Inverse trig functions

[MarkSA]

Hello,

I have a few problems that i'm not sure how to approach.

1) Find the exact value of: sin(2 arctan[sqrt(2)])
If I concern myself with the inner part and Let arctan[sqrt(2)] = x, I know that is equal to: tan(x) = sqrt(2). Where I get stumped is how to figure out what value of tan would equal sqrt(2). It doesn't seem to be the three common (pi/3, pi/4, pi/6) angles. Any ideas?

2) Prove that the derivative of: arcsec(x) is equal to: 1/[x*sqrt(x^2 - 1)]
If I let y = arcsec(x), then sec(y) = x, yes? If I differentiate implicitly then in terms of x, I get: y' = 1/(secytany). I'm not sure where to go after that.

 

[o_O]

1.) Whoops, didn't see that 2 in there. You could use the identity:

\(\displaystyle sin(2 \theta) = 2sin \theta cos \theta = \frac{2tan\theta}{1 + tan^{2}\theta}\)

2.) Initially you defined x = sec(y). And you can use the identity: \(\displaystyle tan^{2}x + 1 = sec^{2}x\) to solve for tany in terms of secy or namely, x.

 

[Loren]

For 1, can't you draw a sketch of an angle in quadrants I and III (tan is positive in these two quadrants) with the ordinate = ?3 and abscissa = 1. Then use the Pythagorean Th. to calculate the length of the hypotenuses? Then from the sketches you can read off the sine of that angle.