sinx(2sinx)=1 **
    sinx=1 or 2sinx=1
    sinx=1 or sinx=0.5

    This does not seem right! You can only use this method if the right hand side of equation ** equals zero. Can you think of another way to do this? Note that there is only 1 sinx term in the original equation.

    Also, do you know what the principal region for sinx is? For part b, I believe the range should be 0°≤x≤180°. And finally, for part c, you need to find a general solution which can "generate" all the possible values of x.

    Please also take note of your expressions. For your working in part b, expressions like "2sin(sinx)=1" and "sin½= 30°,90°" don't seem very right.
    Last edited: May 14, 2006
  4. May 15, 2006 #3


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    Science Advisor

    Please, please, please, learn to write exactly what you mean! I have no idea what "sinx(2sin)" could mean! If you don't want to use Tex or html codes, you could just say sin^2(x)= 1.

    Okay, this at least makes sense but I don't see where you got it from

    sine is a function! It makes no sense to say "2sin/2" or
    "sin= 1/2".
    sin(x)= 1/2. But you still haven't answered the question. What is x?

    I have absolutely no idea what you are doing here!
    The "principal" value is the single value of x between [itex]-\frac{\pi}{2}[/itex] and [itex]\frac{\pi}{2}[/itex] that gives sin(x)= 1/2. Since that is positive, yes, x must be positive. You still haven't said what that is. You can't do (b) until you have done (a).

    Once again, makes no sense! According to my calculator, sin(1/2 (radian))= 0.47942553860420300027328793521557. If you mean arcsin(1/2) or
    sin-1(1/2), say so! In any case, sin(90°)= 1, not 1/2.
    However, you want 2sin2(x)- 1= 0 or [itex]sin(x)= \frac{1}{\sqrt{2}}[/itex]. My calculator gives, as principal value (part (a) again!), x= 45°. You should know that sin(180°- x)= sin(x) so the other value, in [itex]0\le x \le 180[/itex] is 180- 45= 135°.

    sine is periodic with period 360°. Take the two values you got for x between 0 and 180° (the only two solution between 0 and 360°) and add 360n to them.
  5. May 15, 2006 #4
    OOPS!! i guest i made a lot of mistakes.:frown:
    ill go back over my work and reread the chapter, and thanks for checking my answers.

    oh yeah, and it was 0≤x≤180................... i am really sorry i mistyped it.:frown:
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