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Math Help - verifying trig identities

  1. #1
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    verifying trig identities

    How would you verify sec-1/tan=tan/sec+1? also how would you verify 8sin^2cos^2=1-cos4?
    Last edited by derek1008; November 13th 2012 at 02:01 PM.
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    Re: verifying trig identities

    sec is the abbreviation for second

    tan is what one may get while visiting the beach

    sin is a transgression

    cos is the nickname of a famous black comedian

    ... in other words, a trig function without an argument means nothing.


    now, if you meant ...

    \frac{\sec{x} - 1}{\tan{x}} = \frac{\tan{x}}{\sec{x} + 1}

    ... then I recommend multiplying the left side of the equation by \frac{\sec{x}+1}{\sec{x}+1}, then using the Pythagorean identity
    1+\tan^2{x} = \sec^2{x}


    now, can you be more clear with this equation ...

    how would you verify 8sin^2cos^2=1-cos4?
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    Re: verifying trig identities

    The question just has 8(sin^2 x)(cos^2 x)= 1-cos 4x.
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    Re: verifying trig identities

    Quote Originally Posted by derek1008 View Post
    The question just has 8(sin^2 x)(cos^2 x)= 1-cos 4x.
    much better ...

    1 - \cos(4x) =

    2\left(\frac{1-\cos(4x)}{2}\right) =

    2\sin^2(2x) =

    2(2\sin{x}\cos{x})^2 =

    8\sin^2{x}\cos^2{x}
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    Re: verifying trig identities

    Quote Originally Posted by skeeter View Post
    ... in other words, a trig function without an argument means nothing.
    so if we fight over cosines, then it's meaningful? :P
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    Re: verifying trig identities

    Quote Originally Posted by Deveno View Post
    so if we fight over cosines, then it's meaningful? :P
    Deveno, you're a much better mathematician than comedian ... don't quit your day job.
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    Re: verifying trig identities

    but there's a method to my madness: the definition of a function f doesn't really "depend" on its argument.

    for example, we can talk of the "squaring function" f (usually written as f(x) = x2) as [ ]2 (although this is a bit confusing).

    for example (and more to the point in this case):

    the fundamental trig identity can be written as:

    cos2 + sin2 = 1, meaning:

    cos2(x) + sin2(x) is the CONSTANT function 1, no matter WHAT "x" is (and the symbol we use in place of x really doesn't matter, it could be t, or θ, or Y, it's a "dummy variable").

    the equation:

    cos2(x) + sin2(x) = 1 is an equality of two NUMBERS.

    the equation:

    cos2 + sin2 = 1 is an equality between two FUNCTIONS.

    the first means it just happens to be true for some particular x (which turns out to be all of them).

    the second means the two functions (and we sum two functions f+g by summing the values f(x) + g(x) at each point x) sum to a constant function.

    granted, people are used to confusing a function with its value at a point. and your point is well taken that the original poster was probably just "being lazy" in omitting the argument.

    still, it's not entirely wrong.

    (how's THAT for an argument? see what i did there?)
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    Re: verifying trig identities

    sure do ... but I'd still mark off points for leaving out the x (or t or theta). Plato gets upset if one writes \cos{x} instead of \cos(x) ... we all have our quirks.
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    Re: verifying trig identities

    well, there is a reason for these things. for example, in modelling a physical situation, if one writes cos(t), the "t" is there presumably to remind you which UNITS "t" is in (often, in physics for example, one may have equations with respect to several variables at once, but one is considering some of the variables as "variable constants" or parameters chosen before-hand, whereas the independent variable might take on any value in the domain, so you can have cos(ωt), where ω is "fixed" but "t" is a "true variable", which might be important when differentiating).

    of course, trigonometry is peculiar, too, it is one place where cos2(x) means: (cos(x))(cos(x)), and not cos(cos(x)), like it would in other areas.
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    Re: verifying trig identities

    I don't get why you multiplied it by secx+1/ secx+1?
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    Re: verifying trig identities

    Quote Originally Posted by WhatthePatel View Post
    I don't get why you multiplied it by secx+1/ secx+1?
    because (\sec{x}-1)(\sec{x}+1) = \sec^2{x} - 1 = \tan^2{x}
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