2015年4月26日 · Think about double angle formulas until you think of one that might be useful. sin 2 theta = 2 sin theta cos theta That looks helpful. 2 sin theta cos theta = sin 2 theta so sin theta cos theta = 1/2 sin 2 theta And sin 8x cos 8x = 1/2 sin 16x

2017年8月14日 · f'(pi/4) = sqrt(2) We have: f(x) = cosx * cos2x * cos4x * cos8x * cos16x We could use the product rule with 5 products and form a full equation for the derivative, but this is not required, as we just need the value of the derivative when x=pi/4.

2017年1月20日 · The expression is #-sqrt(sin^2(4x)/cos^2(4x))# #=-|tan (4x)|, using sqrt(a^2)=|a|, by convention.practice/definition.

2018年2月8日 · We want to simplify . #cos(x)cos(2x)cos(4x)cos(8x)...cos(2^nx)# The trick is to use the double-angle identity repeatedly

If f(x) = cosx• cos2x •cos4x• cos8x• cos16x , then find derivative at x=π/4 Dear student First check, cos(45) = 1/2^(1/2)Cos(2*45) = 0Cos(4*45) = -1Cos(8*45) = [email protected] 1800-150-456-789

2018年4月14日 · How do you use the power reducing formulas to rewrite the expression #sin^8x# in terms of the first power of cosine?

2015年10月17日 · How do you find the indefinite integral of #int x(cos(8x))^2dx#? Calculus Introduction to Integration Definite and indefinite integrals. 1 Answer

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2017年9月19日 · sqrt(2+sqrt(2+sqrt(2+2cos8theta)))=2costheta As cos2A=2cos^2A-1, we have cos8theta=2cos^2(4theta)-1, cos4theta=2cos^2(2theta)-1 and cos2theta=2cos^2theta-1 i.e. 1 ...

2018年4月2日 · Note: #color(red)((1)intcosec^2thetad(theta)=-cottheta+c#. #color(red)((2)intcosecthetacotthetad(theta)=-cosectheta+c#