What is the value of sin(60)cos(60) ?

The value of sin(60)cos(60) is (sqrt(3))/4

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sin(60)cos(60)

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

\frac{3}{4}

Decimal Notation

0.43301 \dots

Solution steps

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

Use the following trivial identity:

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

sin (6 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{3}{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{3}{2} \cdot \frac{1}{2}

Simplify

\frac{3}{2} \cdot \frac{1}{2} : \frac{3}{4}

\frac{3}{2} \cdot \frac{1}{2}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{3 \cdot 1}{2 \cdot 2}

Multiply:

3 \cdot 1 = 3

= \frac{3}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3}{4}

= \frac{3}{4}

sin (\frac{π}{4}) - sin (0)

\frac{sin ( 0 )}{cos ( 0 )}

\frac{1}{2} cos (0)

sin (\frac{3 π}{4} + \frac{5 π}{6})

tan (- 90 0^{\circ})

What is the value of sin(60)cos(60) ?
The value of sin(60)cos(60) is (sqrt(3))/4