What is the value of sin(90)+cos(60) ?

The value of sin(90)+cos(60) is 3/2

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sin(90)+cos(60)

sin (9 0^{\circ}) + cos (6 0^{\circ})

\frac{3}{2}

Decimal Notation

1.5

Solution steps

sin (9 0^{\circ}) + cos (6 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

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}

= 1 + \frac{1}{2}

Simplify

1 + \frac{1}{2} : \frac{3}{2}

1 + \frac{1}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 \cdot 2 + 1 = 3

1 \cdot 2 + 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 + 1

Add the numbers:

2 + 1 = 3

= 3

= \frac{3}{2}

= \frac{3}{2}

35 tan (0) - 15.022 sec^{2} (0) + 1

cos (arcsec (2))

0.05 \cdot cos (22.91 \cdot 0.07)

2 cos (- \frac{π}{2}) + \frac{π}{2}

- cos (1 6^{\circ})