tentukan bentuk trigonometri ini dengan tabel sudut istimewa 1. sin 210 2. cos 150 3. sin 315 4. cos 240 5. tan 315 6.cot 300 7. sin 750 8. cos 1110 9. tan 570
Matematika
lodriko
Pertanyaan
tentukan bentuk trigonometri ini dengan tabel sudut istimewa
1. sin 210
2. cos 150
3. sin 315
4. cos 240
5. tan 315
6.cot 300
7. sin 750
8. cos 1110
9. tan 570
beserta penjelasannya
1. sin 210
2. cos 150
3. sin 315
4. cos 240
5. tan 315
6.cot 300
7. sin 750
8. cos 1110
9. tan 570
beserta penjelasannya
1 Jawaban
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1. Jawaban asobri68
1. sin 210° = sin (180°+30°)
= sin (180°)*cos(30°)+ cos (180°)*sin(30°)
= (0)*(√3/2)+ (-1)*(1/2)
= 0 - 1/2
= -1/2
2. cos 150° = sin (90° - 150°)
= sin (-60°)
= -sin (60°)
= -√3/2
3. sin 315° = cos (90° - 315°)
= cos (-225°)
= -cos (225°)
= cos (225°)
= cos (180° + 45°)
= cos (180°) cos (40°) - sin (180°) sin (45°)
= (-1) (√2/2) - (0) (√2/2)
= (-√2/2) - (0)
= -√2/2
4. cos 240° = cos (180° + 60°)
= cos (180°) cos (60°) - sin (180°) sin (60°)
= (-1)*(1/2) - (0)*(√3/2)
= (-1/2) - (0)
= -1/2
5. tan 315° = tan (((4+3)/4)180°)
= tan (((4/4)+(3/4))180°)
= tan (180°+(3/4)180°)
= tan (180°) + tan ((3/4)180°)
= (0) + tan (135°)
= tan (135°)
= tan (180° - 45°)
= tan (180°) - tan 45°
= (0) - 1
= -1
6. cos 300° = sin (90° - 300°)
= sin (-210°)
= -sin (210°)
= - (-1/2)
= 1/2
7. sin 750° = sin (((24+1)/6)180°)
= sin (((24/6)+(1/6))180°)
= sin (360°*2+(1/6)180°)
= sin ((1/6)180°)
= sin (180°/6)
= sin (30°)
= 1/2
8. cos 1110 = cos (((36+1)/6)180°)
= cos (((36/6)+(1/6))180°)
= cos (360°*3+(1/6)180°)
= cos ((1/6)180°)
= cos (180°/6)
= cos (30°)
= √3/2
9. tan 570 = tan (((18+1)/6)180°) = tan (((18/6)+(1/6))180°) = tan (180°*3+(1/6)180°) = tan (180°*3) + tan ((1/6)180°) = (0) + tan (30°) = tan (30°) = √3/3