Nagy-Gombás Szilvi
{ Tanár }
megoldása
3 éve
120⁰ = (120 *π)/180 = 2π/3
sin120⁰ = (II. sn.) sin(180⁰ - 120⁰) = sin60⁰ = √ 3 /2
cos120⁰ = (II. sn.) -cos(180⁰ - 120⁰) = -cos60⁰ = -1/2
tg120⁰ = (II. sn.) -tg(180⁰ - 120⁰) = -tg60⁰ = -√ 3
ctg120⁰ = (II. sn.) -ctg(180⁰ - 120⁰) = -ctg60⁰ = -√ 3 /3
135⁰ = (135 *π)/180 = 3π/4
sin135⁰ = (II. sn.) sin(180⁰ - 135⁰) = sin45⁰ = √ 2 /2
cos135⁰ = (II. sn.) -cos(180⁰ - 135⁰) = -cos45⁰ = -√ 2 /2
tg135⁰ = (II. sn.) -tg(180⁰ - 135⁰) = -tg45⁰ = -1
ctg135⁰ = (II. sn.) -ctg(180⁰ - 135⁰) = -ctg45⁰ = -1
150⁰ = (150 *π)/180 = 5π/6
sin150⁰ = (II. sn.) sin(180⁰ - 150⁰) = sin30⁰ = 1/2
cos150⁰ = (II. sn.) -cos(180⁰ - 150⁰) = -cos30⁰ = -√ 3 /2
tg150⁰ = (II. sn.) -tg(180⁰ - 150⁰) = -tg30⁰ = -√ 3 /3
ctg150⁰ = (II. sn.) -ctg(180⁰ - 150⁰) = -ctg30⁰ = -√ 3
180⁰ = π
sin180⁰ = 0
cos180⁰ = -1
tg180⁰ = 0
ctg180⁰ = nincs értelmezve
210⁰ = (210 *π)/180 = 7π/6
sin210⁰ = (III. sn.) -sin(210⁰ - 180⁰) = -sin30⁰ = -1/2
cos210⁰ = (III. sn.) -cos(210⁰ - 180⁰) = -cos30⁰ = -√ 3 /2
tg210⁰ = (III. sn.) tg(210⁰ - 180⁰) = tg30⁰ = √ 3 /3
ctg210⁰ = (III. sn.) ctg(210⁰ - 180⁰) = ctg30⁰ = √ 3
225⁰ = (225 *π)/180 = 5π/4
sin225⁰ = (III. sn.) -sin(225⁰ - 180⁰) = -sin45⁰ = -√ 2 /2
cos225⁰ = (III. sn.) -cos(225⁰ - 180⁰) = -cos45⁰ = -√ 2 /2
tg225⁰ = (III. sn.) tg(225⁰ - 180⁰) = tg45⁰ = 1
ctg225⁰ = (III. sn.) ctg(225⁰ - 180⁰) = ctg45⁰ = 1
240⁰ = (240 *π)/180 = 4π/3
sin240⁰ = (III. sn.) -sin(240⁰ - 180⁰) = -sin60⁰ = -√ 3 /2
cos240⁰ = (III. sn.) -cos(240⁰ - 180⁰) = -cos60⁰ = -1/2
tg240⁰ = (III. sn.) tg(240⁰ - 180⁰) = tg60⁰ = √ 3
ctg240⁰ = (III. sn.) ctg(240⁰ - 180⁰) = ctg60⁰ = √ 3 /3
270⁰ = (270 *π)/180 = 3π/2
sin270⁰ = -1
cos270⁰ = 0
tg270⁰ = nincs értelmezve
ctg270⁰ = 0
300⁰ = (300 *π)/180 = 5π/3
sin300⁰ = (IV. sn.) -sin(360⁰ - 300⁰) = -sin60⁰ = -√ 3 /2
cos300⁰ = (IV. sn.) cos(360⁰ - 300⁰) = cos60⁰ = 1/2
tg300⁰ = (IV. sn.) -tg(360⁰ - 300⁰) = -tg60⁰ = -√ 3
ctg300⁰ = (IV. sn.) -ctg(360⁰ - 300⁰) = -ctg60⁰ = -√ 3 /3
315⁰ = (315 *π)/180 = 7π/4
sin315⁰ = (IV. sn.) -sin(360⁰ - 315⁰) = -sin45⁰ = -√ 2 /2
cos315⁰ = (IV. sn.) cos(360⁰ - 315⁰) = cos45⁰ = √ 2 /2
tg315⁰ = (IV. sn.) -tg(360⁰ - 315⁰) = -tg45⁰ = -1
ctg315⁰ = (IV. sn.) -ctg(360⁰ - 315⁰) = -ctg45⁰ = -1
330⁰ = (330 *π)/180 = 11π/6
sin330⁰ = (IV. sn.) -sin(360⁰ - 330⁰) = -sin30⁰ = -1/2
cos330⁰ = (IV. sn.) cos(360⁰ - 330⁰) = cos30⁰ = √ 3 /2
tg330⁰ = (IV. sn.) -tg(360⁰ - 330⁰) = -tg30⁰ = -√ 3 /3
ctg330⁰ = (IV. sn.) -ctg(360⁰ - 330⁰) = -ctg30⁰ = -√ 3
0
1
Kommentek