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Robert Maddox
Mathematics
15 November, 03:22
Prove:Sinx-2sin3x+sin5x=2sinx (cos4x-cos2x)
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Kennedy Mahoney
15 November, 03:29
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A = sinx - sin3x,
B = - sin3x + sin5x
First A:
The average of x and 3x is 2x, and they (x and 3x, that is) are each a distance of x from this average. That's fancy talk for:
x = 2x-x,
3x = 2x+x.
So, A = sin (2x-x) - sin (2x+x)
Using angle sum formulas:
A = (sin2x cosx - cos2x sinx) - (sin2x cosx + cos2x sinx)
A = - 2 cos2x sinx
Similarly,
B = - sin (4x-x) + sin (4x+x)
= - (sin4x cosx - cos4x sinx) + (sin4x cosx + cos4x sinx)
B = 2 cos4x sinx
Now,
sinx - 2sin3x + sin5x = A+B = - 2 cos2x sinx + 2 cos4x sinx
= 2 sinx (cos4x - cos2x).
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Teagan Sexton
15 November, 03:50
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Sinx-2sin3x+sin5x=2sinx (cos4x-cos2x)
Step-by-step explanation:
sinx - 2sin3x + sin5x = sinx - sin (3x) + sin (5x) - sin (3x)
= 2· cos[ (x+3x) / 2] · sin[ (x-3x) / 2] + 2·cos[ (5x+3x) / 2]· sin[ (5x-3x) / 2]
= 2· cos (2x) ·sin (-x) + 2· cos (4x) · sin (x)
= - 2·cos (2x) ·sinx + 2· cos (4x) ·sinx
= 2·sinx · [ cos (4x) - cos (2x) ]
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