What is the value of (-sin(pi/2))/((1+(cos^2(pi/2)))^{3/2)} ?

The value of (-sin(pi/2))/((1+(cos^2(pi/2)))^{3/2)} is -1

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(-sin(pi/2))/((1+(cos^2(pi/2)))^{3/2)}

\frac{- sin ( \frac{π}{2} )}{( 1 + ( cos ^{2} ( \frac{π}{2} ) ) ) ^{\frac{3}{2}}}

- 1

Solution steps

\frac{- sin ( \frac{π}{2} )}{( 1 + ( cos ^{2} ( \frac{π}{2} ) ) ) ^{\frac{3}{2}}}

= \frac{- sin ( \frac{π}{2} )}{( 1 + cos ^{2} ( \frac{π}{2} ) ) ^{\frac{3}{2}}}

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

= \frac{- 1}{( 1 + 0 ^{2} ) ^{\frac{3}{2}}}

Simplify

= - 1

\frac{sin ( 2 1 ^{\circ} ) + cos ( 2 1 ^{\circ} )}{cos ( 2 4 ^{\circ} )}

tan (\frac{\frac{3 π}{2}}{2}) - csc (\frac{3 π}{2})

- 6 sin (3) (\frac{2 π}{3})

140 cos (3 0^{\circ})

sin (\frac{π}{2} + 2 π)

What is the value of (-sin(pi/2))/((1+(cos^2(pi/2)))^{3/2)} ?
The value of (-sin(pi/2))/((1+(cos^2(pi/2)))^{3/2)} is -1