What is the value of 2cos^2((5pi)/(12))-1 ?

The value of 2cos^2((5pi)/(12))-1 is -(sqrt(3))/2

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2cos^2((5pi)/(12))-1

2 cos^{2} (\frac{5 π}{12}) - 1

- \frac{3}{2}

Decimal Notation

- 0.86602 \dots

Solution steps

2 cos^{2} (\frac{5 π}{12}) - 1

Rewrite using trig identities:

cos (\frac{5 π}{6})

2 cos^{2} (\frac{5 π}{12}) - 1

Use the Double Angle identity:

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

= cos (2 \cdot \frac{5 π}{12})

Simplify:

2 \cdot \frac{5 π}{12} = \frac{5 π}{6}

2 \cdot \frac{5 π}{12}

Multiply fractions:

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

= \frac{5 π 2}{12}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{12}

Cancel the common factor:

2

= \frac{5 π}{6}

= cos (\frac{5 π}{6})

= cos (\frac{5 π}{6})

Use the following trivial identity:

cos (\frac{5 π}{6}) = - \frac{3}{2}

cos (\frac{5 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}

8 tan (\frac{π}{4})

arctan (\frac{4 3}{- 4})

cos (9 2^{\circ})

6 cos (10 0^{\circ})

tan (arcsin (\frac{5}{9}))

What is the value of 2cos^2((5pi)/(12))-1 ?
The value of 2cos^2((5pi)/(12))-1 is -(sqrt(3))/2