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Thread: Trig Identity Help (and solving)

  1. #1
    Junior Member Hasan1's Avatar
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    Exclamation Trig Identity Help (and solving)

    Prove the following identities

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

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

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

    Solve

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

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

    These are the ones I couldn't solve from a worksheet I got in class. Please help
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  2. #2
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    Hello, Hasan1!

    Here are the last two . . .


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

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

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



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

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

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


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

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


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

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

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  3. #3
    Junior Member Hasan1's Avatar
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    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
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  4. #4
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    Here's another solution for this one.

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

    $\displaystyle \tan \alpha - \tan \beta = \frac{{\sin \alpha }}
    {{\cos \alpha }} - \frac{{\sin \beta }}
    {{\cos \beta }} = \frac{{\sin (\alpha - \beta )}}
    {{\cos \alpha \cos \beta }}.$

    Solution:

    $\displaystyle \tan 4x - \tan 2x = 0 \implies \frac{{\sin 2x}}
    {{\cos 4x\cos 2x}} = 0\,\therefore \,\tan 2x = 0.$

    Of course $\displaystyle \cos4x\ne0.$

    The rest follows from Soroban's conclusions.
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  5. #5
    Math Engineering Student
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    Quote Originally Posted by Hasan1 View Post
    2. $\displaystyle \frac{\sin (4x) - \sin (2x)}{\sin (2x)} = \frac{\cos (3x)}{\cos (x)}$
    Just tackle the LHS.

    Apply the following identity

    $\displaystyle \sin \alpha - \sin \beta = 2\sin \frac{{\alpha - \beta }}
    {2} \cdot \cos \frac{{\alpha + \beta }}
    {2}.$

    The conclusion follows quickly.
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  6. #6
    Math Engineering Student
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    Quote Originally Posted by Hasan1 View Post
    3. $\displaystyle \cos (x) + \cos (2x) + \cos (3x) = \cos (2x) (1+ 2 \cos (x)) $
    The same trick as identity #2, apply

    $\displaystyle \cos \alpha + \cos \beta = 2\cos \frac{{\alpha + \beta }}
    {2} \cdot \cos \frac{{\alpha - \beta }}
    {2}.$

    --

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

    And we're done
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  7. #7
    Junior Member Hasan1's Avatar
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    Is there a way to do those identities without the sum to product formula, but the double angle formula instead?
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  8. #8
    Math Engineering Student
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    I think those identities were made to apply the formulae I gave you. It's a quickly way.
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