🌘 Sin N 1 X Sin N 1 X

DetailedSolution. Given if I n = ∫π −π sinnx (1+πx)sinx dx,(1) i f I n = ∫ − π π s i n n x ( 1 + π x) s i n x d x, ( 1) I n = ∫π −π πxsinnx (1+πx)sinx dx.(2) I n = ∫ − π π π x s i n n x ( 1 + π x) s i n x d x. ( 2) On adding Eqs. (i) and (ii), we have. Weknow that cos ( A B) = cos A cos B + sin A sin B Hence A = (n + 1)x ,B = (n + 2)x Hence sin ( + 1) sin ( + 2) +cos ( + 1) cos ( + 2) = cos [ (n + 1)x (n + 2)x ] = cos [ nx + x nx 2x ] = cos [ nx nx x 2 x ] = cos (0 x ) = cos ( x) = cos x = R.H.S. Hence , L.H.S. = R.H.S. Hence proved n\frac{1}{\sin(x)},\nexists n_{1}\in \mathrm{Z}\text{ : }x=\pi n_{1} View solution steps. Steps for Solving Linear Equation. \frac { 1 } { n } = \sin ( x ) Variable n cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by n. 1=n\sin(x) sin(n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x = cos ( ( n + 2 ) x − ( n − 1 ) x ) { ∵ cos ( A − B ) = sin A sin B + cos A cos B } ⇒ = cos ( ( n + 2 − n − 1 ) x ) Solution Verified by Toppr. The given equation is. sin −1x+sin −1(1−x)=cos −1x. ⇒sin −1x+sin −1(1−x)= 2π−sin −1x. ⇒sin −1(1−x)= 2π−2sin −1x (i) Let sin −1x=y. ⇒x=siny. a) Prove the reduction formulaintegrate cos^n(x)dx = 1/n * cos^(n-1) * x sin x + (n-1)/n * integral cos^(n-2)x dx sinA - sin B = 2 cos 1/2 (A+B) sin 1/2 (A-B) sin(x+1)x-sin(x-1)x = 2cos 1/2 ((x+1)x + (x-1)x) sin 1/2 ((x+1)x - (x-1)x) = 2cos 1/2(x²+x+x²-x) sin 1/2(x²+x-x²+x) = 2cos 1/2(2x²) sin 1/2(2x) = 2 cos x² sin x Multiplyboth sides by n n. 1+sin(x) n n = kn 1 + sin ( x) n n = k n. Simplify the left side. Tap for more steps sin(x)+1 = kn sin ( x) + 1 = k n. Subtract 1 1 from both sides of the equation. sin(x) = kn−1 sin ( x) = k n - 1. Take the inverse sine of both sides of the equation to extract x x from inside the sine. Itis the case to consider Laurent series, since both functions have a simple pole in zero. By definition: \frac{z}{e^z-1}=\sum_{n\geq 0}\frac{B_n}{n!}z^n \tag{1} hence: \frac{1}{1-e^{-x}}=\sum_{n\geq 0}\frac{B_n}{n!}(-1)^n x^{n-1} \tag{2} SW0d2bC. Chapter 3 Class 11 Trigonometric Functions Serial order wise Ex Check sibling questions Ex Ex 1 Important Ex 2 Important Ex 3 Important Ex 4 Ex 5 i Important Ex 5 ii Ex 6 Important Ex 7 Ex 8 Important Ex 9 Important Ex 10 You are here Ex 11 Important Ex 12 Ex 13 Important Ex 14 Ex 15 Ex 16 Important Ex 17 Ex 18 Important Ex 19 Ex 20 Ex 21 Important Ex 22 Important Ex 23 Important Ex 24 Ex 25 Ex 10 - Chapter 3 Class 11 Trigonometric Functions Last updated at May 29, 2023 by Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class Transcript E 10 Prove that sin + 1 sin + 2 +cos + 1 cos + 2 =cos Taking We know that cos A B = cos A cos B + sin A sin B Hence A = n + 1x ,B = n + 2x Hence sin + 1 sin + 2 +cos + 1 cos + 2 = cos [ n + 1x n + 2x ] = cos [ nx + x nx 2x ] = cos [ nx nx x 2 x ] = cos 0 x = cos x = cos x = Hence , = Hence proved Chapter 3 Class 11 Trigonometric Functions Serial order wise Ex Ex 1 Important Ex 2 Important Ex 3 Important Ex 4 Ex 5 i Important Ex 5 ii Ex 6 Important Ex 7 Ex 8 Important Ex 9 Important Ex 10 You are here Ex 11 Important Ex 12 Ex 13 Important Ex 14 Ex 15 Ex 16 Important Ex 17 Ex 18 Important Ex 19 Ex 20 Ex 21 Important Ex 22 Important Ex 23 Important Ex 24 Ex 25 Davneet Singh has done his from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo. Question MediumOpen in AppSolutionVerified by TopprThe given equation is ...... i Let Therefore, from i, we get Since, both these values satisfy the given equation. Hence, the solutions of the given equation are .Video ExplanationWas this answer helpful? 00 By l'Hopital's Rule, we can find lim_{x to infty}x sin1/x=1. Let us look at some details. lim_{x to infty}x sin1/x by rewriting a little bit, =lim_{x to infty}{sin1/x}/{1/x} by l'Ho[ital's Rule, =lim_{x to infty}{cos1/xcdot-1/x^2}/{-1/x^2} by cancelling out -1/x^2, =lim_{x to infty}cos1/x=cos0=1 Instead of l'Hopital's Rule, one can use the fundamental trigonometric limit lim_hrarr0sin h/h=1. The limit you are interested in can be written lim_xrarroosin 1/x/1/x. Now, as xrarroo, we know that 1/xrarr0 and we can think of the limit as lim_1/xrarr0sin 1/x/1/x. With h=1/x, this becomeslim_hrarr0sin h/h which is 1. Although it is NOT needed, here's the graph of the function graph{y = x sin1/x [ When you substitute in infinity, oo, you end up with the indeterminate form of oo*0. lim_x->oo xsin1/x=oo*sin1/oo=oo*sin0=oo*0 We still have options though. We now can fall back on L'Hopital's Rule which basically says to take the derivative of the numerator and denominator independently. Do not use the quotient rule. We need to rewrite this function so that is produces an indeterminate in the form oo/oo or 0/0. lim_x->oo sin1/x/x^-1=sin1/x/1/x=sin1/oo/1/oo=sin0/0=0/0 Applying L'Hopital lim_x->oosin1/x'/x^-1' =lim_x->oo-1*x^-2*cos1/x/-1*x^-2 Simplify the previous step =lim_x->oocos1/x=cos1/oo=cos0=1

sin n 1 x sin n 1 x