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Let 0 lt A_(i) pi for i = 1,2,"……"n. Use mathematical induction to prove that sin A_(1) + sin A_(2)+ "….." + sin A_(n) le n sin ((A_(1) + A_(2) + "……" + A_(n))/(n))where n ge 1 is a natural number. [You may use the fact thatp sin x + (1-p) sin y le sin [px+(1-p)y], where 0 le p le 1 and 0 le x , y le pi. |
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Answer» Solution :For `n = 1`,the inequality becomes `sinA_(1)lesin A_(1)`,which is clearlytrue. Let us assume that the inequility holds for `n = K` where k is some positive integer. Then `sinA_(1) + sinA_(2) + "……"+sinA_(n) le k SIN ((A_(1) + A_(2) +"....." + A_(k))/(k))"......"(1)` Adding `sin A_(k+1)` on bothsides, we GET `sin A_(1) + sin A_(2) + "......" +sin A_(k) +suin A_(k+1) le k sin((A_(1) + A_(2) +"......."+A_(k))/(k)) + sin A_(k+1)` Now, `k sin ((A_(1) + A_(2) + "......." + A_(k))/(k))+ sin A_(k+1)` `= (k+1)[(k)/(k+1)sin alpha+(1)/(k+1) sin A_(k+1)]`, where `alpha = (A_(1) + A_(2) + "......" + A_(g))/(k)` ` le (k + 1 )sin{(1+(1)/(k+1)) alpha + (1)/(k+1) A_(k+1)}` {Using `p sin x+ (1-p) sinyle sin[px + (1-p)y]}` `= (k+1)sin((A_(1)+A_(2)+"......."A_(k+1))/(k+1))` Thus, the inequality hold for `n= k +1`. Hence, by the PRINCIPLE of mathematicalinduction, the inequality holds for all `n in N`. |
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