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三⾓形の相互関係 tan θ = sin θ

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三角形の相互関係 ②

三⾓形の相互関係 tan θ = sin θ

cos θ

1 + tan

2

θ = 1 cos

2

θ cos

2

θ + sin

2

θ = 1

> 第3章 図形 計量 > 第1節 三⾓⽐ > 第2講:三⾓⽐ 相互関係

I

例題

のとき,次の値を求めなさい。ただし,

は鋭⾓とする。

tan

θ

= 2

θ

(1)

cos θ

(2)

sin θ

(1) (2)

cosθ = 1 5 1 + tan2θ = 1 から,

cos2θ 1 + 22 = 1

cos2θ 5 = 1

cos2θ cos2θ = 15

であるから,

cosθ > 0

= 5 5

tanθ = sinθ から,

cosθ

= 2× 1 5

= 2 5 5

sinθ = tanθ ×cosθ

を


  使って解いてもよい

cos2θ +sin2θ = 1 3

tanθ = sinθ

cosθ より,

tanθ cosθ = sinθ

cos2θ + sin2θ = 1 に代⼊すると,

cos2θ + (tanθ cosθ)2 = 1 cos2θ + tan2θcos2θ = 1

1 + tan2θ = 1cos2θ

参照

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