What is the value of tan(1395)+sin(1050) ?

The value of tan(1395)+sin(1050) is -3/2

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tan(1395)+sin(1050)

Solution

tan (139 5^{\circ}) + sin (105 0^{\circ})

Solution

- \frac{3}{2}

Decimal Notation

- 1.5

Solution steps

tan (139 5^{\circ}) + sin (105 0^{\circ})

tan (139 5^{\circ}) = tan (13 5^{\circ})

tan (139 5^{\circ})

Rewrite

139 5^{\circ}

18 0^{\circ} \cdot 7 + 13 5^{\circ}

= tan (18 0^{\circ} 7 + 13 5^{\circ})

Apply the periodicity of

tan

tan (x + 18 0^{\circ} \cdot k) = tan (x)

tan (18 0^{\circ} \cdot 7 + 13 5^{\circ}) = tan (13 5^{\circ})

= tan (13 5^{\circ})

= tan (13 5^{\circ}) + sin (105 0^{\circ})

sin (105 0^{\circ}) = sin (33 0^{\circ})

sin (105 0^{\circ})

Rewrite

105 0^{\circ}

36 0^{\circ} \cdot 2 + 33 0^{\circ}

= sin (36 0^{\circ} 2 + 33 0^{\circ})

Apply the periodicity of

sin

sin (x + 36 0^{\circ} \cdot k) = sin (x)

sin (36 0^{\circ} \cdot 2 + 33 0^{\circ}) = sin (33 0^{\circ})

= sin (33 0^{\circ})

= tan (13 5^{\circ}) + sin (33 0^{\circ})

Rewrite using trig identities:

tan (13 5^{\circ}) = - 1

tan (13 5^{\circ})

Rewrite using trig identities:

\frac{sin ( 13 5 ^{\circ} )}{cos ( 13 5 ^{\circ} )}

tan (13 5^{\circ})

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( 13 5 ^{\circ} )}{cos ( 13 5 ^{\circ} )}

= \frac{sin ( 13 5 ^{\circ} )}{cos ( 13 5 ^{\circ} )}

Use the following trivial identity:

sin (13 5^{\circ}) = \frac{2}{2}

sin (13 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 (13 5^{\circ}) = - \frac{2}{2}

cos (13 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{\frac{2}{2}}{- \frac{2}{2}}

Simplify

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

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

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{\frac{2}{2}}{\frac{2}{2}}

Divide fractions:

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

= - \frac{2 \cdot 2}{2 2}

Cancel the common factor:

2

= - \frac{2}{2}

Cancel the common factor:

2

= - 1

= - 1

Rewrite using trig identities:

sin (33 0^{\circ}) = - \frac{1}{2}

sin (33 0^{\circ})

Rewrite using trig identities:

sin (18 0^{\circ}) cos (15 0^{\circ}) + cos (18 0^{\circ}) sin (15 0^{\circ})

sin (33 0^{\circ})

Write

sin (33 0^{\circ})

sin (18 0^{\circ} + 15 0^{\circ})

= sin (18 0^{\circ} + 15 0^{\circ})

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (18 0^{\circ}) cos (15 0^{\circ}) + cos (18 0^{\circ}) sin (15 0^{\circ})

= sin (18 0^{\circ}) cos (15 0^{\circ}) + cos (18 0^{\circ}) sin (15 0^{\circ})

Use the following trivial identity:

sin (18 0^{\circ}) = 0

sin (18 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}

= 0

Use the following trivial identity:

cos (15 0^{\circ}) = - \frac{3}{2}

cos (15 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{3}{2}

Use the following trivial identity:

cos (18 0^{\circ}) = (- 1)

cos (18 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}

= (- 1)

Use the following trivial identity:

sin (15 0^{\circ}) = \frac{1}{2}

sin (15 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{1}{2}

= 0 \cdot (- \frac{3}{2}) + (- 1) \frac{1}{2}

Simplify

= - \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

Subtract the numbers:

- 2 - 1 = - 3

= - 3

= \frac{- 3}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{3}{2}

= - \frac{3}{2}

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Frequently Asked Questions (FAQ)

What is the value of tan(1395)+sin(1050) ?
The value of tan(1395)+sin(1050) is -3/2