Prove - sin5x-2sin3x+sinx /cos5x-cosx = tanx
Asked by prashant.jain | 24th May, 2010, 12:43: PM
We have explained the steps, please also refer to the solution in book, and apply the steps we listed below.
(sin5x-2sin3x+sinx) /(cos5x-cosx) =
In the numerator we can write for sin5x + sinx = 2 sin3x cos2x, using the result,
sinA + sinB = 2sin((A+B)/2)cos((A-B)/2), and here A = 5x and B = x.
Similarly for the denominator we can write, cos5x - cosx = -2 sin3x sin2x,
cosA - cos B = -2sin((A+B)/2)sin((A-B)/2), and here A = 5x and B = x.
After this step, the next step is to take out sin3x common in numerator and denominator which cancels out leaving,
Now make the final substitution using the trignometric results,
sin2A = 2sinAcosA and cos2A = cos2A - sin2A = 1 - 2sin2A.
After this substitution rest step is just algebraic manipulations to get the final result.
Answered by | 24th May, 2010, 06:34: PM
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