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Thread: Trig identities

  1. June 5th 2010, 05:19 AM #1
    RogueDemon
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    Trig identities

    14. a) Which equations are not identities? Justify your answers.
    b) For those equations that are identities, state any restrictions on the variables.

    i) (1 - cos^2x)(1 - tan^2x) = \frac{sin^2x - 2sin^4x}{1 - sin^2x}

    Left Side
    = (1 - cos^2x)(1 - tan^2x)
    = 1 - tan^2x - cos^2x + (-cos^2x)(-tan^2x)
    = 1 - cos^2x - tan^2x + (cos^2x)(tan^2x)
    = sin^2x - tan^2x + sin^2x
    = (cos^2x)(tan^2x) - tan^2x + (cos^2x)(tan^2x)
    = tan^2x(cos^2x - 1 + cos^2x)
    = tan^2x(-sin^2x + cos^2x) Edit: Was an error made here?
    = tan^2x(-1)
    = -tan^2x

    Right Side:
    = \frac{sin^2x - 2sin^4x}{1 - sin^2x}
    = \frac{sin^2x - 2(sin^2x)(sin^2x)}{cos^2x}
    = \frac{sin^2x(1 -2(1)(1))}{cos^2x}
    = \frac{sin^2x(1 -2)}{cos^2x}
    = \frac{sin^2x(-1)}{cos^2x}
    = -tan^2x

    Left Side = Right Side
    Therefore, this equation is an identity.

    When I showed this solution to my teacher, I was told that had I used identities that we haven't learned yet. She said that on the exam, we were to only be able to use reciprocal identities and the Pythagorean identity sin^2x + cos^2x = 1. She also told me that using these identities, I can make the LS and RS equal however with different values. Is there another way of proving this equation to be an identity?

    Also, how do I know whether or not an equation will eventually end up as LS = RS? The only reason I know it with the few assigned questions is because there is an answer at the back of the book. Though on an exam I won't have access to these answers. Is it just trial and error? Because if it is, I can be going on for a pretty long time just trying to solve one question.
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  2. June 5th 2010, 06:05 AM #2
    skeeter
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    (1-\cos^2{x})(1-\tan^2{x}) =<br />

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

    \sin^2{x} - \frac{\sin^4{x}}{\cos^2{x}} =<br />

    \frac{\sin^2{x}\cos^2{x}}{\cos^2{x}} - \frac{\sin^4{x}}{\cos^2{x}} =

    \frac{\sin^2{x}\cos^2{x} - \sin^4{x}}{\cos^2{x}} =

    \frac{\sin^2{x}(1 - \sin^2{x}) - \sin^4{x}}{1 - \sin^2{x}} =

    \frac{\sin^2{x} - \sin^4{x} - \sin^4{x}}{1 - \sin^2{x}} =

    \frac{\sin^2{x} - 2\sin^4{x}}{1 - \sin^2{x}}
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  3. June 5th 2010, 06:10 AM #3
    e^(i*pi)
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    Quote Originally Posted by RogueDemon View Post
    14. a) Which equations are not identities? Justify your answers.
    b) For those equations that are identities, state any restrictions on the variables.

    i) (1 - cos^2x)(1 - tan^2x) = \frac{sin^2x - 2sin^4x}{1 - sin^2x}

    Left Side
    = (1 - cos^2x)(1 - tan^2x)
    = 1 - tan^2x - cos^2x + (-cos^2x)(-tan^2x)
    = 1 - cos^2x - tan^2x + (cos^2x)(tan^2x)
    = sin^2x - tan^2x + sin^2x
    = (cos^2x)(tan^2x) - tan^2x + (cos^2x)(tan^2x)
    = tan^2x(cos^2x - 1 + cos^2x)
    = tan^2x(-sin^2x + cos^2x) Edit: Was an error made here?
    = tan^2x(-1)
    = -tan^2x

    Right Side:
    = \frac{sin^2x - 2sin^4x}{1 - sin^2x}
    = \frac{sin^2x - 2(sin^2x)(sin^2x)}{cos^2x}
    = \frac{sin^2x(1 -2(1)(1))}{cos^2x}
    = \frac{sin^2x(1 -2)}{cos^2x}
    = \frac{sin^2x(-1)}{cos^2x}
    = -tan^2x

    Left Side = Right Side
    Therefore, this equation is an identity.

    When I showed this solution to my teacher, I was told that had I used identities that we haven't learned yet. She said that on the exam, we were to only be able to use reciprocal identities and the Pythagorean identity sin^2x + cos^2x = 1. She also told me that using these identities, I can make the LS and RS equal however with different values. Is there another way of proving this equation to be an identity?

    Also, how do I know whether or not an equation will eventually end up as LS = RS? The only reason I know it with the few assigned questions is because there is an answer at the back of the book. Though on an exam I won't have access to these answers. Is it just trial and error? Because if it is, I can be going on for a pretty long time just trying to solve one question.
    When proving identities it is good practice to only change one side and manipulate it to get the other.

    <br /> (1 - cos^2x)(1 - tan^2x) = \frac{sin^2x - 2sin^4x}{1 - sin^2x}<br />

    I have put my steps as spoilers along with explanation.

    From the LHS

    Use FOIL

    Spoiler:
    1-\tan^2(x) - \cos^2(x) + \sin^2(x)


    Multiply through by \frac{\cos^2(x)}{\cos^2(x)} as this is the LCD of the expression.

    Spoiler:
    \frac{\cos^2(x) - \sin^2(x) - \cos^4(x) + \sin^2(x) \cos^2(x)}{\cos^2(x)}


    Using the identity \sin^2(x) + \cos^2(x) = 1 rewrite any cos terms as sin.

    Spoiler:
    \frac{(1-\sin^2(x)) - \sin^2(x) - (1-\sin^2(x))(1-\sin^2(x)) + \sin^2(x)(1-\sin^2(x))}{1- \sin^2(x)}


    Expand the brackets

    Spoiler:
    \frac{1-2\sin^2(x) - 1 + 2\sin^2(x) - \sin^4(x) + \sin^2(x) - \sin^4(x)}{1-\sin^2(x)}


    Collect like terms and simplify

    Spoiler:
    \frac{\sin^2(x) - 2\sin^4(x)}{1-\sin^2(x)}
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  4. June 5th 2010, 07:02 AM #4
    Soroban
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    Hello, RogueDemon!

    We should work on one side only
    . . and try to make it equal the other side.

    14. a) Which equations are not identities? Justify your answers.
    . . .b) For those equations that are identities, state any restrictions on the variables.

    i)\;(1 - \cos^2\!x)(1 - \tan^2\!x) \:=\: \frac{\sin^2\!x - 2\sin^4\!x}{1 - \sin^2\!x}

    I started with the left side:


    . . (1-\cos^2\!x)(1-\tan^2\!x). . . . . . . . . .Replace 1-\cos^2\!x with \sin^2\!x

    . . . . . =\;\sin^2\!x(1 - \tan^2\!x) . . . . . . . . Replace \tan^2\!x with \frac{\sin^2\!x}{\cos^2\!x}

    . . . . . =\;\sin^2\!x\left(1-\frac{\sin^2\!x}{\cos^2\!x}\right) . . . . . . . Subtract the fractions

    . . . . . =\;\sin^2\!x\left(\frac{\cos^2\!x - \sin^2\!x}{\cos^2\!x}\right). . . . . Replace \cos^2\!x with 1-\sin^2\!x

    . . . . . =\;\sin^2\!x\left(\frac{[1-\sin^2\!x]-\sin^2\!x}{\cos^2\!x}\right). . Combine like terms

    . . . . . =\;\sin^2\!x\left(\frac{1 - 2\sin^2\!x}{\cos^2\!x}\right). . . . . . .Multiply

    . . . . . =\;\frac{\sin^2\!x - 2\sin^4\!x}{\cos^2\!x} . . . . . . . . . .Replace \cos^2\!x with 1-\sin^2\!x

    . . . . . =\;\frac{\sin^2\!x-2\sin^4\!x}{1-\sin^2\!x}

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  5. June 5th 2010, 03:46 PM #5
    Ackbeet
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    There is an error in the OP. For the LHS, you cannot go from

    =\tan^{2}(x)(-\sin^{2}(x)+\cos^{2}(x))
    =\tan^{2}(x)(-1),

    because it is only when the \sin^{2}(x) and the \cos^{2}(x) have the same sign, that they sum to 1.

    Regards.
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