calculus problem: trig

[warm karma]

can someone help me figure out how to solve this?

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[elder stone]

Note that cos^2(x)=1-sin^2(x)

[elder stone]

You can then find the solution

Also sin(sin^-1(x))=x

[warm karma]

im a little confused how i would do that

cause this is inverse trig right

[elder stone]

You just need to simplify the expression no?

[warm karma]

i thought i would first try to find the solution for arcsin(2x) first

[elder stone]

Do you mean the domain ?

Well cos(sin-1(2x))=sqrt(cos^2(sin-1(2x))=sqrt(1-sin^2(sin-1(2x))

[elder stone]

Then use this and you’ll have simplified your expression

[unkempt reef]

if you have trig functions of trig functions like f(g^-1(x)) one thing you can do is draw a triangle where the ratio is like g = x/1
for example
here theta is inverse sine of 2x
then cos is just the adjacent side

i prefer this method

[warm karma]

Yup this is easier

[elder stone]

Yeah it’s easier you’re right

[warm karma]

thank you this is really helpful

[lyric dagger]

@stable epoch has given 1 rep to @unkempt reef

[warm karma]

Did you figure out the answer brodie

[warm karma]

yeah