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

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

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-pi^2cos(pi(2))

- π^{2} cos (π (2))

- π^{2}

Decimal Notation

- 9.86960 \dots

Solution steps

- π^{2} cos (π (2))

= - π^{2} cos (π 2)

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^{2} \cdot 1

Simplify

= - π^{2}

- 2 (- 2 sec^{2} (0) + 3 sec^{4} (0))

2 sec (3 0^{\circ})

sin (\frac{60}{2})

cos (12 0^{\circ}) + tan (31 5^{\circ}) - sin (21 0^{\circ})

60 sin (2 5^{\circ})