What is the value of cos^2(3pi) ?

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cos^2(3pi)

cos^{2} (3 π)

1

Solution steps

cos^{2} (3 π)

Rewrite using trig identities:

cos (3 π) = - 1

cos (3 π)

cos (3 π) = cos (π)

cos (3 π)

Rewrite

3 π

2 π + π

= cos (2 π + π)

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

cos (2 π + π) = 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

sin^{2} (\frac{1}{4} π)

cot (6 2^{\circ})

cos (\frac{13}{12})

cos (\frac{19 π}{30}) cos (\frac{π}{5}) - sin (\frac{19 π}{30}) sin (\frac{π}{5})

sec (85 5^{\circ})