What is the value of cos^2(67.5)-sin^2(67.5) ?

The value of cos^2(67.5)-sin^2(67.5) is -(sqrt(2))/2

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cos^2(67.5)-sin^2(67.5)

cos^{2} (67. 5^{\circ}) - sin^{2} (67. 5^{\circ})

- \frac{2}{2}

Decimal Notation

- 0.70710 \dots

Solution steps

cos^{2} (67. 5^{\circ}) - sin^{2} (67. 5^{\circ})

Rewrite using trig identities:

cos (13 5^{\circ})

cos^{2} (67. 5^{\circ}) - sin^{2} (67. 5^{\circ})

Use the Double Angle identity:

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

= cos (2 \cdot 67. 5^{\circ})

Simplify

= cos (13 5^{\circ})

= cos (13 5^{\circ})

Use the following trivial identity:

cos (13 5^{\circ}) = - \frac{2}{2}

cos (13 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}

- 2 sin (1)

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

30 tan (3 9^{\circ})

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

arccos (cos (- \frac{5 π}{6}))

What is the value of cos^2(67.5)-sin^2(67.5) ?
The value of cos^2(67.5)-sin^2(67.5) is -(sqrt(2))/2