# I'm not 100% sure as to how to "solve" "compounded" inverse trig functions. For example: tan(sin^(-1)2/3)I know that I start off by drawing the triangle for (sin^(-1)2/3) and that should show...

I'm not 100% sure as to how to "solve" "compounded" inverse trig functions. For example:

tan(sin^(-1)2/3)

I know that I start off by drawing the triangle for (sin^(-1)2/3) and that should show up in the first quandrant, but after that, I have no clue as to how do I get the answer for tan.

I thought that I was supposed to use the pythagorem theory to get the missing length, but then I've seen some answers, in which a triangle is drawn for the other trig function (in this case tan).

Can someone tell me the steps, in clear, easy to understand english? This is the only thing I have trouble with.

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### 2 Answers

I've interpreded your sin^-1 2/3 to be arcsin( 2/3 )

The way to solve this is to be aware that there are inverse trig to trig identities.

The relevent one here is tan(asin x) = x / sqrt( 1 - x^2)

so tan( asin( 2/3 ) ) = (2/3) / sqrt( 1 - 4/9 )

= (2/3) / (sqrt5 / 3) = 2 / sqrt5

**= 2*sqrt5 / 5**

To determine tan (sin ^(-1) (2/3))).

Here sin ^(-1)x is the arcsine x which is the sine ratio of an angle formed by the arc length x of a unit radius circle.

We write arcsinx = y for a function for which x = siny.

x = sin y for two values of y.

When y is in in first quadrant siny = x

When 180-y is in 2nd quadrant also , sin (180-y) = sin y = x.

If siny = x, cos y = sqrt (1-sin^2y) = sqrt(1-x^2).

Therefore tany = x/sqrt(1-x^2).

Therefore tan (arcsinx) = tan (sin inverse x) = x/sqrt(1-x^2).

So in the example, x = 2/3.

Therefore tan arcsin(2/3) = (2/3)/ sqrt{1-(2/3)^2} .

tan {arcsin (2/3)} = (2/3)/sqrt{1-4/9}.

tan{arcsin(2/3)} = (2/3)/sqrt{5/9}.

tan {arc(2/3)} = (2/3)*3/ sqrt5.

tan{arcsi(2/3)} = 2/sqrt5.

tan {arc sin (2/3)} = (2sqrt5)/5.