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October 17th, 2013, 08:45 AM   #1
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Stumped simplifying this trig expression

Need help simplifying! The answer key in my book swears I can simplify this further, but I don't see how!

sin(arcsec(-5))cos(arcsec(-5))

There is an exact answer, but I can't arrive there unless I know the angle in radians... but this is not a common angle and I have no clue how to go about this.
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October 17th, 2013, 09:34 AM   #2
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Re: Stumped simplifying this trig expression

I get -2?(6)/25. Construct a right triangle with hypotenuse 5, adjacent side 1 and opposite side ?(5 - 1) = 2?(6). The cosine of this triangle is 1/5 and the sine is 2?(6)/5, so we start with 2?(6)/5 * 1/5 = 2?(6)/25. To account for the sign, we have that arcsec(5) = arccos(1/5), but we have arcsec(-5) = arccos(-1/5), so cos(arccos(-1/5)) = -1/5. The range of the arccos function is [0, ?], hence sin(arccos(-1/5)) is positive, giving us 2?(6)/5 * -1/5 = -2?(6)/25.

One could initially assume the identity arcsec(x) = arccos(1/x) and proceed in a fashion similar to the above.

http://en.wikipedia.org/wiki/Inverse_tr ... _functions
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