What is the value of sin^3(270)+csc^2(495) ?

The value of sin^3(270)+csc^2(495) is 1

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sin^3(270)+csc^2(495)

sin^{3} (27 0^{\circ}) + csc^{2} (49 5^{\circ})

1

Solution steps

sin^{3} (27 0^{\circ}) + csc^{2} (49 5^{\circ})

Rewrite using trig identities:

sin (27 0^{\circ}) = 2 sin (13 5^{\circ}) cos (13 5^{\circ})

= (2 sin (13 5^{\circ}) cos (13 5^{\circ}))^{3} + csc^{2} (49 5^{\circ})

Simplify

= 8 sin^{3} (13 5^{\circ}) cos^{3} (13 5^{\circ}) + csc^{2} (49 5^{\circ})

csc (49 5^{\circ}) = csc (13 5^{\circ})

= 8 sin^{3} (13 5^{\circ}) cos^{3} (13 5^{\circ}) + csc^{2} (13 5^{\circ})

Use the following trivial identity:

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

Use the following trivial identity:

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

Rewrite using trig identities:

csc (13 5^{\circ}) = 2

= 8 (\frac{2}{2})^{3} (- \frac{2}{2})^{3} + (2)^{2}

Simplify

8 (\frac{2}{2})^{3} (- \frac{2}{2})^{3} + (2)^{2} : 1

= 1

cos (53. 1^{\circ})

\frac{12.8 sin ( 5 8 ^{\circ} )}{11.4}

cos^{2} (5 3^{\circ})

cos (- \frac{1}{3})

sec (\frac{19 π}{6})

What is the value of sin^3(270)+csc^2(495) ?
The value of sin^3(270)+csc^2(495) is 1