What is the value of 1-2sin^2(157.5) ?

The value of 1-2sin^2(157.5) is (sqrt(2))/2

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1-2sin^2(157.5)

1 - 2 sin^{2} (157. 5^{\circ})

\frac{2}{2}

Decimal Notation

0.70710 \dots

Solution steps

1 - 2 sin^{2} (157. 5^{\circ})

Rewrite using trig identities:

sin (157. 5^{\circ}) = sin (22. 5^{\circ})

sin (157. 5^{\circ})

Use the basic trigonometric identity:

sin (x) = sin (18 0^{\circ} - x)

= sin (18 0^{\circ} - 157. 5^{\circ})

Simplify

= sin (22. 5^{\circ})

= 1 - 2 sin^{2} (22. 5^{\circ})

Rewrite using trig identities:

cos (4 5^{\circ})

1 - 2 sin^{2} (22. 5^{\circ})

Use the Double Angle identity:

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

= cos (2 \cdot 22. 5^{\circ})

Simplify

= cos (4 5^{\circ})

= cos (4 5^{\circ})

Use the following trivial identity:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\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{2}{2}

= \frac{2}{2}

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

arcsin (\frac{7}{12})

sin (\frac{15}{17})

arctan (\frac{15}{20})

cos (7 6^{\circ})

What is the value of 1-2sin^2(157.5) ?
The value of 1-2sin^2(157.5) is (sqrt(2))/2