Prove that csc(2x+pie/5)=2 is equal to sin(2x+pie/5)=1/2?
csc(2x+pie/5)=2 is equal to sin(2x+pie/5)=1/2 Have some π csc(2x+π/5)=2 is equal to sin(2x+π/5)=1/2 Let y = (2x+π/5) rewrite it as csc(y ) = 2 by definition, csc(A) = 1 / sin(A) so 1 / sin(y) = 2 multiply both sides by sin(y) 1 = 2 * sin(y) divide by sides by 2 1/2 = sin(y) substitute back in for y 1/2 = sin(2x+π/5)
Which is greater, sin10, sin12, or sin13?
I am writing this answer as I didnot find the right answer in previous answers.Since, the degree sign(°) is missing, it means that the angles are in radians.Now,3π < 10 < 7π/2 {So, 10 lies in third quadrant}7π/2 < 12 < 4π {So, 12 lies in forth quadrant}4π < 13 < 9π/2 {So, 13 lies in first quadrant}Since, sine function is positive in 1st quadrant but negative in 3rd and 4th quadrant, it is clear that Sin13 is the greatest…
Prove that if n3 is even then n is even, for all integers n.?
Proof by Contrapositive: (If n is odd, then n^3 is odd.) Suppose that n is odd. Writing n = 2k + 1 for some integer k, n^3 = 8k^3 + 12k^2 + 6k + 1 = 2(4k^3 + 6k^2 + 3k) + 1. Since 4k^3 + 6k^2 + 3k is an integer, we conclude that n^3 is odd. I hope this helps!
Does the sequence [1*3*5*…*(2n-1)]/[2*4*6*…*(2n)... converge?
Rewrite this as [1 * 3 * 5 *…* (2n-1)]/[2 * 4 * 6 *…* (2n)] = (2n)! / [2 * 4 * 6 * … * (2n)]^2 = (2n)! / [2^n * (1 * 2 * 3 * … * n)]^2 = (2n)! / [2^n * n!]^2. Using Stirling's Formula lim(n→∞) n! / [(n/e)^n √(2πn)] = 1, lim(n→∞) (2n)! / [2^n * n!]^2 = lim(n→∞) [(2n/e)^(2n) √(2π*2n)] / {2^(2n) * [(n/e)^n √(2πn)]^2} = lim(n→∞) √(πn) / (πn) = lim(n→∞) 1/√(πn) = 0. I hope this helps!
What is the integration of sin3x cos5x?
Integrate sin3x cos5x;Let A=3x and B= 5x;sin(A+B)=sinA*cosB + cosA * sinB;sin(A-B)=sinA*cosB - cosA * sinB; Adding vertically both identities:sin(A+B) + sin(A-B) = 2 sinA*cosB ; Now replace A +B with 8x and A-B with (-2x);(1/2)*[sin 8x - sin (2x)] to integrate;∫(1/2)[sin 8x - sin (2x)] dx = (- 1/16)cos8x + (1/4)cos2x + C; C integration Constant for initial conditions.
How do u prove whether a stochastic matrix is regular or not?
A regular stochastic matrix A will have at least one A^n where n = 1, 2, 3 ... power that has all of its elements positive and greater than zero. The easy way usually is to multiple the matrix by itself until you find one of powers with all positive elements. This can be a lengthy process. The more surefire method is to use PERRON FROBENIUS THEOREM and make sure the eigenvalues satisfy the condition of PFT. But any method that can demonstrated that all the elements of some power of A are all positive and not zero works.