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

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cos^2(-pi)

cos^{2} (- π)

1

Solution steps

cos^{2} (- π)

Rewrite using trig identities:

cos (- π) = - 1

cos (- π)

Use the following property:

cos (- x) = cos (x)

cos (- π) = cos (π)

= cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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

= (- 1)^{2}

Simplify

(- 1)^{2} : 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

= 1

cos (\frac{11 π}{9})

cos (\frac{240}{2})

arcsin (\frac{2}{2})

tan (13 0^{\circ})

sin (2) - sin (1)