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F(x)=(sin4x + sin3x + sin2x ) ÷ (cos4x + cos3x + cos2x)
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Best Answer
Recall that
sinA + sinB = 2sin[(A + B)/2] cos[(A - B)/2]
cosA + cosB = 2cos[(A + B)/2] cos[(A - B)/2]
sin(2A) = 2sinA cosA
cos(2A) = 2cos²A - 1
sinx + sin2x + sin3x + sin4x
= (sin4x + sinx) + (sin3x + sin2x)
= 2sin(5x/2) cos(3x/2) + 2sin(5x/2) cos(x/2)
= 2sin(5x/2) [ cos(3x/2) + cos(x/2) ]
cosx + cos2x + cos3x + cos4x
= (cos4x + cosx) + (cos3x + cos2x)
= 2cos(5x/2) cos(3x/2) + 2cos(5x/2) cos(x/2)
= 2cos(5x/2) [ cos(3x/2) + cos(x/2) ]
Thus
(sinx + sin2x + sin3x + sin4x) / (cosx + cos2x + cos3x + cos4x)
= { 2sin(5x/2) [ cos(3x/2) + cos(x/2) ] } / { 2cos(5x/2) [ cos(3x/2) + cos(x/2) ] }
= sin(5x/2) / cos(5x/2)
= 2sin(5x/2) cos(5x/2) / [ 2cos²(5x/2) ]
= 2sin(5x/2) cos(5x/2) / [ 2cos²(5x/2) - 1 + 1]
= 2sin(5x/2) cos(5x/2) / [ 2cos²(5x/2) - 1 + 1]
= sin(5x) / [ cos(5x) + 1]
= sin(5x) / (1 + cos(5x))
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