What is the value of cos^2(180)-sin^2(180) ?

The value of cos^2(180)-sin^2(180) is 1

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cos^2(180)-sin^2(180)

cos^{2} (18 0^{\circ}) - sin^{2} (18 0^{\circ})

1

Solution steps

cos^{2} (18 0^{\circ}) - sin^{2} (18 0^{\circ})

Use the following trivial identity:

cos (18 0^{\circ}) = (- 1)

cos (18 0^{\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}

= (- 1)

Use the following trivial identity:

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

sin (18 0^{\circ})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= (- 1)^{2} - 0^{2}

Simplify

(- 1)^{2} - 0^{2} : 1

(- 1)^{2} - 0^{2}

Apply rule

0^{a} = 0

0^{2} = 0

= (- 1)^{2} - 0

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

= 1

tan (7 3^{\circ})

arctan (\frac{12}{16})

sec (arcsin (\frac{20}{21}))

3 cos (\frac{2 π}{3})

tan (37 5^{\circ})

What is the value of cos^2(180)-sin^2(180) ?
The value of cos^2(180)-sin^2(180) is 1