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Show that for any real number x, `sin^3 2x cos 6x + cos^3 2x sin 6x = (3/4) sin 8x`
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`sin^3 2x cos 6x + cos^3 2x sin 6x`
`= [(3sin2x-sin6x)/4]cos 6x+[(cos6x+3cos2x)/4]sin6x`
`= (1/4)(3sin2x cos6x-sin6x cos6x+cos6x sin6x+3cos2x sin6x)`
`= (1/4)[3(sin2x cos6x + cos2x sin6x)]`
`= (3/4) sin8x`
So the answer is obtained as required.
The trigonometric expressions used here is;
`sin3A = 3sinA-sin^3A`
`cos3A = 4cos^3A-3cosA`
`sin(A+B) = sinAcosB+cosAsinB`
Posted by jeew-m on September 20, 2013 at 11:00 AM (Answer #1)
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