What is the value of cos^2(pi/(12))-sin^2(pi/(12)) ?

The value of cos^2(pi/(12))-sin^2(pi/(12)) is (sqrt(3))/2

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cos^2(pi/(12))-sin^2(pi/(12))

cos^{2} (\frac{π}{12}) - sin^{2} (\frac{π}{12})

\frac{3}{2}

Decimal Notation

0.86602 \dots

Solution steps

cos^{2} (\frac{π}{12}) - sin^{2} (\frac{π}{12})

Rewrite using trig identities:

cos (\frac{π}{6})

cos^{2} (\frac{π}{12}) - sin^{2} (\frac{π}{12})

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

= cos (2 \cdot \frac{π}{12})

Simplify:

2 \cdot \frac{π}{12} = \frac{π}{6}

2 \cdot \frac{π}{12}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{12}

Cancel the common factor:

2

= \frac{π}{6}

= cos (\frac{π}{6})

= cos (\frac{π}{6})

Use the following trivial identity:

cos (\frac{π}{6}) = \frac{3}{2}

cos (\frac{π}{6})

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}

= \frac{3}{2}

= \frac{3}{2}

sin (π (1))

sin (\frac{π}{2}) + \frac{π}{2} cos (\frac{π}{2})

\frac{10}{sin ( 5 5 ^{\circ} )}

sin (\frac{π}{11})

sec (- 5)

What is the value of cos^2(pi/(12))-sin^2(pi/(12)) ?
The value of cos^2(pi/(12))-sin^2(pi/(12)) is (sqrt(3))/2