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Math Help - Proving Indentity

  1. #1
    milkywayio
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    Proving Indentity

    i have no idea how to do these four problems please help me!!!

    1. cos 2 x = cotx - tanx
    sinx cosx

    2. (cosx-sinx)^2 = 1 - sin 2 x


    3. secx - sinxtanx = cosx


    4. cos(x+y)cosy + sin(x+y) sin y = cosx
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  2. #2
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    Quote Originally Posted by milkywayio
    i have no idea how to do these four problems please help me!!!

    1. cos 2 x = cotx - tanx
    sinx cosx
    For Left Hand Side
    \frac{\cos 2x}{\sin x\cos x}
    Express as,
    \frac{2\cos 2x}{2\sin x\cos x}
    Double angle for,
    \frac{2\cos 2x}{\sin 2x}
    Thus,
    2\cot 2x

    For Right Hand Side,
    \cot x-\tan x
    Use reciprocal identities to get,
    \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}
    Add fractions,
    \frac{\cos^2x-\sin^2x}{\sin x\cos x}
    Use double angle for cosine,
    \frac{\cos 2x}{\sin x\cos x}
    Express as,
    \frac{2\cos 2x}{2\sin x\cos x}
    Use double angle for sine,
    \frac{2\cos 2x}{\sin 2x}
    Use identities,
    2\cot 2x
    Thus, the left hand and the right hand give the same result
    Q.E.D.
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  3. #3
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    Quote Originally Posted by milkywayio

    2. (cosx-sinx)^2 = 1 - sin 2 x
    Open the Left hand side remember to use FOIL,
    (\cos x-\sin x)^2=\cos^2x-2\sin x\cos x+\sin^2x
    Notice, that
    \cos^2x+\sin^2x=1 and -2\sin x \cos x=-2\sin x
    Thus,
    (\cos x-\sin x)^2=1-\sin 2x
    Q.E.D.
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  4. #4
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    Quote Originally Posted by milkywayio
    3. secx - sinxtanx = cosx
    Express tangent through sine and cosine,
    \sec x-\sin x\cdot \frac{\sin x}{\cos x}
    Express, secant through cosine,
    \frac{1}{\cos x}-\sin x\cdot \frac{\sin x}{\cos x}
    Thus, add fractions
    \frac{1-\sin^2x}{\cos x}
    Notice that 1-\sin^2x=\cos^2x
    Thus,
    \frac{\cos^2x}{\cos x}=\cos x
    Thus,
    \sec x-\tan x\sin x=\cos x
    Q.E.D.
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  5. #5
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    Quote Originally Posted by milkywayio
    4. cos(x+y)cosy + sin(x+y) sin y = cosx
    Call, z=x+y
    Thus,
    \cos z\cos y+\sin z\sin y
    Notice that this is the cosine for difference of two angles, thus,
    \cos (z-y)
    Substitute that value of z thus,
    \cos (x+y-y)=\cos x
    Thus,
    \cos (x+y)\cos y + \sin (x+y) \sin y=\cos x
    Q.E.D.
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