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Question Number 120727 by 77731 last updated on 02/Nov/20

	Answered by TANMAY PANACEA last updated on 02/Nov/20
	$N_r =2+2cosx−2cosnx−cos(n−1)x−cos(n+1)x =2+2cosx−2cosnx−2cosnxcosx =2(1+cosx)−2cosnx(1+cosx) =2(1+cosx)(1−cosnx) (N_r /D_r )=((2(1+cosx)(1−cosnx))/(1−cos2x))=((2(1+cosx)(1−cosnx))/(2sin^2 x)) I_n −I_(n−2) =∫_0 ^π (((1+cosx)/(sin^2 x)))({1−cosnx−1+cos(n−2)x} dx =∫_0 ^π (((1+cosx)/(sin^2 x)))×2sin(n−1)xsinx =∫_0 ^π ((1+cosx)/(sinx))×sin(n−1)xdx =∫_0 ^π ((2cos^2 (x/2))/(2sin(x/2)cos(x/2)))×sin(n−1)xdx =∫_0 ^π cot(x/2)×sin(n−1)xdx I_n −I_(n−2) =∫_0 ^π cot(x/2)sin(n−1)xdx wait...$
	$N_{r} = 2 + 2 c o s x - 2 c o s n x - c o s (n - 1) x - c o s (n + 1) x$ $= 2 + 2 c o s x - 2 c o s n x - 2 c o s n x c o s x$ $= 2 (1 + c o s x) - 2 c o s n x (1 + c o s x)$ $= 2 (1 + c o s x) (1 - c o s n x)$ $\frac{N_{r}}{D_{r}} = \frac{2 (1 + c o s x) (1 - c o s n x)}{1 - c o s 2 x} = \frac{2 (1 + c o s x) (1 - c o s n x)}{2 {s i n}^{2} x}$ $I_{n} - I_{n - 2} = \int_{0}^{π} (\frac{1 + c o s x}{{s i n}^{2} x}) ({1 - c o s n x - 1 + c o s (n - 2) x} d x$ $= \int_{0}^{π} (\frac{1 + c o s x}{{s i n}^{2} x}) \times 2 s i n (n - 1) x s i n x$ $= \int_{0}^{π} \frac{1 + c o s x}{s i n x} \times s i n (n - 1) x d x$ $= \int_{0}^{π} \frac{2 {c o s}^{2} \frac{x}{2}}{2 s i n \frac{x}{2} c o s \frac{x}{2}} \times s i n (n - 1) x d x$ $= \int_{0}^{π} c o t \frac{x}{2} \times s i n (n - 1) x d x$ $I_{n} - I_{n - 2} = \int_{0}^{π} c o t \frac{x}{2} s i n (n - 1) x d x$ $w a i t . . .$
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