Results 1 to 5 of 5

Thread: Proving Indentity

  1. #1
    milkywayio
    Guest

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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
    $\displaystyle \frac{\cos 2x}{\sin x\cos x}$
    Express as,
    $\displaystyle \frac{2\cos 2x}{2\sin x\cos x}$
    Double angle for,
    $\displaystyle \frac{2\cos 2x}{\sin 2x}$
    Thus,
    $\displaystyle 2\cot 2x$

    For Right Hand Side,
    $\displaystyle \cot x-\tan x$
    Use reciprocal identities to get,
    $\displaystyle \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}$
    Add fractions,
    $\displaystyle \frac{\cos^2x-\sin^2x}{\sin x\cos x}$
    Use double angle for cosine,
    $\displaystyle \frac{\cos 2x}{\sin x\cos x}$
    Express as,
    $\displaystyle \frac{2\cos 2x}{2\sin x\cos x}$
    Use double angle for sine,
    $\displaystyle \frac{2\cos 2x}{\sin 2x}$
    Use identities,
    $\displaystyle 2\cot 2x$
    Thus, the left hand and the right hand give the same result
    Q.E.D.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by milkywayio

    2. (cosx-sinx)^2 = 1 - sin 2 x
    Open the Left hand side remember to use FOIL,
    $\displaystyle (\cos x-\sin x)^2=\cos^2x-2\sin x\cos x+\sin^2x$
    Notice, that
    $\displaystyle \cos^2x+\sin^2x=1$ and $\displaystyle -2\sin x \cos x=-2\sin x$
    Thus,
    $\displaystyle (\cos x-\sin x)^2=1-\sin 2x$
    Q.E.D.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by milkywayio
    3. secx - sinxtanx = cosx
    Express tangent through sine and cosine,
    $\displaystyle \sec x-\sin x\cdot \frac{\sin x}{\cos x}$
    Express, secant through cosine,
    $\displaystyle \frac{1}{\cos x}-\sin x\cdot \frac{\sin x}{\cos x}$
    Thus, add fractions
    $\displaystyle \frac{1-\sin^2x}{\cos x}$
    Notice that $\displaystyle 1-\sin^2x=\cos^2x$
    Thus,
    $\displaystyle \frac{\cos^2x}{\cos x}=\cos x$
    Thus,
    $\displaystyle \sec x-\tan x\sin x=\cos x$
    Q.E.D.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by milkywayio
    4. cos(x+y)cosy + sin(x+y) sin y = cosx
    Call, $\displaystyle z=x+y$
    Thus,
    $\displaystyle \cos z\cos y+\sin z\sin y$
    Notice that this is the cosine for difference of two angles, thus,
    $\displaystyle \cos (z-y)$
    Substitute that value of $\displaystyle z$ thus,
    $\displaystyle \cos (x+y-y)=\cos x$
    Thus,
    $\displaystyle \cos (x+y)\cos y + \sin (x+y) \sin y=\cos x$
    Q.E.D.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving indentity..
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Feb 21st 2010, 08:00 AM
  2. Somebody plz try out proving this indentity....
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: Feb 10th 2010, 04:42 AM
  3. i need help proving the indentity
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Nov 5th 2009, 08:49 PM
  4. trigo indentity
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Aug 5th 2007, 06:19 AM
  5. Trig indentity
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: May 2nd 2006, 01:30 PM

Search Tags


/mathhelpforum @mathhelpforum