What is the value of (sin^2(pi/2))/2 ?

The value of (sin^2(pi/2))/2 is 1/2

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(sin^2(pi/2))/2

\frac{sin ^{2} ( \frac{π}{2} )}{2}

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\frac{sin ^{2} ( \frac{π}{2} )}{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

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

Simplify

= \frac{1}{2}

2 cos^{2} (16 5^{\circ}) - 1

arccot (\frac{3}{2})

sin (1 5^{\circ}) cos (4 5^{\circ}) + cos (1 5^{\circ}) sin (4 5^{\circ})

cos (5 0^{\circ}) cos (2 0^{\circ}) + sin (5 0^{\circ}) sin (2 0^{\circ})

\frac{tan ( 4 5 ^{\circ} ) + tan ( 3 0 ^{\circ} )}{1 - tan ( 4 5 ^{\circ} ) tan ( 3 0 ^{\circ} )}