What is the value of sin^2(-90) ?

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sin^2(-90)

sin^{2} (- 9 0^{\circ})

1

Solution steps

sin^{2} (- 9 0^{\circ})

Rewrite using trig identities:

sin (- 9 0^{\circ}) = - 1

sin (- 9 0^{\circ})

Use the following property:

sin (- x) = - sin (x)

sin (- 9 0^{\circ}) = - sin (9 0^{\circ})

= - sin (9 0^{\circ})

Use the following trivial identity:

sin (9 0^{\circ}) = 1

sin (9 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}

= 1

= - 1

= (- 1)^{2}

Simplify

(- 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

400 cos (2 5^{\circ}) + 700 cos (8 0^{\circ})

arctan (\frac{40}{70})

arcsec (1.632)

sin (27 5^{\circ})

\frac{1}{cos ( 2 )}