Results 1 to 8 of 8

Math Help - Trig Identity Help (and solving)

  1. #1
    Junior Member Hasan1's Avatar
    Joined
    Dec 2007
    Posts
    27

    Exclamation Trig Identity Help (and solving)

    Prove the following identities

    1.  \sin (\Pi + x) + \cos ( \frac{\Pi}{2} - x) + \tan (\frac{\Pi}{2} + x)  = -\cot x

    2.  \frac{\sin (4x) - \sin (2x)}{\sin (2x)} = \frac{\cos (3x)}{\cos (x)}

    3.  \cos (x) + \cos (2x) + \cos (3x) = \cos (2x) (1+ 2 \cos (x))

    Solve

    1.  \cos^2 (2x) + 2 \cos (2x) + 1 = 0     [-\Pi , \Pi]

    2.  \tan (4x) - \tan (2x) = 0 [0,\Pi]

    These are the ones I couldn't solve from a worksheet I got in class. Please help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,751
    Thanks
    651
    Hello, Hasan1!

    Here are the last two . . .


    1)\;\;\cos^2(2x) + 2\cos(2x) + 1 \:= \:0\qquad[-\pi,\:\pi]
    The left side is a square . . .

    . . \bigg[\cos(2x) + 1\bigg]^2 \:=\:0\quad\Rightarrow\quad\cos(2x) + 1 \:=\:0\quad\Rightarrow\quad\cos(2x) \:=\:-1

    Therefore: . 2x \:=\:-\pi,\:\pi\quad\Rightarrow\quad\boxed{ x \:=\:-\frac{\pi}{2},\:\frac{\pi}{2}}



    2)\;\;\tan(4x) - \tan(2x) \:=\: 0\qquad[0,\,\pi]
    This one gets messy . . .

    We know the identity: . \tan2\theta \:=\:\frac{2\tan\theta}{1-\tan^2\!\theta}
    Apply it to \tan4x

    We have: . \frac{2\tan2x}{1-\tan^2\!2x} - \tan(2x) \:=\:0


    Multiply by (1-\tan^2\!2x)\!:\;\;2\tan2x - \tan2x(1 - \tan^2\!2x) \:=\:0

    . . 2\tan2x -\tan2x + \tan^3\!2x \:=\:0\quad\Rightarrow\quad\tan^3\!2x + \tan2x \:=\:0


    Factor: . \tan2x\left[\tan^2\!2x + 1\right] \:=\:0

    The equation: . \tan^2\!2x + 1 \:=\:0 has no real roots.
    So we have: . \tan2x\:=\:0\quad\Rightarrow\quad 2x \:=\:0,\:\pi,\:2\pi\quad\Rightarrow\quad \boxed{x \:=\:0,\:\frac{\pi}{2},\:\pi}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Hasan1's Avatar
    Joined
    Dec 2007
    Posts
    27
    that second solving question has a nice solution, thanks!

    I guess I never saw the perfect square trinomial in that first one though

    thanks again
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Here's another solution for this one.

    Quote Originally Posted by Hasan1 View Post
    2.  \tan (4x) - \tan (2x) = 0 [0,\Pi]
    Previous knowledge:

    \tan \alpha - \tan \beta = \frac{{\sin \alpha }}<br />
{{\cos \alpha }} - \frac{{\sin \beta }}<br />
{{\cos \beta }} = \frac{{\sin (\alpha - \beta )}}<br />
{{\cos \alpha \cos \beta }}.

    Solution:

    \tan 4x - \tan 2x = 0 \implies \frac{{\sin 2x}}<br />
{{\cos 4x\cos 2x}} = 0\,\therefore \,\tan 2x = 0.

    Of course \cos4x\ne0.

    The rest follows from Soroban's conclusions.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by Hasan1 View Post
    2.  \frac{\sin (4x) - \sin (2x)}{\sin (2x)} = \frac{\cos (3x)}{\cos (x)}
    Just tackle the LHS.

    Apply the following identity

    \sin \alpha - \sin \beta = 2\sin \frac{{\alpha - \beta }}<br />
{2} \cdot \cos \frac{{\alpha + \beta }}<br />
{2}.

    The conclusion follows quickly.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by Hasan1 View Post
    3.  \cos (x) + \cos (2x) + \cos (3x) = \cos (2x) (1+ 2 \cos (x))
    The same trick as identity #2, apply

    \cos \alpha + \cos \beta = 2\cos \frac{{\alpha + \beta }}<br />
{2} \cdot \cos \frac{{\alpha - \beta }}<br />
{2}.

    --

    The last one is easy. Expand \sin(\pi+x) & observe that \cos \left( {\frac{\pi }<br />
{2} - x} \right) = \sin x. (This last identity is a fact that you actually should know.)

    And we're done
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member Hasan1's Avatar
    Joined
    Dec 2007
    Posts
    27
    Is there a way to do those identities without the sum to product formula, but the double angle formula instead?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    I think those identities were made to apply the formulae I gave you. It's a quickly way.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig Identity
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: February 17th 2011, 01:21 AM
  2. [SOLVED] Solving a trig identity
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: February 3rd 2011, 08:54 AM
  3. solving this identity
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: December 29th 2009, 02:24 PM
  4. Replies: 6
    Last Post: March 18th 2009, 04:10 AM
  5. trig identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 29th 2008, 10:09 PM

Search Tags


/mathhelpforum @mathhelpforum