What is the value of sin(45)-cos(60) ?

The value of sin(45)-cos(60) is (sqrt(2)-1)/2

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sin(45)-cos(60)

sin (4 5^{\circ}) - cos (6 0^{\circ})

\frac{2 - 1}{2}

Decimal Notation

0.20710 \dots

Solution steps

sin (4 5^{\circ}) - cos (6 0^{\circ})

Use the following trivial identity:

sin (4 5^{\circ}) = \frac{2}{2}

sin (4 5^{\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{2}{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{2}{2} - \frac{1}{2}

Simplify

\frac{2}{2} - \frac{1}{2} : \frac{2 - 1}{2}

\frac{2}{2} - \frac{1}{2}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 - 1}{2}

= \frac{2 - 1}{2}

- 40 sin (1) - 1

e^{6} \cdot 3^{2} - cos (3) ln (3) - 4

\frac{120}{tan ( 3 5 ^{\circ} )}

790 (9.81 \cdot cos (14. 3^{\circ}) - \frac{51.38 9 ^{2}}{298})

arcsin (\frac{14}{15})

What is the value of sin(45)-cos(60) ?
The value of sin(45)-cos(60) is (sqrt(2)-1)/2