What is the value of sqrt(2)cos(pi/2)-(pi/2) ?

The value of sqrt(2)cos(pi/2)-(pi/2) is -pi/2

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sqrt(2)cos(pi/2)-(pi/2)

2 cos (\frac{π}{2}) - (\frac{π}{2})

- \frac{π}{2}

Decimal Notation

- 1.57079 \dots

Solution steps

2 cos (\frac{π}{2}) - (\frac{π}{2})

= 2 cos (\frac{π}{2}) - \frac{π}{2}

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

= 2 \cdot 0 - \frac{π}{2}

Simplify

= - \frac{π}{2}

2 (sin (3 0^{\circ})) - tan (4 5^{\circ})

3 + 2 sin (\frac{π}{2})

sin (arcsin (\frac{12}{13}) + arcsin (\frac{4}{5}))

\frac{9810 sin ( 12 0 ^{\circ} )}{sin ( 9 0 ^{\circ} )}

e^{2 s i n (\frac{7 π}{6})}

What is the value of sqrt(2)cos(pi/2)-(pi/2) ?
The value of sqrt(2)cos(pi/2)-(pi/2) is -pi/2