What is the value of 2cos^2((3pi)/8)-1 ?

The value of 2cos^2((3pi)/8)-1 is -(sqrt(2))/2

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2cos^2((3pi)/8)-1

2 cos^{2} (\frac{3 π}{8}) - 1

- \frac{2}{2}

Decimal Notation

- 0.70710 \dots

Solution steps

2 cos^{2} (\frac{3 π}{8}) - 1

Rewrite using trig identities:

cos (\frac{3 π}{4})

2 cos^{2} (\frac{3 π}{8}) - 1

Use the Double Angle identity:

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

= cos (2 \cdot \frac{3 π}{8})

Simplify:

2 \cdot \frac{3 π}{8} = \frac{3 π}{4}

2 \cdot \frac{3 π}{8}

Multiply fractions:

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

= \frac{3 π 2}{8}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{6 π}{8}

Cancel the common factor:

2

= \frac{3 π}{4}

= cos (\frac{3 π}{4})

= cos (\frac{3 π}{4})

Use the following trivial identity:

cos (\frac{3 π}{4}) = - \frac{2}{2}

cos (\frac{3 π}{4})

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{2}{2}

= - \frac{2}{2}

sin (- 8)

- 3 π + sin^{2} (- \frac{3 π}{3}) - 8

arcsin (0.15)

\frac{cos ( π )}{sin ( π )}

csc (- 40 5^{\circ})

What is the value of 2cos^2((3pi)/8)-1 ?
The value of 2cos^2((3pi)/8)-1 is -(sqrt(2))/2