What is the value of 2cos^2(15)-1 ?

The value of 2cos^2(15)-1 is (sqrt(3))/2

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2cos^2(15)-1

2 cos^{2} (1 5^{\circ}) - 1

\frac{3}{2}

Decimal Notation

0.86602 \dots

Solution steps

2 cos^{2} (1 5^{\circ}) - 1

Rewrite using trig identities:

cos (3 0^{\circ})

2 cos^{2} (1 5^{\circ}) - 1

Use the Double Angle identity:

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

= cos (2 \cdot 1 5^{\circ})

Simplify

= cos (3 0^{\circ})

= cos (3 0^{\circ})

Use the following trivial identity:

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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}

2 sin (6 0^{\circ})

arctan (tan (\frac{10 π}{11}))

sin (1 9^{\circ})

cos^{2} (18 0^{\circ})

sec (arcsin (- \frac{1}{2}))