Question

is there an easier way to turn a trig function into an algebraic expression?

Answer 1

Use a triangle to write \[\sin (2\sin^{-1} x)\]

Answer 2

@satellite73

Answer 3

no real easy way. no

Answer 4

I never seem to get it..

Answer 5

sorry typo there

Answer 6

\[\sin(2\sin^{-1}(x))=2\sin(\sin^{-1}(x))\cos(\sin^{-1}(x))\]

Answer 7

the first part is sort of obvious,
\[\sin(\sin^{-1}(x))=x\]

Answer 8

yes i understand that, but after that step is confusing

Answer 9

they you need to know that
\[\cos(\sin^{-1}(x))=\sqrt{1-x^2}\]

Answer 10

which is true by any number of facts
a triangle works nicely

Answer 11

or by using \(\sin^2(\theta)+\cos^2(\theta)=1\implies \cos(\theta)=\sqrt{1-\sin^2(\theta)}\implies \cos(\sin^{-1}(x))=\sqrt{1-x^2}\)