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cos2x*cos4x*cos6x...

cos2x*cos4x*cos6x

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f(x)=cos x*cos2x*cos4x*cos8x.cos16x then find f'((pi)/(4))

If f(x)=cos x*cos2x*cos4x*cos8x.cos16x, then f'((pi)/(4)) is :

Evaluate the following integrals: int cos2x cos 4x cos6x dx .

The equation 8cos x*cos2x*cos4x=(sin6x)/(sin x) has a solution if

cos2x-cos8x+cos6x=1

If g(x)=cos x cos2x cos4x cos8x then g'((pi)/(4))=

If f(x)=cos x cos2x cos4x cos8x cos16x then

lim_(x->0)(1-cos x-cos2x+cos x*cos2x)/(x^(4))