F.3簡單三角數問題

1.
5sin θ - 2cos θ / sinθ
=5sin θ / sin θ - 2cos θ / sin θ
=5 - 2 / tan θ
=5 - 2 / 3
=13 / 3
2.
sin θ
=√(1 - cos^2 θ)
=√[1 - (2/3)^2]
=√5 / 3
tan θ
=sin θ / cos θ
=(√5 / 3) / (2 / 3)
=√5 / 2
3.
L.H.S.
=sin θ tan (90° - θ) + 3 cosθ
=sin θ cos θ / sin θ + 3 cosθ
=cos θ + 3 cosθ
=4cos θ
=4sin(90- θ)
=R.H.S.
4.
L.H.S.
=sin^2 θ - sin^2 (90° - θ)
=sin^2 θ - cos^2 θ
=1 - cos^2 θ - cos^2 θ
=1 - 2 cos^2 θ
=R.H.S.

1.
(5sin θ - 2cos θ)/ sinθ
=5-2/3
=13/3

2.
sin^2 θ+cos^2 θ=1
sin^2 θ=1-cos^2 θ
sin^2 θ=1-(2/3)^2
sin^2 θ=5/9
sin θ=[5^(1/2)]/3

tan θ =sin θ /cos θ
tan θ= {[5^(1/2)]/3}/(2/3)
tan θ=[5^(1/2)]/2

3
left=sinθ /tanθ +3cosθ
left=cosθ +3cosθ
left=4cosθ
left=4sin(90-θ )
left=right

4
left=sin^2θ-cos^2θ
left=1-cos^2θ-cos^2θ
left=1-2cos^2θ
left=right

1 tan θ =3, sin θ = 3/√10, cosθ = 1/√10

5sin θ - 2cos θ / sinθ

= 15/√10 - (2/√10) * ( √10/3)

= 15/√10 - 2/3

2 cos θ = 2/3, sin θ =√(3^2-2^2)/ 3 = √5/3

tan θ = √5/2

3 sin θ tan (90° - θ) + 3 cosθ

= sin θ/ tan θ + 3 cosθ

= sin θ * (cosθ/sin θ) + 3 cosθ

= 4cosθ

= 4sin(90 - θ)

4 sin^2 θ - sin^2 (90° - θ)

= sin^2 θ - cos^2 θ

= 1 - cos^2 θ - cos^2 θ

= 1 - 2cos^2 θ