1. sin 60° per 1 + cos 60° 2. 1 - cos 60° per sin 60° 3. sin 60° - 2 sin 30°.cos 30° 4.cosec 30° + codec 60° + codec 90° per cos 0° + cos 30° + cos 60°
Matematika
Ezequiel1
Pertanyaan
1. sin 60° per 1 + cos 60°
2. 1 - cos 60° per sin 60°
3. sin 60° - 2 sin 30°.cos 30°
4.cosec 30° + codec 60° + codec 90° per cos 0° + cos 30° + cos 60°
2. 1 - cos 60° per sin 60°
3. sin 60° - 2 sin 30°.cos 30°
4.cosec 30° + codec 60° + codec 90° per cos 0° + cos 30° + cos 60°
1 Jawaban
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1. Jawaban asobri68
1. sin 60° /1 + cos 60° = (√3/2)/(1+(1/2))
= √3/(2*(3/2))
= √3/3
2. 1 - cos 60° / sin 60° = (1 - 1/2)/(√3/2)
= (2*(-1/2+1))/√3
= (2*1/2)/√3
= 1/√3
= (1*√3)/(√3*√3)
= √3/3
3. sin 60° - 2 sin 30°.cos 30° = sin 60° - 2*(sin(2*30))/2
= sin 60° - 2*(sin 60°)/2
= sin 60° - sin 60°
= 0
4. cosec 30° + cosec 60° + cosec 90° per cos 0° + cos 30° + cos 60°
cosec 30° = 1/sin 30° = 1/(1/2) = 2
cosec 60° = 1/sin 60° = 1/(√3/2) = 2/√3 = (2*√3)/(√3*√3) = 2√3/3
cosec 90° = 1/sin 90° = 1/1 = 1
cos 0° = 1
cos 30° = √3/2
cos 60° = 1/2
=> cosec 30° + cosec 60° + cosec 90° per cos 0° + cos 30° + cos 60°
=> (2 + 2√3/3 + 1)/(1 + √3/2 + 1/2)
=> (3+2√3/3)/(1+(√3+1)/2)
=> (3+2√3/3)/((3+√3)/2)
=> ((3+2√3)/3)*((3+√3)/2)/((3+√3)/2)*((3+√3)/2)
=> (9+(2√3/3))/((3+√3)/2)
=> (9+(2√3)*2)/(3*(3+√3))
=> (2*((2√3)+9)))/(3*(3+√3))
=> (2*((2√3)+9)))*(3-√3)/(3*(3+√3))*(3-√3)
=> (4-16√3)/18
=> 6(7-√3)/18
=> (7-√3)/3