Question

Find the value of Sin90. Cos0 + sin88. Cos2 + sin 86. Cos 4+....+ sin2. cos88 + sin0. Cos 90

Answer 1

20+sin50.cos5

20+sin50.sin50 (or)

20+cos50.cos50

Answer 2

Answer:

23

Step-by-step explanation:

let A = sin90cos0+sin88cos2+sin86cos4+.......+sin4cos86+sin2cos88+sin0cos90

suppose B=0,2,4,6,......,90

therefore 90=a+(n-1)d=0+(n-1)2

solving for n=46.....

middle terms are 23rd and 24th terms=44 and 46

therefore comparing B to A

middle terms of A are sin46cos44 and cos46sin44

therefore A=sin90cos0+sin88cos2+sin86cos4+.......+sin46cos44+sin44cos46+.......+sin4cos86+sin2cos88+sin0cos90

rearranging A = C =(sin90cos0+sin0cos90)+(sin88cos2+sin2cos88)+(sin86cos4+sin4cos86)+.....+(sin46cos44+sin44cos46)

C = sin(90+0) + sin(88+2) + sin(86+4) +.....+ sin(46+44)

=sin90+sin90+sin90+.......+sin90

=1+1+1+.......+1

pls keep in mind that A has 46 terms so combing terms once gives 23 terms so in C 1 is repeated 23 times

therefore A=C=23