Question

http://webwork.math.ttu.edu/wwtmp/equations/10/59c780daa010831640837edb24b5801.png

Answer 1

\[\cos(2\sin^{-1}(x))\]double angle formuila for this one

Answer 2

\[\cos(2\theta)=2\cos(\theta)\sin(\theta)\]

Answer 3

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

Answer 4

now
\[\sin(\sin^{-1}(x))=x\]
so the only question is what is
\[\cos(\sin^{-1}(x))\]

Answer 5

do you know that one?

Answer 6

no.. haha not at all trig absolutely blows my mind

Answer 7

this is like saying "if sin(x) = something, what is cosine of x"

Answer 8

solve for the other side via pythagoras and get
\[\sqrt{1-x^2}\]

Answer 9

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

Answer 10

Oh ok, that actually makes sense

Answer 11

all in all you should end up with
\[2x\sqrt{1-x^2}\]

Answer 12

you probably had problem that said "if
\[\sin(\theta)=\frac{3}{5}\] find
\[\cos(\theta)\] and you can do that by drawing a triangle and finding the missing side. this is the same thing, only with x instead of 3/5

Answer 13

well I tried plugging 2xsqrt(1-x^2) in and it said its still not right...

Answer 14

let me see if i made a mistake.

Answer 15

ok

Answer 16

yeah because i am an idiot. it is
\[\sin(2\theta)=2\cos(\theta)\sin(\theta)\]
\[\cos(2\theta)=1-2\sin^2(\theta)\] so it is even easier!

Answer 17

type there
\[1-2x^2\]

Answer 18

now i will try to write it correctly
\[\cos(2\sin^{-1}(x))=1-2sin^2(\sin^{-1}(x))=1-2x^2\]

Answer 19

There we go man that got it thanks a ton! I'm still not entirely sure i understand it. What should i do to get better at working with trig expressions like these

Answer 20

we can do the next one too and i will try not to make a stupid mistake

Answer 21

haha ok

Answer 22

practice

Answer 23

it is just understanding that you are supposed to use the double angle formula, and replace theta by whatever is in there