What is the value of -picos(pi(2))pi ?

The value of -picos(pi(2))pi is -pi^2

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-picos(pi(2))pi

- π cos (π (2)) π

- π^{2}

Decimal Notation

- 9.86960 \dots

Solution steps

- π cos (π (2)) π

= - π cos (π 2) π

Rewrite using trig identities:

cos (π 2) = 1

cos (π 2)

cos (π 2) = cos (0)

cos (π 2)

Rewrite

π 2

2 π + 0

= cos (2 π + 0)

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

cos (2 π + 0) = cos (0)

= cos (0)

= cos (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

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}

= 1

= 1

= - π 1 π

Simplify

- π 1 π : - π^{2}

- π 1 π

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

ππ = π^{1 + 1}

= - 1 \cdot π^{1 + 1}

Add the numbers:

1 + 1 = 2

= - 1 \cdot π^{2}

Multiply:

1 \cdot π^{2} = π^{2}

= - π^{2}

= - π^{2}

sin (0.28)

arccos (\frac{8}{78})

1 - 2 cos (π)

arctan (\frac{2.55}{9.807} + 0.5)

cos (113 5^{\circ})