What is the value of sin^2(30)cos^2(60) ?

The value of sin^2(30)cos^2(60) is 1/16

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sin^2(30)cos^2(60)

sin^{2} (3 0^{\circ}) cos^{2} (6 0^{\circ})

\frac{1}{16}

Decimal Notation

0.0625

Solution steps

sin^{2} (3 0^{\circ}) cos^{2} (6 0^{\circ})

Use the following trivial identity:

sin (3 0^{\circ}) = \frac{1}{2}

sin (3 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}

= \frac{1}{2}

Use the following trivial identity:

cos (6 0^{\circ}) = \frac{1}{2}

cos (6 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}

= \frac{1}{2}

= (\frac{1}{2})^{2} (\frac{1}{2})^{2}

Simplify

(\frac{1}{2})^{2} (\frac{1}{2})^{2} : \frac{1}{16}

(\frac{1}{2})^{2} (\frac{1}{2})^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

(\frac{1}{2})^{2} (\frac{1}{2})^{2} = (\frac{1}{2})^{2 + 2}

= (\frac{1}{2})^{2 + 2}

Add the numbers:

2 + 2 = 4

= (\frac{1}{2})^{4}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{4}}{2 ^{4}}

Apply rule

1^{a} = 1

1^{4} = 1

= \frac{1}{2 ^{4}}

2^{4} = 16

= \frac{1}{16}

= \frac{1}{16}

sin (\frac{π}{60})

- sec (3 0^{\circ})

cos (1.517)

arcsin (- 0.173)

arcsin (0.695)

What is the value of sin^2(30)cos^2(60) ?
The value of sin^2(30)cos^2(60) is 1/16