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受欢迎的三角函数 >

((9arccos(cos(1))+4)^{3/2}-8)/(27)

解答

\frac{( 9 arccos ( cos ( 1 ) ) + 4 ) ^{\frac{3}{2}} - 8}{27}

解答

\frac{13 \cdot 1 3 ^{\frac{1}{2}} - 8}{27}

+1

十进制

82.49

求解步骤

\frac{( 9 arccos ( cos ( 1 ) ) + 4 ) ^{\frac{3}{2}} - 8}{27}

使用三角恒等式改写:

arccos (cos (1)) = 1

arccos (cos (1))

对于

0 \leq x \leq π, a rccos (cos (x)) = x

0 \leq 1 \leq π

= 1

= \frac{( 9 \cdot 1 + 4 ) ^{\frac{3}{2}} - 8}{27}

化简

\frac{( 9 \cdot 1 + 4 ) ^{\frac{3}{2}} - 8}{27} : \frac{13 \cdot 1 3 ^{\frac{1}{2}} - 8}{27}

\frac{( 9 \cdot 1 + 4 ) ^{\frac{3}{2}} - 8}{27}

(9 \cdot 1 + 4)^{\frac{3}{2}} = 1 3^{\frac{3}{2}}

(9 \cdot 1 + 4)^{\frac{3}{2}}

数字相乘:

9 \cdot 1 = 9

= (9 + 4)^{\frac{3}{2}}

数字相加:

9 + 4 = 13

= 1 3^{\frac{3}{2}}

= \frac{1 3 ^{\frac{3}{2}} - 8}{27}

1 3^{\frac{3}{2}} = 13 \cdot 1 3^{\frac{1}{2}}

1 3^{\frac{3}{2}}

1 3^{\frac{3}{2}} = 1 3^{1 + \frac{1}{2}}

= 1 3^{1 + \frac{1}{2}}

使用指数法则:

x^{a + b} = x^{a} x^{b}

= 1 3^{1} \cdot 1 3^{\frac{1}{2}}

整理后得

= 13 \cdot 1 3^{\frac{1}{2}}

= \frac{13 \cdot 1 3 ^{\frac{1}{2}} - 8}{27}

= \frac{13 \cdot 1 3 ^{\frac{1}{2}} - 8}{27}

流行的例子

sin (9.6)

\frac{sin ( 6 0 ^{\circ} )}{11}

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- cos (π) - \frac{1}{2} sin (2) π^{2}

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