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Math Help - Trigonometric function...

  1. #1
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    Trigonometric function...

    Proved the following identities:
    1. tan(3x/2)-tan(x/2)=(2sinx)/(cosx+cos2x)
    2. cos4a. tan2a-sin4a=(2tan a)/(tan^2 a -1)
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  2. #2
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    Re: Trigonometric function...

    Wouldn't it be better for you to proe them? Do you know identities for trig functions of multiple or fractional arguments?

    sin(2x)= 2sin(x)cos(x) and cos(2x)= cos^2(x)- sin^2(x)

    In the second, we can replace sin^2(x) by 1- cos^2(x) to get cos(2x)= cos^2(x)- (1- cos^2(x))= 2cos^2(x)- 1
    so that cos^2(x)= (1+ cos(2x))/2, cos(x)= \pm \sqrt{(1/2)(1+ cos(2x))} and letting \theta= 2x that becomes cos(\theta/2)= \pm\sqrt{(1/2)(1+ cos(2\theta))}. Similarly, sin(2x)= 2sin(x)cos(x) and sin(\theta/2)= \pm\sqrt{(1/2)(1- cos(\theta))}.

    Those, together with the fact that tan(x)= \frac{sin(x)}{cos(x)}, are what you need.
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    Re: Trigonometric function...

    Quote Originally Posted by HallsofIvy View Post
    Wouldn't it be better for you to proe them? Do you know identities for trig functions of multiple or fractional arguments?

    sin(2x)= 2sin(x)cos(x) and cos(2x)= cos^2(x)- sin^2(x)

    In the second, we can replace sin^2(x) by 1- cos^2(x) to get cos(2x)= cos^2(x)- (1- cos^2(x))= 2cos^2(x)- 1
    so that cos^2(x)= (1+ cos(2x))/2, cos(x)= \pm \sqrt{(1/2)(1+ cos(2x))} and letting \theta= 2x that becomes cos(\theta/2)= \pm\sqrt{(1/2)(1+ cos(2\theta))}. Similarly, sin(2x)= 2sin(x)cos(x) and sin(\theta/2)= \pm\sqrt{(1/2)(1- cos(\theta))}.

    Those, together with the fact that tan(x)= \frac{sin(x)}{cos(x)}, are what you need.
    yeah, I know that, I just don't know how to get tan(3x/2)
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    Re: Trigonometric function...

    Use the half angle identity for tangent, then everything will be in terms of tan(3x).

    Then use the angle sum identity for tangent using tan(x + 2x).
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    Re: Trigonometric function...

    Quote Originally Posted by Prove It View Post
    Use the half angle identity for tangent, then everything will be in terms of tan(3x).

    Then use the angle sum identity for tangent using tan(x + 2x).
    but I still can't get rid of the tan^2 (3x/2)
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    Re: Trigonometric function...

    Quote Originally Posted by Trefoil2727 View Post
    but I still can't get rid of the tan^2 (3x/2)
    Prove It just told you how to do this. Tell you what...Use the 1/2 angle part and post what you've got. That will make things easier for us to know how to help you.

    -Dan
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    Re: Trigonometric function...

    Quote Originally Posted by topsquark View Post
    Prove It just told you how to do this. Tell you what...Use the 1/2 angle part and post what you've got. That will make things easier for us to know how to help you.

    -Dan
    tan(3x/2)-tan(x/2)=(2sinx)/(cosx+cos2x)
    LHS=(tan3x)(1-tan^2 3x/2)-(tanx)(1-tan^2 x/2)
    =((tanx+tan2x)/(1-tanxtan2x))(1-tan^2 3x/2)-(tanx)(1-tan^2 x/2)..
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    Re: Trigonometric function...

    The second one is quite simple
    LHS = cos 4a tan 2a - sin 2a = ( cos 4a sin 2a ) / cos 2a - sin 2a = ( cos 4a sin 2a - sin 4a cos 2a )/ cos 2a = - sin ( 4a - 2a )/ cos 2a = - tan 2a
    RHS = ( 2 tan a )/ ( tan^2 a - 1 ) = - ( 2 tan a )/ ( 1 - tan^2 a ) = - tan 2a
    hence LHS = RHS
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    Re: Trigonometric function...

    Quote Originally Posted by ibdutt View Post
    The second one is quite simple
    LHS = cos 4a tan 2a - sin 2a = ( cos 4a sin 2a ) / cos 2a - sin 2a = ( cos 4a sin 2a - sin 4a cos 2a )/ cos 2a = - sin ( 4a - 2a )/ cos 2a = - tan 2a
    RHS = ( 2 tan a )/ ( tan^2 a - 1 ) = - ( 2 tan a )/ ( 1 - tan^2 a ) = - tan 2a
    hence LHS = RHS
    why ( cos 4a sin 2a - sin 4a cos 2a )/ cos 2a = - sin ( 4a - 2a )/ cos 2a = - tan 2a ?
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    Re: Trigonometric function...

    remember the formula
    sin ( A-B ) = sin A cosB - cos A sin B
    Thus cos 4a sin 2a - sin 4a cos 2a = - ( sin 4a cos 2a - cos 4a sin 2a ) = - sin ( 4a - 2a ) = - sin 2a
    also we know that sin A / cos A = tan A
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    Re: Trigonometric function...

    Trigonometric function...-doc1.jpg
    Attached Thumbnails Attached Thumbnails Trigonometric function...-doc1.png  
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    Re: Trigonometric function...

    Trigonometric function...-28-jun-13.png
    Attached Thumbnails Attached Thumbnails Trigonometric function...-26-jun-13.png  
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    Re: Trigonometric function...

    Quote Originally Posted by ibdutt View Post
    Click image for larger version. 

Name:	28 Jun 13.png 
Views:	3 
Size:	17.0 KB 
ID:	28683
    I've solved that already, left the 1st question..
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  14. #14
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    Re: Trigonometric function...

    Trigonometric function...-untitled.jpg
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