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Math Help - Trig Identities Help? Thanks!

  1. #1
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    Trig Identities Help? Thanks!

    Ok, I have a couple of questions on my homework that i just can't figure out how they work....


    "Verify that each equation is an identity"

    Cos^4(s)-sin^4(s) = cos2(s)

    I'm a little confused as the how this could equal each other? I get the double angle identity (cos^2(x)-sin^2(x)=cos2x) But how would powers to the 4th simplify down?

    And this one...
    (sin4t/4) = cos^3(t)sin(t)-sin^3(t)cos(t)

    I don't really even know where to begin on this one...We are working with double-angle identities and half-angle identities....

    Thanks to anyone who can help!!
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  2. #2
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    Quote Originally Posted by Justinram22 View Post
    Ok, I have a couple of questions on my homework that i just can't figure out how they work....


    "Verify that each equation is an identity"

    Cos^4(s)-sin^4(s) = cos2(s)

    I'm a little confused as the how this could equal each other? I get the double angle identity (cos^2(x)-sin^2(x)=cos2x) But how would powers to the 4th simplify down?

    And this one...
    (sin4t/4) = cos^3(t)sin(t)-sin^3(t)cos(t)

    I don't really even know where to begin on this one...We are working with double-angle identities and half-angle identities....

    Thanks to anyone who can help!!
    \cos^4{s} - \sin^4{s} = (\cos^2{s} + \sin^2{s})(\cos^2{s} - \sin^2{s})

     = 1(\cos^2{s} - \sin^2{s})

     = \cos^2{s} - \sin^2{s}

     = \cos{2s}.
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  3. #3
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    Quote Originally Posted by Justinram22 View Post
    Ok, I have a couple of questions on my homework that i just can't figure out how they work....


    "Verify that each equation is an identity"

    Cos^4(s)-sin^4(s) = cos2(s)

    I'm a little confused as the how this could equal each other? I get the double angle identity (cos^2(x)-sin^2(x)=cos2x) But how would powers to the 4th simplify down?

    And this one...
    (sin4t/4) = cos^3(t)sin(t)-sin^3(t)cos(t)

    I don't really even know where to begin on this one...We are working with double-angle identities and half-angle identities....

    Thanks to anyone who can help!!
    I assume for the second one, the left hand side is \frac{\sin{4t}}{4}...


    \cos^3{t}\sin{t} - \sin^3{t}\cos{t} = \sin{t}\cos{t}(\cos^2{t} - \sin^2{t})

     = \frac{1}{2}\cdot 2\sin{t}\cos{t}\cos{2t}

     = \frac{1}{2}\sin{2t}\cos{2t}

     = \frac{1}{4}\cdot 2\sin{2t}\cos{2t}

     = \frac{1}{4}\sin{4t}

     = \frac{\sin{4t}}{4}.
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    Thanks you!!!

    Could i ask one more? Sorry, i thought i had this one but can't seem to get it...

    cos2(y) = ((1-tan^2(y))/(1+tan^2(y))

    Thanks again!

    (Am i correct in thinking that you have to turn 1+tan^2(y) into sec^2(y))?
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    Quote Originally Posted by Justinram22 View Post
    Thanks you!!!

    Could i ask one more? Sorry, i thought i had this one but can't seem to get it...

    cos2(y) = ((1-tan^2(y))/(1+tan^2(y))

    Thanks again!

    (Am i correct in thinking that you have to turn 1+tan^2(y) into sec^2(y))?
    Use the identity \tan^2{x} + 1 = \sec^2{x}.


    \frac{1 - \tan^2{y}}{1 + \tan^2{y}} = \frac{1 - (\sec^2{y} - 1)}{\sec^2{y}}

     = \frac{2 - \sec^2{y}}{\sec^2{y}}

     = \frac{2}{\sec^2{y}} - 1

     = 2\cos^2{y} - 1

     = \cos^2{y} + \cos^2{y} - 1

     = \cos^2{y} - (1 - \cos^2{y})

     = \cos^2{y} - \sin^2{y}

     = \cos{2y}.
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