What is the value of tan^2(pi/4)+2cos((8pi)/3)-sin((13pi)/6) ?

The value of tan^2(pi/4)+2cos((8pi)/3)-sin((13pi)/6) is -1/2

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tan^2(pi/4)+2cos((8pi)/3)-sin((13pi)/6)

tan^{2} (\frac{π}{4}) + 2 cos (\frac{8 π}{3}) - sin (\frac{13 π}{6})

- \frac{1}{2}

Decimal Notation

- 0.5

Solution steps

tan^{2} (\frac{π}{4}) + 2 cos (\frac{8 π}{3}) - sin (\frac{13 π}{6})

cos (\frac{8 π}{3}) = cos (\frac{2 π}{3})

cos (\frac{8 π}{3})

Rewrite

\frac{8 π}{3}

2 π + \frac{2 π}{3}

= cos (2 π + \frac{2 π}{3})

Apply the periodicity of

cos

cos (x + 2 π) = cos (x)

cos (2 π + \frac{2 π}{3}) = cos (\frac{2 π}{3})

= cos (\frac{2 π}{3})

= tan^{2} (\frac{π}{4}) + 2 cos (\frac{2 π}{3}) - sin (\frac{13 π}{6})

sin (\frac{13 π}{6}) = sin (\frac{π}{6})

sin (\frac{13 π}{6})

Rewrite

\frac{13 π}{6}

2 π + \frac{π}{6}

= sin (2 π + \frac{π}{6})

Apply the periodicity of

sin

sin (x + 2 π) = sin (x)

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

= sin (\frac{π}{6})

= tan^{2} (\frac{π}{4}) + 2 cos (\frac{2 π}{3}) - sin (\frac{π}{6})

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

Use the following trivial identity:

cos (\frac{2 π}{3}) = - \frac{1}{2}

cos (\frac{2 π}{3})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} 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}

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}

= 1^{2} + 2 (- \frac{1}{2}) - \frac{1}{2}

Simplify

1^{2} + 2 (- \frac{1}{2}) - \frac{1}{2} : - \frac{1}{2}

1^{2} + 2 (- \frac{1}{2}) - \frac{1}{2}

Apply rule

1^{a} = 1

1^{2} = 1

= 1 + 2 (- \frac{1}{2}) - \frac{1}{2}

Remove parentheses:

(- a) = - a

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

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

2 \cdot \frac{1}{2}

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 1 - 1 - \frac{1}{2}

1 - 1 = 0

= - \frac{1}{2}

= - \frac{1}{2}

arcsin (\frac{0.43}{0.82})

4 sin^{2} (\frac{π}{3})

arctan (\frac{5}{- 7})

sin (60 \circ)

cos (6 0^{\circ}) \cdot 8

What is the value of tan^2(pi/4)+2cos((8pi)/3)-sin((13pi)/6) ?
The value of tan^2(pi/4)+2cos((8pi)/3)-sin((13pi)/6) is -1/2