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Trigonometric Identities

  1. Mar 13, 2005 #1
    I'm having problems with it at school lately, im not gonna layout every single problem and ask for help. Im just wondering if there is a better approach to it rather than trying to solve one side in order to get it to equal the other side. For example, sin^2 x + cos^2 x = -cos^2 x - sin^2 x. Something like that. It bothers me how i just can't seem to be able to solve them. It's just like proves, where you either know it or you dont. Are there any easier way to solve them other than memorizing the trig identities?

    here's 1 that i jsut cant seem to solve for:

    sin^2 x / 1 - cos^2 x = 1 + cos^2 x / sin^2 x

    we are trying to show that they are equal.
  2. jcsd
  3. Mar 13, 2005 #2
    [tex]sin^{2}x + cos^{2}x = 1[/tex]

    You can rearrange that to [tex]sin^{2}x = 1 - cos^{2}x[/tex]

    You will need to memorize the trig identities, unless you want to derive them everytime, which I wouldn't suggest.

    EDIT: actually, i'm not sure if those are equal. is that supposed to be a "plus" on the right side of the equation?
  4. Mar 13, 2005 #3
    is never true! It implies [tex]\sin^2{x} + \cos^2{x} = 0 \Longrightarrow \sin{x} = 0 \ \mbox{and} \ \cos{x} = 0[/tex] which has no solutions.

    There are some easy ways to derive trig identities, but these mainly use the exponential forms of the trig functions. If you like I can demonstrate, but unless you are familiar with Euler's formula, [tex]e^{ix} = \cos{x} + i\sin{x}[/tex], it probably won't help you right now.

    Some identities are obvious if you draw a unit circle and define [tex]\sin{ \theta }, \ \cos{\theta }[/tex] to be the y- and x- coordinates of the point on the circle at angle [tex] \theta [/tex], respectively. For example, by the Pythagorean theorem doing this immediately leads to [tex] \sin^2{x} + \cos^2{x} = 1[/tex], and another obvious one is [tex]\tan{x} = \frac{\sin{x}}{\cos{x}}[/tex].

    The equation that you put in your post,

    [tex]\frac{\sin^2{x}}{1} - \cos^2{x} = 1 + \frac{\cos^2{x}}{\sin^2{x}}[/tex]

    is not true in general.

    Are you sure you didn't mean

    [tex] \frac{\sin^2{x}}{1-\cos^2{x}} = \frac{1 - \cos^2{x}}{\sin^2{x}}[/tex]

    ? That can be proved using only identities trivially derived from [tex]\sin^2{x} + \cos^2{x} = 1[/tex]
    Last edited: Mar 13, 2005
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