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posted by  laughsalot on 9/28/2008 2:10:38 PM  |  status: Live  

verifying identies please help

Course Textbook Chapter Problem
Trigonometry College Trigonometry 6th edition, Aufmann, Barker, Nation 3 N/A
Question Details:
 1.       sinx cos3x-cosx sin3x=1/4sin4x
 
2.        cosx + cos4x + cos7x  = cot 4x
               sinx  + sin 4x + sin 7x
             
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Oracle
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posted by Grace on 9/28/2008 2:37:23 PM  |  status: Live
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laughsalot's comment:
"thanks much"
Response Details:
Remember to only post one question at a time (board rules).

1) Take out common factors:
sinxcosx[cos2x - sin2x] = (1/4)sin(4x)

double angle identity:
sin(2x) = 2sin(x)cos(x)
So, sin(2x)/2 = sin(x)cos(x)


double angle identity:
cos(2x) = cos2x - sin2x
So,
=


We already know that:
sin(2u) = sin(u)cos(u)
if we let u = 2x:
sin[2(2x)] = sin(2x)cos(2x)
sin(4x) = sin(2x)cos(2x)



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Scholar
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posted by laughsalot on 9/28/2008 2:40:34 PM  |  status: Live
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Response Details:
Thanks so much... Im new and did not know....thanks for the help
(Cramster SME)
posted by WSAN on 9/29/2008 4:26:23 PM  |  status: Live
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Response Details:
2)  we know that cos x + cos 7x = 2 cos ( 7x+x)/2*cos(7x-x)/2 = 2 cos 4x cos 3x
                  and  sin x + sin 7x = 2 sin ( 7x+x)/2* cos( 7x -x)/ 2 = 2 sin 4x cos 3x
        cosx + cos4x + cos7x  
        sinx  + sin 4x + sin 7x               =  =
 
SWAN
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