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cosx*cos2x*cos4x.....cos(2^(n-1)x)=(sin2...

`cosx*cos2x*cos4x.....cos(2^(n-1)x)=(sin2^n x)/(2^nsinx)AAn in N`

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Knowledge Check

  • A : cos 24^(@) cos 48^(@) cos 96^@ cos 168^(@) = 1//16 R : cos x cos 2 x cos 4x ........ cos (2^n x)=( sin (2^(n+1)x))/(2^(n+1) sin x )

    A
    A is true , R is true and R is correct explanation of A
    B
    A is true , R is true and R is not correct explanation of A
    C
    A is true , R is false
    D
    A is false , R is true

  • If cos""(x)/(2).cos""(x)/(2^(2))...cos""(x)/(2^(n))=(sinx)/(2^(n)sin""(x)/(2^(n))) , then (1)/(2)tan""(x)/(2)+(1)/(2^(2))tan""(x)/(2^(2))+...+(1)/(2^(n))tan""(x)/(2^(n)) is

    A
    `cotx-cot""(x)/(2^(n))`
    B
    `(1)/(2^(n))cot((x)/(2^(n)))-cotx`
    C
    `(1)/(2^(n))tan((x)/(2^(n)))-tanx`
    D
    `(1)/(2)cotx-(1)/(2^(n))cot((x)/(2^(n)))`

  • If f(x) = cos x cos 2x cos 2^2 x cos^(2^3) x .....cos 2^(n-1) x and n gt 1 then f^(1)(pi/2) is

    A
    1
    B
    0
    C
    `-1`
    D
    `(-1)^(n-1)`

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    If f(x) = cos x\ cos 2x\ cos 2^2\ x\ cos 2^3 x\ ....cos2^(n-1) x and n gt 1, then f'(pi/2) is

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    We have f (x) lim_(n to oo) cos (x)/(2) cos (x)/(2^(2)) cos (x)/(2^(3)) cos (x)/(2^(4)) …… …. cos (x)/(2^(n)) = ("sin" x)/(2^(n) "sin" (x)/(2^(n))) using the identity lim_(n to oo) sum_(k=1)^(n) (1)/(2^(2k)) sec^(2) ((x)/(2^(k))) equals

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