Drawing a right triangle to simpliy the given expressions

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Johnyi
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1. Homework Statement
Sin(sec^-1(sqrt(x^2+16)/4))

2. Homework Equations



3. The Attempt at a Solution
I did the math and ended up getting x^2-1 as the opposite, but the answers on the back of the book say other wise.
 
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For future reference, trigonometry is usually precalculus material. Posting in the appropriate forum may be a good idea.

Assuming a is the adjacent side, b the opposite, and c the hypotenuse, you need to use

sec x = c / a (so arcsec c / a = x)
b2 = c2 - a2
sin x = b / c

Carefully resolve.
 
My solution: b^2=(x^2+16/16) + 1

I don't know what to do from there!
 
Johnyi said:
My solution: b^2=(x^2+16/16) + 1

I don't know what to do from there!
This doesn't make any sense to me.

Draw a right triangle, with one acute angle labeled θ. Label the two sides and hypotenuse so that sec θ = √(x2 + 16)/4. With appropriate restrictions, this equation is equivalent to θ = sec-1[√(x2 + 16)/4].

After labeling the sides, find sinθ, and you're done.