Question

1.oblicz wartość wyrażenia:
a.)sin²13° + cos²13° +tg13° × ctg13°
b.)sin35° × cos55° + cos35° × sin55°
c.)2(sin²25° + cos²25°) / 2 tg25° × ctg25°
d.)3 tg20° × tg 70°
e.)tg 35°× tg45° × tg55°
f.)sin²17° + sin²73°
g.)tg18°/ctg72° - ctg33°/tg57°
h.)sin15°/cos75° + cos75°/sin15°


z góry dziękuje!!
nagradzam najlepszą!!!

Answer 1

A)0,050625+0,948676+0,231*4,331=1,999762
b)0,574*0,574+0,819*0,819=0,329476+0,670761=1,000273
c)2(0,178929+0,820836)/2*0,466*2,145=1,99953/1,99914=1,000195084
d)3*0,364*2,747=1,092*2,747=2,999724
e)0,700+1,000*1,428=0,700+1,428=2,128
f)0,085264+0,913936=0,9992
g)0,325/0,325-1,540/1,428=1-1,078431373=-0,078431373
h)0,259/0,259+0,259/0,259=1+1=2

Answer 2

A) sin²13° + cos²13° + tg13° × ctg13° =
= 1 + sin13° / cos13° × cos13° /sin13° =
= 1 +1 = 2


b) sin35° × cos55° + cos35° × sin55° =
= sin35° × sin(90°-55°) + sin(90°-35°) × sin55° =
= sin35° × sin35° + sin55° × sin55° =
= sin²35° + sin²55° = sin²35° + cos²(90°-55°) =
= sin²35° + cos²35° = 1


c) 2(sin²25° + cos²25°) / 2 tg25° × ctg25° =
= 2×1 / (2 sin25°/cos25° × cos25°/sin25°) =
= 2 / 2 = 1


d) 3 tg20° × tg 70° =
= 3 tg20° × ctg(90°-70°) =
= 3 tg20° × ctg20° =
= 3 × 1 = 3


e) tg 35°× tg45° × tg55° =
= ctg(90°-35°) × 1 × tg55° =
= ctg55° × tg55° = 1


f) sin²17° + sin²73° = sin²17° + cos²(90°-73°) =
= sin²17° + cos²17° = 1


g) tg18°/ctg72° - ctg33°/tg57° =
= ctg(90°-18°) × tg72° - tg(90°-33°) × ctg57° =
= ctg 72° × tg72° - tg57° × ctg57° = 1 - 1 = 0

h) sin15°/cos75° + cos75°/sin15° =
= sin15°/sin(90°-75°) + cos75°/cos(90°-15°) =
= sin15°/sin15° + cos75°/cos75° = 1 +1 =2

Answer 3

A)sin²13 + cos²13 + tg13* ctg 13 = 1 + 1 = 2
korzystamy z sin²α + cos²α = 1 oraz tgα*cygα = 1
b)
sin35*cos 55 + cos35 *sin55= sin35*sin35 + cos35*cos35 = 1
bo sin²α = cos²α = 1 oraz
cos 55 = cos (90-35) = sin 35
sin 55 = sin(90- 35) = cos 35
c)
= 1 , bo sin²25 + cos²25 = 1 oraz tg 25*ctg 25 = 1
d)
3tg 20 *tg 70 = 3 tg20*ctg20 = 3*1 = 3
bo tg 70 = tg(90-20) = ctg 20
e)
tg35*tg45*tg55 = tg35*ctg35*1 = 1*1 = 1
bo tg 45 = 1, tg55 = tg(90-35) = ctg 35
f)
sin²17 + sin²73 = sin²17 + cos²17 = 1
bo sin 73 = sin(90-17) = cos 17
g)
tg18/ctg72 -ctg33/tg57 = tg18/tg18 - ctg33/ctg33 = 1-1 = 0
bo ctg 72 = ctg(90-18) = tg18
tg 57 = tg(90-33) = ctg33
h)
sin15/cos75 +cos 75/sin15 = sin15/sin15 + cos75/cos75 = 1+1 =2
bo cos 75 = cos(90-15) = sin15
oraz sin15 = sin(90-75) = cos 75
wszędzie w stopniach.