What is the value of sin((3pi)/2)+tan(pi) ?

The value of sin((3pi)/2)+tan(pi) is -1

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sin((3pi)/2)+tan(pi)

sin (\frac{3 π}{2}) + tan (π)

- 1

Solution steps

sin (\frac{3 π}{2}) + tan (π)

tan (π) = tan (0)

tan (π)

Rewrite

π

π + 0

= tan (π + 0)

Apply the periodicity of

tan

tan (x + π) = tan (x)

tan (π + 0) = tan (0)

= tan (0)

= sin (\frac{3 π}{2}) + tan (0)

Rewrite using trig identities:

sin (\frac{3 π}{2}) = 2 sin (\frac{3 π}{4}) cos (\frac{3 π}{4})

sin (\frac{3 π}{2})

Write

sin (\frac{3 π}{2})

sin (2 \cdot \frac{3 π}{4})

= sin (2 \cdot \frac{3 π}{4})

Use the Double Angle identity:

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

= 2 sin (\frac{3 π}{4}) cos (\frac{3 π}{4})

= 2 sin (\frac{3 π}{4}) cos (\frac{3 π}{4}) + tan (0)

Use the following trivial identity:

sin (\frac{3 π}{4}) = \frac{2}{2}

sin (\frac{3 π}{4})

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{2}{2}

Use the following trivial identity:

cos (\frac{3 π}{4}) = - \frac{2}{2}

cos (\frac{3 π}{4})

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{2}{2}

Use the following trivial identity:

tan (0) = 0

tan (0)

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}

= 0

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

Simplify

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

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

Remove parentheses:

(- a) = - a

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

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

2 \cdot \frac{2}{2} \cdot \frac{2}{2}

Multiply fractions:

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

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

Cancel the common factor:

2

= \frac{2 2}{2}

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

= - 1 + 0

Add/Subtract the numbers:

- 1 + 0 = - 1

= - 1

= - 1

tan (arcsec (6))

\frac{8}{15} \cdot 0.58 \cdot tan (4 5^{\circ}) \cdot 2 \cdot 9.81 \cdot 0. 1^{\frac{5}{2}}

- cos (2) (\frac{π}{4})

\frac{2 - 2 sin ( \frac{π}{4} )}{cos ( \frac{π}{4} )}

arcsin (\frac{π}{32}) - sin^{2} (\frac{π}{32})

What is the value of sin((3pi)/2)+tan(pi) ?
The value of sin((3pi)/2)+tan(pi) is -1