Question

HELPPP!!!

Find (cos1°+cos2°+…+cos44°)/(sin1°+sin2°+…+sin44°)

Answer 1

cos1 + cos44 + ...+ cos22 + cos23
------------------------------
sin1 + sin 44 +...+sin22 + sin23

2cos(45)cos(43)+....+2cos(45)cos(1)
------------------------------
2sin(45)cos(43)+...+2sin(45)cos1

1

Answer 2

\( 1+ \sqrt{2} \) ?

Answer 3

tan 45/2

Answer 4

not 1, sorry

Answer 5

as sinc + sind = 2sin(c+d)/2cos(c-d)/2
and cosc + cosd = 2cos(c+d)/2*cos(c-d)/2

Answer 6

http://openstudy.com/study#/updates/4f0c9d15e4b084a815fc728d
when u done with this do have a look

Answer 7

tan45 = (2tan45/2)/(1 - tan^2(45/2)
1 -x^2 = 2x
note. x = tan45/2

Answer 8

Ishaan \( \tan \frac {\pi}{8} \) doesn't seem to be the correct answer.

Answer 9

i did wrong, i know :-(

Answer 10

cot 45/2 must be it

Answer 11

x^2 + 2x - 1 + 1 -1 = 0
(x + 1)^2 = 2
x = -1 + sqrt2


x was tan 45/2

henc, cot 45/2 must be 1/(-1+ sqrt2)

Answer 12

hence*

Answer 13

actually it;s

2cos(45/2)cos(43/2)+....+2cos(45/2)cos(1/2)
------------------------------
2sin(45/2)cos(43/2)+...+2sin(45/2)cos1/2

cos(45/2) ( cos43/2 + .. + cos1/2)
------------------------------
sin(45/2)( cos43/2 + ...+cos1/2)

which gets down to cot 45/2

Answer 14

i didn't add (1/2) part in my first post, sorry i am lazy...

but now i think you can follow my answer

Answer 15

kwenisha??