an= cos(n*pi /2) Thank you so much! Log On Algebra: Sequences of numbers, series and how to sum them ...

If the series is geometric find the sum as well: [cos(n*pi/3)] / n factorial at n =1 to infinity Thank you s Algebra -> Sequences-and-series -> SOLUTION: Determine if the following series are convergent of divergent.

On WolframAlpha, it says the x int's are npi/2 - pi/4, where n is an element of the integers. I do not know how they arrived at that answer. I know the amplitude is 4, the period is pi, but when I solve bx - h = 0 I get pi/2, but not pi/2 - pi/4.

me the two results: cot(x) = -1; x= -PI/4 + nPI; n€Z cot(x) = 0; x= PI/2 + nPI; n€Z Where am I doing or thinking wrong? Thank you very much RB-----sin(x)cos(x) + cos^2(x) = 0 cos*(sin + cos) = 0 cos = 0 x = pi/2 + n*pi, n = any integer-----sin + cos = 0 tan + 1 = 0 tan = …

You can put this solution on YOUR website! cos (sin x) = 1 The answer is npi, n=o, plus or minus 1. plus or minus 2, etc.

At 45 degrees, or pi/4, sin x and cos x are equal and their product is 1/4. Each is sqrt (2)/2. One must be negative. Second and fourth quadrants are where this happens. Cosine is negative in the second, and sine negative in the fourth. I would say 3pi/4 and 7 pi/4. In the problem above, a and d give negative results but not the product.

You can put this solution on YOUR website! Why is cos(n*pi) = (–1)^n true, when n could be any integer?

If z^n is real, then sin npi/3 = 0 , equivalently n pi /3 must be multiple of pi. We see that when n = 3 , n pi /3 = pi. Hence,the smallest positive integer n such that z^n is real is 3 and we have z^3 = 2^3(cos pi + i sin pi) = 8(-1+ i * 0) = -8. Note that if z^n is imaginary then cos n …

You can put this solution on YOUR website! z = 1cis(180 + 2n*pi) pi= 0,1,2,3... 6th roots = 1cis(30 + npi/3) ...

Cosine 1 (1 radian) is 1 unit less than the 2*pi circumference, which is 6.28. That would make cos theta=cos 1, be 5.28. At both 1 and at 5.28, the cosine of theta is cos 1. The former is excluded by the interval given.