What is the value of sin(45)+cos(45) ?

The value of sin(45)+cos(45) is square root of 2

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sin(45)+cos(45)

sin (4 5^{\circ}) + cos (4 5^{\circ})

2

Decimal Notation

1.41421 \dots

Solution steps

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

cos (4 5^{\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{2}{2}

= \frac{2}{2} + \frac{2}{2}

Simplify

\frac{2}{2} + \frac{2}{2} : 2

\frac{2}{2} + \frac{2}{2}

Apply rule

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

= \frac{2 + 2}{2}

Add similar elements:

2 + 2 = 22

= \frac{2 2}{2}

Divide the numbers:

\frac{2}{2} = 1

= 2

= 2

arccos (0.2)

arctan (\frac{3}{3})

cos (12 0^{\circ}) + cos (4 5^{\circ})

cos (11 5^{\circ})

2 sin (2 π)

What is the value of sin(45)+cos(45) ?
The value of sin(45)+cos(45) is square root of 2