What is the value of 1-2sin^2((7pi)/8) ?

The value of 1-2sin^2((7pi)/8) is (sqrt(2))/2

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1-2sin^2((7pi)/8)

1 - 2 sin^{2} (\frac{7 π}{8})

\frac{2}{2}

Decimal Notation

0.70710 \dots

Solution steps

1 - 2 sin^{2} (\frac{7 π}{8})

Rewrite using trig identities:

sin (\frac{7 π}{8}) = sin (\frac{π}{8})

sin (\frac{7 π}{8})

Use the basic trigonometric identity:

sin (x) = sin (π - x)

= sin (π - \frac{7 π}{8})

Simplify:

π - \frac{7 π}{8} = \frac{π}{8}

π - \frac{7 π}{8}

Convert element to fraction:

π = \frac{π 8}{8}

= \frac{π 8}{8} - \frac{7 π}{8}

Since the denominators are equal, combine the fractions:

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

= \frac{π 8 - 7 π}{8}

Add similar elements:

8 π - 7 π = π

= \frac{π}{8}

= sin (\frac{π}{8})

= 1 - 2 sin^{2} (\frac{π}{8})

Rewrite using trig identities:

cos (\frac{π}{4})

1 - 2 sin^{2} (\frac{π}{8})

Use the Double Angle identity:

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

= cos (2 \cdot \frac{π}{8})

Simplify:

2 \cdot \frac{π}{8} = \frac{π}{4}

2 \cdot \frac{π}{8}

Multiply fractions:

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

= \frac{π 2}{8}

Cancel the common factor:

2

= \frac{π}{4}

= cos (\frac{π}{4})

= cos (\frac{π}{4})

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{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}

sec (\frac{17 π}{3})

2 sin (- \frac{π}{3}) cos (- \frac{π}{3})

csch (0)

cos (100 π)

cos (40 0^{\circ}) sec (4 0^{\circ})

What is the value of 1-2sin^2((7pi)/8) ?
The value of 1-2sin^2((7pi)/8) is (sqrt(2))/2