What is the value of sin(300)+cos(390)+tan(540) ?

The value of sin(300)+cos(390)+tan(540) is 0

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sin(300)+cos(390)+tan(540)

Solution

sin (30 0^{\circ}) + cos (39 0^{\circ}) + tan (54 0^{\circ})

Solution

0

Solution steps

sin (30 0^{\circ}) + cos (39 0^{\circ}) + tan (54 0^{\circ})

cos (39 0^{\circ}) = cos (3 0^{\circ})

cos (39 0^{\circ})

Rewrite

39 0^{\circ}

36 0^{\circ} + 3 0^{\circ}

= cos (36 0^{\circ} + 3 0^{\circ})

Apply the periodicity of

cos

cos (x + 36 0^{\circ}) = cos (x)

cos (36 0^{\circ} + 3 0^{\circ}) = cos (3 0^{\circ})

= cos (3 0^{\circ})

= sin (30 0^{\circ}) + cos (3 0^{\circ}) + tan (54 0^{\circ})

tan (54 0^{\circ}) = tan (0)

tan (54 0^{\circ})

Rewrite

54 0^{\circ}

18 0^{\circ} \cdot 3 + 0

= tan (18 0^{\circ} 3 + 0)

Apply the periodicity of

tan

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

tan (18 0^{\circ} \cdot 3 + 0) = tan (0)

= tan (0)

= sin (30 0^{\circ}) + cos (3 0^{\circ}) + tan (0)

Rewrite using trig identities:

sin (30 0^{\circ}) = 2 sin (15 0^{\circ}) cos (15 0^{\circ})

sin (30 0^{\circ})

Write

sin (30 0^{\circ})

sin (2 \cdot 15 0^{\circ})

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

= 2 sin (15 0^{\circ}) cos (15 0^{\circ}) + cos (3 0^{\circ}) + tan (0)

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}

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

cos (3 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:

tan (0) = 0

tan (0)

tan (x)

periodicity table with

18 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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 0

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

Simplify

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

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

Remove parentheses:

(- a) = - a

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

2 \cdot \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{2}

2 \cdot \frac{1}{2} \cdot \frac{3}{2}

Multiply fractions:

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

= \frac{1 \cdot 3 \cdot 2}{2 \cdot 2}

Cancel the common factor:

2

= \frac{1 \cdot 3}{2}

Multiply:

1 \cdot 3 = 3

= \frac{3}{2}

= - \frac{3}{2} + \frac{3}{2} + 0

Add similar elements:

- \frac{3}{2} + \frac{3}{2} = 0

= 0

= 0

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Frequently Asked Questions (FAQ)

What is the value of sin(300)+cos(390)+tan(540) ?
The value of sin(300)+cos(390)+tan(540) is 0