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If \(\cos 4x = \frac15,\) what is the value of \(\sin 6x \sin 2x?\)

 Oct 26, 2019
 #1
avatar+8863 
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\(If\ cos\ 4x=\frac{1}{5}, what\ is\ the\ value\ of\ sin\ 6x\times sin\ 2x?\)

 

Hello Guest!

 

\(cos \ 4x=\frac{1}{5}\\ 4x=arc\ cos\frac{1}{5}=78.4630...°\\ \color{blue}x=19.6157602418...°\)

 

\(sin\ 6x\times sin\ 2x=0.56=\frac{14}{25}\)

laugh  !

 Oct 26, 2019
 #2
avatar+8863 
+1

\(If\ cos\ 4x=\frac{1}{5}, what\ is\ the\ value\ of\ sin\ 6x\times sin\ 2x? \)

 

\(cos\ 4x=8\ cos^4\ x-8\ cos^2\ x+1=\frac{1}{5}\)

\(8\ cos^4\ x-8\ cos^2\ x+1=\frac{1}{5}\)

\(cos^4\ x-cos^2\ x=-\frac{4}{5\cdot 8}\ \small (Bartsch\ B.u.Z. K\ddot oln\ 1980)\\ cos^4\ x-cos^2\ x=-\frac{1}{10}\\ (cos^2\ x+ cos\ x)(cos^2\ x-cos\ x)=-\frac{1}{10}\)

\(sin\ a\times sin\ b=\frac{1}{2}(cos(a-b)-cos(a+b))\)

       6x          2x                 6x    2x             6x   2x 

\(sin\ 6x\times sin\ 2x=\frac{1}{2}(cos(6x-2x)-cos(6x+2x))\\ sin\ 6x\times sin\ 2x=\frac{1}{2}(cos(4x)-cos(8x))\)

                                                a                  b

\(\frac{1}{2}(cos\ a-cos\ b)= -sin\frac{a+b}{2}sin\frac{a-b}{2}\)

\(sin\ 6x\times sin\ 2x= -sin\frac{4x+8x}{2}\times sin\frac{4x-8x}{2}\\ sin\ 6x\times sin\ 2x= -sin\ 6x\times sin (-2x) \)

 

I will continue.

laugh  !

 Oct 26, 2019

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