What is the value of sin(pi+i) ?

The value of sin(pi+i) is i(1-e^2)/(2e)

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sin(pi+i)

sin (π + i)

i \frac{1 - e ^{2}}{2 e}

Solution steps

sin (π + i)

Rewrite using trig identities:

sin (π) cosh (1) + i cos (π) sinh (1)

= sin (π) cosh (1) + i cos (π) sinh (1)

Use the following trivial identity:

sin (π) = 0

Rewrite using trig identities:

cosh (1) = \frac{e ^{2} + 1}{2 e}

Use the following trivial identity:

cos (π) = (- 1)

Rewrite using trig identities:

sinh (1) = \frac{e ^{2} - 1}{2 e}

= 0 \cdot \frac{e ^{2} + 1}{2 e} + i (- 1) \frac{e ^{2} - 1}{2 e}

Simplify

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

= i \frac{1 - e ^{2}}{2 e}

arctan (\frac{4}{30})

3 + 2 sin (\frac{π}{6})

1 - 2 sin^{2} (\frac{5 π}{12})

4 sin (\frac{1}{4})

\frac{tan ( \frac{π}{4} ) + tan ( \frac{π}{6} )}{1 - tan ( \frac{π}{4} ) tan ( \frac{π}{6} )}