# How do you evaluate #sin(sin^-1(10/3))#?

##### 1 Answer

For Real valued

#sin(sin^(-1)(10/3)) = 10/3#

as we would hope.

#### Explanation:

If

But...

Note that:

#e^(i theta) = cos theta + i sin theta#

Hence we find:

#cos theta = (e^(i theta) + e^(-i theta))/2#

#sin theta = (e^(i theta) - e^(-i theta))/(2i)#

We can use these to extend the definition of

#cos z = (e^(iz) + e^(-iz))/2#

#sin z = (e^(iz) - e^(-iz))/(2i)#

So what values of

Suppose:

#10/3 = sin z = (e^(iz) - e^(-iz))/(2i)#

Multiplying both ends by

#e^(iz) - e^(-iz) = 20/3i#

Multiplying through by

#(e^(iz))^2-20/3i (e^(iz)) - 1 = 0#

Using the quadratic formula, we find:

#e^(iz) = (20/3i+-sqrt(-400/9+4))/2#

#color(white)(e^(iz)) = 10/3i+-sqrt(91)/3i#

#color(white)(e^(iz)) = 1/3(10+-sqrt(91))i#

Hence:

#z = 1/i(ln(1/3(10+-sqrt(91))i))+2kpi#

#color(white)(z) = 2kpi-i (ln(1/3(10+-sqrt(91))i))#

Rather than spend time figuring out which of these might be the best candidate for the principal value of

Then:

#sin z = sin (2kpi-i (ln(1/3(10+-sqrt(91))i)))#

#color(white)(sin z) = sin (-i (ln(1/3(10+-sqrt(91))i)))#

#color(white)(sin z) = (e^(ln(1/3(10+-sqrt(91))i))-e^(-ln(1/3(10+-sqrt(91))i)))/(2i)#

#color(white)(sin z) = (1/3(10+-sqrt(91))i-1/(1/3(10+-sqrt(91))i))/(2i)#

#color(white)(sin z) = (1/3(10+-sqrt(91))i+1/3(10 color(white)(.)bar("+")color(white)(.) sqrt(91))i)/(2i)#

#color(white)(sin z) = 10/3#