F.4 M2 trigo

1)
20b)1 / tan A/2 = tan (π/4 + B/2)tan (π/2 - A/2) = tan (π/4 + B/2)∵
0 < (π/2 - A/2) < π/2 for 0 < A/2 < π/2
and
π/4 < (π/4 + B/2) < 3π/4 for 0 < B/2 < π/2 ∴ π/4 < (π/2 - A/2) , (π/4 + B/2) < π/2 , hence (π/2 - A/2) = (π/4 + B/2) A + B = π/2
2)
25d)cos72° + cos144° + cos216° + cos288°
= cos72° + cos144° + cos(360 - 216)° + cos(360 - 288)°
= cos72° + cos144° + cos144° + cos72°
= 2(cos72° + cos144°)Hence
- 1 = 2(cos72° + cos144°)
2(cos72° + cos 2(72°)) + 1 = 0
2(cos72° + 2cos² 72° - 1) + 1 = 0
4cos² 72° + 2cos 72° - 1 = 0
cos72° = [- 2 ± √(2² + 4*4)] / 8
cos72° = (- 1 ± √5)/4
cos72° = (√5 - 1)/4 since cos72° > 0.
26a)In △ABD , by sine formula ,
BD / sin(α + β)
= AD / sin∠ ABD
= AD / sin∠ ACD ..... (∠ ABD = ∠ ACD )
= AC / sin∠ADC ...... ( By sine formula in △ADC)
= AC / sin90°
= AC
= 2ri.e. BD = 2r sin(α + β)
b)BD = 2r sin(α + β)
BD = 2r (sinα cosβ + cosα sinβ)
BD = 2r (CD/AC * AB/AC + AD/AC * BC/AC)
BD = AC (CD/AC * AB/AC + AD/AC * BC/AC)
AC * BD = AB * CD + AD * BC

3c)cos² (2π/7) + cos² (4π/7) + cos² (6π/7) = ( cos (2π/7) + cos (4π/7) + cos(6π/7) )²
- 2 ( cos (2π/7) cos (4π/7) + cos (4π/7) cos(6π/7) + cos(6π/7) cos (2π/7) ) = (- 1/2)²
- 2 (1/2)( cos(6π/7) + cos(2π/7) + cos(10π/7) + cos(2π/7)
+ cos(8π/7) + cos(4π/7) ) = 1/4
- ( cos(2π/7) + cos(4π/7) + cos(6π/7) )
- ( cos(2π/7) + cos(10π/7) + cos(8π/7) ) = 1/4
- ( cos(2π/7) + cos(4π/7) + cos(6π/7) )
- ( cos(2π/7) + cos(4π/7) + cos(6π/7) ) = 1/4 + 1/2 + 1/2= 5/4

pls visit below webpage for the solutions

20b.
http://upload.lsforum.net/users/public/h5497820bh102.bmp

25d,
http://upload.lsforum.net/users/public/d179625dm102.bmp

26ab
http://upload.lsforum.net/users/public/c5103226abr102.jpg

3c
http://upload.lsforum.net/users/public/w286853cf102.bmp