ŞEVVAL58 21:23 10 May 2011 #1
(cot2a.tana) / tan
2-1=? (-1/2)
sin
2(a-3∏/2).(1-tan
2a).tan(∏/4+a) / cos
2(∏ /2 -a)=? (cosec
2a +2cota)
(cosma-cosna) / (sinna-sinma)=? tan(m+n/2 a)
sin
4∏/8 + cos
4∏/8=? (3/4)
(4cosx/3) .(cos∏+x /3) .(cos∏-x /3)=? (cosx)
nolur yardım edin

hasim 15:08 11 May 2011 #2
1)(1/tan2a).tana/(tan²-1)
[(1-tan²a)/(2.tana)] . tana /(tan²-1)=-1/2
2)cos²a . (1-tan²a) . (1+tana)/(1-tana) . 1/sin²a
(1+tana)²/tan²a =[(1+tana)/tana]² = (cota+1)²=cot²+2cota+1=2cota+(1/sin²a)
3)
x=(m+n)a/2
y=(m-n)a/2 olsun
[cos(x+y)-cos(x-y)]/[sin(-y+x)-sin(-y-x)] =[-2.sinx.siny]/[2.(sin-y) .cosx] =tanx
4)2cos²pi/8 =1+cospi/4 ise cos²pi/8 =(2+√2)/4 bulunur
=[sin²pi/8]2+[cos²pi/8]2
=[(1-cos²pi/8)]2+[cos²pi/8]2
=[(2-√2)/4]2+[(2+√2)/4]2
=12/16=3/4
ŞEVVAL58 19:21 11 May 2011 #3
çok teşekkür ederim çok sağol