Calculus
posted by CHRIS .
what is the limit n to infinite of
cos1*cos(1/2)*cos(1/4)*cos(1/8)*cos(1/16)*...*cos(1/2^n)

Since sinc(1) = cos(1/2)*cos(1/4)*cos(1/8)*cos(1/16)*...*cos(1/2^n)
the limit here is cos(1)sinc(1) = cos(1)sin(1) = 0.4546...
You can read about sinc(x) in various places. This function doesn't usually pop up in introductory calculus courses.
Respond to this Question
Similar Questions

trig
Reduce the following to the sine or cosine of one angle: (i) sin145*cos75  cos145*sin75 (ii) cos35*cos15  sin35*sin15 Use the formulae: sin(a+b)= sin(a) cos(b) + cos(a)sin(b) and cos(a+b)= cos(a)cos(b)  sin(a)sin)(b) (1)The quantity … 
precal
Simplify the given expression........? (2sin2x)(cos6x) sin 2x and cos 6x can be expressed as a series of terms that involve sin x or cos x only, but the end result is not a simplification. sin 2x = 2 sinx cosx cos 6x = 32 cos^6 x 48 
Mathematics  Trigonometric Identities
Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y … 
Math  Solving Trig Equations
What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x)  1 cos(x) (+/)\sqrt{1  cos^2(x)} = 2cos^2(x)  1 cos^2(x)(1  cos^2(x)) 
maths
Find the roots of z^6 + 1 and hence resolve z^6 + 1into read quadratic factors; deduce that cos3x = 4[cos(x) cos(pi/6)][(cos(x) cos(pi/2)][(cos(x) cos(5pi/6)] 
maths
Find the roots of z^6 + 1 and hence resolve z^6 + 1into read quadratic factors; deduce that cos3x = 4[cos(x) cos(pi/6)][(cos(x) cos(pi/2)][(cos(x) cos(5pi/6)] 
TRIG!
Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1  (3/4)sin^2 2x work on one side only! Responses Trig please help!  Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 … 
Calc.
Differentiate. y= (cos x)^x u= cos x du= sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x1) * (sin x) =  x sin(x)cos^(x1)(x) (dy/dx)(dx/du)= … 
calculus
Differentiate. y= (cos x)^x u= cos x du= sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x1) * (sin x) =  x sin(x)cos^(x1)(x) (dy/dx)(dx/du)= … 
math
Determine exact value of cos(cos^1(19 pi)). is this the cos (a+b)= cos a cos b sina sin b?