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Find sin x expressed by a and b

  1. Sep 13, 2016 #1
    1. The problem statement, all variables and given/known data

    Firstly, sorry for the probably weird title. I have no idea how to title this problem, but hopefully my explanation is better. =)

    Given $$cos x = \frac {2\sqrt{ab}} {a + b},$$ where x is in the first quadrant and a + b ≠ 0, ab > 0.
    Calculate sin x expressed by a and b.

    2. Relevant equations

    Sine formula
    Pythagorean theorem

    3. The attempt at a solution

    Now I've come up with what I think is a solution, but for some reason it feels weird, and like I'm not thinking correctly when it comes to the task.

    I used Pythagorean theorem combined with the information is given in the cos-equation.
    If ##2\sqrt{ab}## is a side, B, of a triangle, and ##a + b## is the hypotenuse, then the last side, C, of the triangle should be:

    $$C = \sqrt {(a+b)^2-(2\sqrt{ab})^2}$$

    Using this in the sine formula, I get the following:

    $$sin x = \frac {\sqrt{(a+b)^2-(2\sqrt{ab})^2}} {a+b}$$

    So the question is, does this sound like a reasonable answer to this problem, or am I misunderstanding what I'm supposed to do?

    Thanks for any help. =)
     
  2. jcsd
  3. Sep 13, 2016 #2

    Math_QED

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    Use the identity ##\cos^2 x+ \sin^2 x= 1 ##. This is much easier.

    Note that this formula is derived directly from the Pythagorean theorem.

    Edit: Your answer seems correct. Redo the exercise with the formula I recommended and notice that it's easier.
     
    Last edited: Sep 13, 2016
  4. Sep 13, 2016 #3
    Thanks, I have a tendency to redo my assignments wrongly because I felt the original, and often correct, answer was wrong.

    I'm a bit unsure as to how to use the ##cos^2 x + sin^2 x## part. For instance, for the next part of this task, I have to find the exact value of ##sin 2x## and ##cos 2x## given that the values of ##a = 4##, and ##b =1##.

    For ##sin 2x##, I found a formula that says ##sin 2x = 2 sin x cos x##, where I just added the variables I already found, and ended up with ##\frac {24}{25}##.

    However, for ##cos 2x##, the formula says ##cos 2x = 2 cos^2 x - 1##, and so I'm a bit confused as to how both of those would work out with the variables I have. How exactly should I use the variables I have when the formula is either ##cos 2x## or ##cos^2 x##?
     
  5. Sep 13, 2016 #4

    SammyS

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    The expression under the radical can be simplified in: ##\ \displaystyle sin x = \frac {\sqrt{(a+b)^2-(2\sqrt{ab})^2}} {a+b}##
     
  6. Sep 13, 2016 #5
    Would that make it: ##sin x = \frac {\sqrt {(a + b)^2 - 4ab}} {a + b}##?

    Or maybe it can be shortened into: ##sin x = \frac {\sqrt {a^2 - 2ab + b^2}} {a + b}##
     
  7. Sep 13, 2016 #6

    SammyS

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    Perhaps go further and factor a2 - 2ab + b2 .

    Added in Edit:
    I didn't notice earlier, but this is just simplifying what you called C .
     
    Last edited: Sep 13, 2016
  8. Sep 14, 2016 #7

    Math_QED

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    You can simplify things even further in your expression for ##\sin x ## and end up with a nice expression.
    ##a^2 -2ab + b^2 = \dots ##

    Why are you hesistant to use ##\cos^2 x + \sin^2 x = 1##. Just solve it for ##\sin x##:

    ##\sin x = \pm\sqrt{1 - \cos^2 x}##

    Now, as x is in the first quadrant, pick the plus sign as solution as ##\sin x ## is positive whenever ## x \in [0, \pi]##

    For ##\cos (2x)##, you know that ##\cos (2x) = 2\cos^2 x - 1## You are given ##\cos x## in terms of a and b, thus substitute that in ##2\cos^2 x - 1##
     
    Last edited: Sep 14, 2016
  9. Sep 14, 2016 #8
     
  10. Sep 14, 2016 #9

    Math_QED

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    Thus, I'll go into detail and write out every single step. Your work here is to understand every single step. That really is the only way to learn mathematics. If you have any question, don't hesitate to ask.

    We have ##\cos^2 x + \sin^2 x = 1##
    ##\Rightarrow \sin^2 x = 1 - \cos^2 x##
    ##\Rightarrow \sin x = \pm\sqrt{1 - \cos^2 x}##

    Given that ##x \in [0, \frac{\pi}{2}]## (first quadrant), we pick the + sign (as ## \sin x ## is positive there)

    ##\Rightarrow \sin x = \sqrt{1 - \cos^2 x}## (1)

    But, we have ##\cos x = \frac{2\sqrt{ab}}{a+b}##. Substituting this in (1) we get:

    ##\sin x = \sqrt{1 - (\frac{2\sqrt{ab}}{a+b})^2} = \sqrt{\frac{(a+b)^2 - 4ab}{(a+b)^2}}##
    ##= \sqrt{\frac{a^2 - 2ab + b^2}{(a+b)^2}} = \frac{\sqrt{a^2 - 2ab + b^2}}{|a+b|}##

    Since we are given that ##x \in [0, \frac{\pi}{2}]##, we obtain, because ##\cos x = \frac{2\sqrt{ab}}{a+b} ## must be positive that ##a + b > 0##

    And thus, we are left with :

    ##\sin x = \frac{\sqrt{a^2 - 2ab + b^2}}{a+b}##

    Now, you can still factor ##a^2 - 2ab + b^2## and end up with a nice expression without square root.

    I'm sure you can take care of the problems with the ##\sin (2x)## and ##\cos (2x)## yourself now.
     
    Last edited: Sep 14, 2016
  11. Sep 14, 2016 #10
    This is really detailed, and well explained. This makes a lot more sense than the notes our lecturer uses. =)

    I understand what you mean with ##cos^2 x + sin^2 x = 1##, and it is a lot easier. There's only one part of the equation I'm a bit confused with in your method:

    ##sin x = \sqrt {1 - (\frac {2 \sqrt {ab}} {a+b})^2 } = \sqrt { \frac {(a+b)^2-4ab} {(a+b)^2}}##

    Where did the ##1 -## go? It seems to change into ##(a+b)^2##, but I fail to see exactly where you get that from? The end result is the same that I got with my, more lengthy, version.

    In the end, I factor ##a^2 - 2ab + b^2## and end up with ##sin x = \frac {a - b} {a + b}##

    And using this, I was able to do the ##sin(2x)## and ##cos(2x)## problem as well.

    Thank you again for your time. I find calculus challenging, but rewarding. =)
     
  12. Sep 14, 2016 #11

    Math_QED

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    Well,

    ##1 - (\frac {2 \sqrt{ab}} {a+b})^2 = 1 - \frac{4ab}{(a+b)^2} = \frac{(a+b)^2}{(a+b)^2} - \frac{4ab}{(a+b)^2} = \frac{(a+b)^2 - 4ab}{(a+b)^2}##

    Well, this is almost correct. It's a common mistake.

    What is ##\sqrt{(a-b)^2}## equal to?

    Very good.

    I'm glad I could help.
     
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