What is the value of ((9arccos(cos(1))+4)^{3/2}-8)/(27) ?

The value of ((9arccos(cos(1))+4)^{3/2}-8)/(27) is (13*13^{1/2}-8)/(27)

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

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

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

Decimal Notation

82.49

Solution steps

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

Rewrite using trig identities:

arccos (cos (1)) = 1

arccos (cos (1))

For

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}

Simplify

\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}}

Multiply the numbers:

9 \cdot 1 = 9

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

Add the numbers:

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}}

Apply exponent rule:

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

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

Refine

= 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}

sin (9.6)

\frac{tan ( 1 )}{2}

sin (- \frac{π}{2}) + 1

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

cos (- \frac{1}{4} π)

What is the value of ((9arccos(cos(1))+4)^{3/2}-8)/(27) ?
The value of ((9arccos(cos(1))+4)^{3/2}-8)/(27) is (13*13^{1/2}-8)/(27)