Prove - sin5x-2sin3x+sinx /cos5x-cosx = tanx

Asked by prashant.jain | 24th May, 2010, 12:43: PM

Expert Answer:

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,

(1-cos2x)/sin2x,

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.

Regards,

Team,

TopperLearning.

 

Answered by  | 24th May, 2010, 06:34: PM

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