What is the value of sin((49pi)/6)+tan((41pi)/4) ?

The value of sin((49pi)/6)+tan((41pi)/4) is 3/2

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sin((49pi)/6)+tan((41pi)/4)

sin (\frac{49 π}{6}) + tan (\frac{41 π}{4})

\frac{3}{2}

Decimal Notation

1.5

Solution steps

sin (\frac{49 π}{6}) + tan (\frac{41 π}{4})

sin (\frac{49 π}{6}) = sin (\frac{π}{6})

sin (\frac{49 π}{6})

Rewrite

\frac{49 π}{6}

2 π \cdot 4 + \frac{π}{6}

= sin (2 π 4 + \frac{π}{6})

Apply the periodicity of

sin

sin (x + 2 π \cdot k) = sin (x)

sin (2 π \cdot 4 + \frac{π}{6}) = sin (\frac{π}{6})

= sin (\frac{π}{6})

= sin (\frac{π}{6}) + tan (\frac{41 π}{4})

tan (\frac{41 π}{4}) = tan (\frac{π}{4})

tan (\frac{41 π}{4})

Rewrite

\frac{41 π}{4}

π \cdot 10 + \frac{π}{4}

= tan (π 10 + \frac{π}{4})

Apply the periodicity of

tan

tan (x + π \cdot k) = tan (x)

tan (π \cdot 10 + \frac{π}{4}) = tan (\frac{π}{4})

= tan (\frac{π}{4})

= sin (\frac{π}{6}) + tan (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{6}) = \frac{1}{2}

sin (\frac{π}{6})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} 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:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

= \frac{1}{2} + 1

Simplify

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

\frac{1}{2} + 1

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}

75 tan (2 5^{\circ})

\frac{1 - cos ( 4 5 ^{\circ} )}{1 + cos ( 4 5 ^{\circ} )}

e^{2} sin (0)

sin (arcsin (\frac{π}{3}))

cos (\frac{π}{4}) - 1

What is the value of sin((49pi)/6)+tan((41pi)/4) ?
The value of sin((49pi)/6)+tan((41pi)/4) is 3/2