Question 595833
{{{sin(2*sin^-1(5/13))}}}
Broadly speaking this expression is a reference to the sin of twice some angle. This suggests that the identity to use is the one for sin(2x):
sin(2x) = 2*sin(x)*cos(x)<br>
Applying this pattern to your expression we get:
{{{2*sin(sin^-1(5/13))*cos(sin^-1(5/13))}}}<br>
{{{sin(sin^-1(5/13))}}} is "the sine of the angle whose sin is 5/13". So it obviously has a value of 5/13.<br>
{{{cos(sin^-1(5/13))}}} is "the cosine of the angle whose sin is 5/13". Its value is not so obvious. To find it you can use the Pythagorean identity:
{{{sin^2(x) + cos^2(x) = 1}}}
Substituting in our sin value this becomes:
{{{(5/13)^2 + cos^2(x) = 1}}}
To solve for cos(x) we start by simplifying:
{{{25/169 + cos^2(x) = 1}}}
Subtract the fraction:
{{{cos^2(x) = 1 - 25/169}}}
{{{cos^2(x) = 169/169 - 25/169}}}
{{{cos^2(x) = 144/169}}}
Find the square root of each side (discarding the negative square root) we get:
{{{cos(x) = 12/13}}}<br>
Now that we have values for {{{sin(sin^-1(5/13))}}} and {{{cos(sin^-1(5/13))}}} we can substitute them into {{{2*sin(sin^-1(5/13))*cos(sin^-1(5/13))}}}:
{{{2*(5/13)*(12/13)}}}
Simplifying we get:
{{{(10/13)*(12/13)}}}
{{{120/169}}}
This fraction does not reduce so
{{{sin(2*sin^-1(5/13)) = 120/169}}}