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Homework Help: Please help me with this Trigonometry Question?

  1. Nov 10, 2009 #1
    Determine an exact value of b such that csc(6b+π/8)=sec(2b-π/8).

    Just to clarify, π is pi.

    This question deals with equivalent trigonometric expressions.

    I have absolutely no idea how to solve this problem.

    Please provide me with explanations. Thanks in advance.
  2. jcsd
  3. Nov 10, 2009 #2


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    Can you see how you could perhaps make use of the fact that:

    [tex]sin(\theta) \equiv cos(\frac{\pi}{2}-\theta)[/tex]
  4. Nov 10, 2009 #3
    Sorry, I still don't understand it. Can you please go further into explaining it?
  5. Nov 10, 2009 #4


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    Using the definition of csc(x) and sec(x), can you see how the equation can be rearranged to the following?:


    Now can you see how the identity that i posted above could help?
  6. Nov 13, 2009 #5
    Sorry I still don't understand how you got this from the original equation. Also, how do I proceed with the rest of the question?
  7. Nov 13, 2009 #6


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    [tex]csc(x)=\frac{1}{sin(x)}[/tex] and [tex]sec(x)=\frac{1}{cos(x)}[/tex] right? Ive used those identities to get the equation to the form i posted above.

    Then the identity:
    sin(\theta) \equiv cos(\frac{\pi}{2}-\theta)

    Tells me that i can now change the equation once more to:


    Now can you see what to do?
  8. Nov 13, 2009 #7


    Staff: Mentor

    Well, you have me stumped, too. It's not intuitively obvious to the casual observer that
    [tex] sin(6b+\frac{\pi}{8})=sin(\frac{\pi}{2}-(2b-\frac{\pi}{8})) [/tex]

    from the hint that [itex]sin(\theta) = cos(\pi/2 - \theta)[/itex]

    From the original equation you get
    [tex]\frac{1}{sin(6b + \pi/8)}~=~\frac{1}{cos(2b - \pi/8)}[/tex]

    The equation above comes directly from the definition of the csc and sec functions.

    By cross-multiplying, we get to
    [tex]sin(6b + \pi/8)~=cos(2b - \pi/8)[/tex]

    The key in the following step is to recognize that 6b + pi/8 and 2b - pi/8 can be written as the sum and difference, respectively, of the same two quantities: 4b and 2b + pi/8. If you're not convinced add 4b and 2b + pi/8 and see what you get, and then subtract 2b + pi/8 from 4b and see what you get.

    If you're still with me, what we have now is
    [tex]sin(4b + (2b + \pi/8))~=cos(4b - (2b + \pi/8))[/tex]

    Now, write a new equation that uses the sum rule for sine and cosine, and you should get an equation that is identically true. Since each equation is equivalent to the preceding equation, all the way back to the first one you started with, that means that it must be identically true as well, and you're done.

    This is a standard technique in proving trig (or other) identities: working with one or both sides until you get an identity. If each equation along the way is equivalent to the preceding one, then your first equation is an identity as well.

    Note that this doesn't work if your steps aren't reversible (i.e., that a subsequent equation is merely implied by the previous one, rather than being equivalent to it). For example
    x = - 2
    x2 = 4
    x = 2, or x = -2

    The first equation is NOT equivalent to the last one, since x = 2 is a solution of the last one (obviously), but the first. The step of squaring both sides is not a reversible step, and can bring in new solutions that aren't solutions of the beginning equation.
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