What is the value of [sin^2(45)cos^2(45)]^{3/2} ?

The value of [sin^2(45)cos^2(45)]^{3/2} is 1/8

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[sin^2(45)cos^2(45)]^{3/2}

[sin^{2} (4 5^{\circ}) cos^{2} (4 5^{\circ})]^{\frac{3}{2}}

\frac{1}{8}

Solution steps

[sin^{2} (4 5^{\circ}) cos^{2} (4 5^{\circ})]^{\frac{3}{2}}

= (sin^{2} (4 5^{\circ}) cos^{2} (4 5^{\circ}))^{\frac{3}{2}}

Use the following trivial identity:

sin (4 5^{\circ}) = \frac{2}{2}

Use the following trivial identity:

cos (4 5^{\circ}) = \frac{2}{2}

= (\frac{2}{2})^{2} (\frac{2}{2})^{2}^{\frac{3}{2}}

Simplify

(\frac{2}{2})^{2} (\frac{2}{2})^{2}^{\frac{3}{2}} : \frac{1}{8}

= \frac{1}{8}

cos (11 1^{\circ}) + sin (11 1^{\circ})

\frac{cos ( 3 0 ^{\circ} )}{sin ( 6 0 ^{\circ} ) + cos ( 4 5 ^{\circ} )}

6 (cos (\frac{π}{2}) + i sin (\frac{π}{2}))

cos^{2} (0) - 3 sin (0)

arccos (\frac{( 20 ) ^{2} + ( 45 ) ^{2} - ( 53 ) ^{2}}{2 \cdot 20 \cdot 45})

What is the value of [sin^2(45)cos^2(45)]^{3/2} ?
The value of [sin^2(45)cos^2(45)]^{3/2} is 1/8