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人気のある三角関数 >

170=(90^2)/(16)sin(x)cos(x)

解

170 = \frac{9 0 ^{2}}{16} sin (x) cos (x)

解

x = \frac{0.73637 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.73637 \dots}{2} + πn

+1

度

x = 21.09552 \dots^{\circ} + 18 0^{\circ} n, x = 68.90447 \dots^{\circ} + 18 0^{\circ} n

解答ステップ

170 = \frac{9 0 ^{2}}{16} sin (x) cos (x)

辺を交換する

\frac{9 0 ^{2}}{16} sin (x) cos (x) = 170

三角関数の公式を使用して書き換える

\frac{9 0 ^{2}}{16} sin (x) cos (x)

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}

= \frac{sin ( 2 x )}{2} \cdot \frac{9 0 ^{2}}{16}

\frac{sin ( 2 x )}{2} \cdot \frac{9 0 ^{2}}{16} = 170

簡素化

\frac{sin ( 2 x )}{2} \cdot \frac{9 0 ^{2}}{16} : \frac{2025 sin ( 2 x )}{8}

\frac{sin ( 2 x )}{2} \cdot \frac{9 0 ^{2}}{16}

\frac{9 0 ^{2}}{16} = \frac{3 ^{4} \cdot 5 ^{2}}{2 ^{2}}

\frac{9 0 ^{2}}{16}

因数

9 0^{2} : 3^{4} \cdot 2^{2} \cdot 5^{2}

因数

90 = 3^{2} \cdot 2 \cdot 5

= (3^{2} \cdot 2 \cdot 5)^{2}

指数の規則を適用する:

(ab)^{c} = a^{c} b^{c}

= 2^{2} \cdot 5^{2} (3^{2})^{2}

簡素化

(3^{2})^{2} : 3^{4}

(3^{2})^{2}

指数の規則を適用する:

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

= 3^{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= 3^{4}

= 3^{4} \cdot 2^{2} \cdot 5^{2}

因数

16 : 2^{4}

因数

16 = 2^{4}

= \frac{3 ^{4} \cdot 2 ^{2} \cdot 5 ^{2}}{2 ^{4}}

キャンセル

\frac{2 ^{2} \cdot 3 ^{4} \cdot 5 ^{2}}{2 ^{4}} : \frac{3 ^{4} \cdot 5 ^{2}}{2 ^{2}}

\frac{2 ^{2} \cdot 3 ^{4} \cdot 5 ^{2}}{2 ^{4}}

指数の規則を適用する:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{2}}{2 ^{4}} = \frac{1}{2 ^{4 - 2}}

= \frac{3 ^{4} \cdot 5 ^{2}}{2 ^{4 - 2}}

数を引く:

4 - 2 = 2

= \frac{3 ^{4} \cdot 5 ^{2}}{2 ^{2}}

= \frac{3 ^{4} \cdot 5 ^{2}}{2 ^{2}}

= \frac{3 ^{4} \cdot 5 ^{2}}{2 ^{2}} \cdot \frac{sin ( 2 x )}{2}

分数を乗じる:

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

= \frac{sin ( 2 x ) \cdot 3 ^{4} \cdot 5 ^{2}}{2 \cdot 2 ^{2}}

2 \cdot 2^{2} = 2^{3}

2 \cdot 2^{2}

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{2} = 2^{1 + 2}

= 2^{1 + 2}

数を足す:

1 + 2 = 3

= 2^{3}

= \frac{3 ^{4} \cdot 5 ^{2} sin ( 2 x )}{2 ^{3}}

sin (2 x) \cdot 3^{4} \cdot 5^{2} = 2025 sin (2 x)

sin (2 x) \cdot 3^{4} \cdot 5^{2}

3^{4} = 81

= 5^{2} \cdot 81 sin (2 x)

5^{2} = 25

= 81 \cdot 25 sin (2 x)

数を乗じる:

81 \cdot 25 = 2025

= 2025 sin (2 x)

= \frac{2025 sin ( 2 x )}{2 ^{3}}

2^{3} = 8

= \frac{2025 sin ( 2 x )}{8}

\frac{2025 sin ( 2 x )}{8} = 170

以下で両辺を乗じる:

8

\frac{2025 sin ( 2 x )}{8} = 170

以下で両辺を乗じる:

8

\frac{8 \cdot 2025 sin ( 2 x )}{8} = 170 \cdot 8

簡素化

2025 sin (2 x) = 1360

2025 sin (2 x) = 1360

以下で両辺を割る

2025

2025 sin (2 x) = 1360

以下で両辺を割る

2025

\frac{2025 sin ( 2 x )}{2025} = \frac{1360}{2025}

簡素化

sin (2 x) = \frac{272}{405}

sin (2 x) = \frac{272}{405}

三角関数の逆数プロパティを適用する

sin (2 x) = \frac{272}{405}

以下の一般解

sin (2 x) = \frac{272}{405}

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

2 x = arcsin (\frac{272}{405}) + 2 πn, 2 x = π - arcsin (\frac{272}{405}) + 2 πn

2 x = arcsin (\frac{272}{405}) + 2 πn, 2 x = π - arcsin (\frac{272}{405}) + 2 πn

解く

2 x = arcsin (\frac{272}{405}) + 2 πn : x = \frac{arcsin ( \frac{272}{405} )}{2} + πn

2 x = arcsin (\frac{272}{405}) + 2 πn

以下で両辺を割る

2

2 x = arcsin (\frac{272}{405}) + 2 πn

以下で両辺を割る

2

\frac{2 x}{2} = \frac{arcsin ( \frac{272}{405} )}{2} + \frac{2 πn}{2}

簡素化

x = \frac{arcsin ( \frac{272}{405} )}{2} + πn

x = \frac{arcsin ( \frac{272}{405} )}{2} + πn

解く

2 x = π - arcsin (\frac{272}{405}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{272}{405} )}{2} + πn

2 x = π - arcsin (\frac{272}{405}) + 2 πn

以下で両辺を割る

2

2 x = π - arcsin (\frac{272}{405}) + 2 πn

以下で両辺を割る

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{272}{405} )}{2} + \frac{2 πn}{2}

簡素化

x = \frac{π}{2} - \frac{arcsin ( \frac{272}{405} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{272}{405} )}{2} + πn

x = \frac{arcsin ( \frac{272}{405} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{272}{405} )}{2} + πn

10進法形式で解を証明する

x = \frac{0.73637 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.73637 \dots}{2} + πn

グラフ

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