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人気のある三角関数 >

10tan(θ)-4.9((10)/(13cos(θ)))^2=0

解

10 tan (θ) - 4.9 (\frac{10}{13 cos ( θ )})^{2} = 0

解

θ = \frac{0.61858 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.61858 \dots}{2} + πn

+1

度

θ = 17.72110 \dots^{\circ} + 18 0^{\circ} n, θ = 72.27889 \dots^{\circ} + 18 0^{\circ} n

解答ステップ

10 tan (θ) - 4.9 (\frac{10}{13 cos ( θ )})^{2} = 0

サイン, コサインで表わす

- (\frac{10}{13 cos ( θ )})^{2} \cdot 4.9 + 10 tan (θ)

基本的な三角関数の公式を使用する:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - (\frac{10}{13 cos ( θ )})^{2} \cdot 4.9 + 10 \cdot \frac{sin ( θ )}{cos ( θ )}

簡素化

- (\frac{10}{13 cos ( θ )})^{2} \cdot 4.9 + 10 \cdot \frac{sin ( θ )}{cos ( θ )} : \frac{- 2.89940 \dots + 10 sin ( θ ) cos ( θ )}{cos ^{2} ( θ )}

- (\frac{10}{13 cos ( θ )})^{2} \cdot 4.9 + 10 \cdot \frac{sin ( θ )}{cos ( θ )}

(\frac{10}{13 cos ( θ )})^{2} \cdot 4.9 = \frac{2.89940 \dots}{cos ^{2} ( θ )}

(\frac{10}{13 cos ( θ )})^{2} \cdot 4.9

(\frac{10}{13 cos ( θ )})^{2} = \frac{1 0 ^{2}}{1 3 ^{2} cos ^{2} ( θ )}

(\frac{10}{13 cos ( θ )})^{2}

指数の規則を適用する:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 0 ^{2}}{( 13 cos ( θ ) ) ^{2}}

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

(13 cos (θ))^{2} = 1 3^{2} cos^{2} (θ)

= \frac{1 0 ^{2}}{1 3 ^{2} cos ^{2} ( θ )}

= 4.9 \cdot \frac{1 0 ^{2}}{1 3 ^{2} cos ^{2} ( θ )}

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 0 ^{2} \cdot 4.9}{1 3 ^{2} cos ^{2} ( θ )}

1 0^{2} \cdot 4.9 = 490

1 0^{2} \cdot 4.9

1 0^{2} = 100

= 100 \cdot 4.9

数を乗じる:

100 \cdot 4.9 = 490

= 490

= \frac{490}{1 3 ^{2} cos ^{2} ( θ )}

1 3^{2} = 169

= \frac{490}{169 cos ^{2} ( θ )}

元を10進法形式に変換する

\frac{490}{169} = 2.89940 \dots

= \frac{2.89940 \dots}{cos ^{2} ( θ )}

10 \cdot \frac{sin ( θ )}{cos ( θ )} = \frac{10 sin ( θ )}{cos ( θ )}

10 \cdot \frac{sin ( θ )}{cos ( θ )}

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ( θ ) \cdot 10}{cos ( θ )}

= - \frac{2.89940 \dots}{cos ^{2} ( θ )} + \frac{10 sin ( θ )}{cos ( θ )}

以下の最小公倍数:

cos^{2} (θ), cos (θ) : cos^{2} (θ)

cos^{2} (θ), cos (θ)

最小公倍数 (LCM)

cos^{2} (θ)

または以下のいずれかに現れる因数で構成された式を計算する:

cos (θ)

= cos^{2} (θ)

LCMに基づいて分数を調整する

該当する分母を乗じてLCMに変えるために

必要な量で各分子を乗じる

cos^{2} (θ)

\frac{sin ( θ ) \cdot 10}{cos ( θ )}

の場合

:

分母と分子に以下を乗じる:

cos (θ)

\frac{sin ( θ ) \cdot 10}{cos ( θ )} = \frac{sin ( θ ) \cdot 10 cos ( θ )}{cos ( θ ) cos ( θ )} = \frac{sin ( θ ) \cdot 10 cos ( θ )}{cos ^{2} ( θ )}

= - \frac{2.89940 \dots}{cos ^{2} ( θ )} + \frac{sin ( θ ) \cdot 10 cos ( θ )}{cos ^{2} ( θ )}

分母が等しいので, 分数を組み合わせる:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 2.89940 \dots + sin ( θ ) \cdot 10 cos ( θ )}{cos ^{2} ( θ )}

= \frac{- 2.89940 \dots + 10 sin ( θ ) cos ( θ )}{cos ^{2} ( θ )}

\frac{- 2.89940 \dots + 10 cos ( θ ) sin ( θ )}{cos ^{2} ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 2.89940 \dots + 10 cos (θ) sin (θ) = 0

三角関数の公式を使用して書き換える

- 2.89940 \dots + 10 cos (θ) sin (θ)

2倍角の公式を使用:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= - 2.89940 \dots + 10 \cdot \frac{sin ( 2 θ )}{2}

- 2.89940 \dots + 10 \cdot \frac{sin ( 2 θ )}{2} = 0

10 \cdot \frac{sin ( 2 θ )}{2} = 5 sin (2 θ)

10 \cdot \frac{sin ( 2 θ )}{2}

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ( 2 θ ) \cdot 10}{2}

数を割る:

\frac{10}{2} = 5

= 5 sin (2 θ)

- 2.89940 \dots + 5 sin (2 θ) = 0

2.89940 \dots

を右側に移動します

- 2.89940 \dots + 5 sin (2 θ) = 0

両辺に

2.89940 \dots

を足す

- 2.89940 \dots + 5 sin (2 θ) + 2.89940 \dots = 0 + 2.89940 \dots

簡素化

5 sin (2 θ) = 2.89940 \dots

5 sin (2 θ) = 2.89940 \dots

以下で両辺を割る

5

5 sin (2 θ) = 2.89940 \dots

以下で両辺を割る

5

\frac{5 sin ( 2 θ )}{5} = \frac{2.89940 \dots}{5}

簡素化

sin (2 θ) = 0.57988 \dots

sin (2 θ) = 0.57988 \dots

三角関数の逆数プロパティを適用する

sin (2 θ) = 0.57988 \dots

以下の一般解

sin (2 θ) = 0.57988 \dots

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

2 θ = arcsin (0.57988 \dots) + 2 πn, 2 θ = π - arcsin (0.57988 \dots) + 2 πn

2 θ = arcsin (0.57988 \dots) + 2 πn, 2 θ = π - arcsin (0.57988 \dots) + 2 πn

解く

2 θ = arcsin (0.57988 \dots) + 2 πn : θ = \frac{arcsin ( 0.57988 \dots )}{2} + πn

2 θ = arcsin (0.57988 \dots) + 2 πn

以下で両辺を割る

2

2 θ = arcsin (0.57988 \dots) + 2 πn

以下で両辺を割る

2

\frac{2 θ}{2} = \frac{arcsin ( 0.57988 \dots )}{2} + \frac{2 πn}{2}

簡素化

θ = \frac{arcsin ( 0.57988 \dots )}{2} + πn

θ = \frac{arcsin ( 0.57988 \dots )}{2} + πn

解く

2 θ = π - arcsin (0.57988 \dots) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( 0.57988 \dots )}{2} + πn

2 θ = π - arcsin (0.57988 \dots) + 2 πn

以下で両辺を割る

2

2 θ = π - arcsin (0.57988 \dots) + 2 πn

以下で両辺を割る

2

\frac{2 θ}{2} = \frac{π}{2} - \frac{arcsin ( 0.57988 \dots )}{2} + \frac{2 πn}{2}

簡素化

θ = \frac{π}{2} - \frac{arcsin ( 0.57988 \dots )}{2} + πn

θ = \frac{π}{2} - \frac{arcsin ( 0.57988 \dots )}{2} + πn

θ = \frac{arcsin ( 0.57988 \dots )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( 0.57988 \dots )}{2} + πn

10進法形式で解を証明する

θ = \frac{0.61858 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.61858 \dots}{2} + πn

グラフ

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