What is the value of cos^2(22.5)-sin^2(22.5) ?

The value of cos^2(22.5)-sin^2(22.5) is (sqrt(2))/2

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cos^2(22.5)-sin^2(22.5)

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

\frac{2}{2}

Decimal Notation

0.70710 \dots

Solution steps

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

Rewrite using trig identities:

cos (4 5^{\circ})

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

Use the Double Angle identity:

cos^{2} (x) - 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}

cos (4 5^{\circ}) cos (3 0^{\circ}) + sin (4 5^{\circ}) sin (3 0^{\circ})

sin (\frac{5 π}{12}) - sin (\frac{π}{12})

cos (8 7^{\circ})

cos (2 \frac{π}{3})

4 - 2 arctan (4)

What is the value of cos^2(22.5)-sin^2(22.5) ?
The value of cos^2(22.5)-sin^2(22.5) is (sqrt(2))/2