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October 17th, 2013, 09:45 AM  #1 
Newbie Joined: Mar 2013 Posts: 24 Thanks: 0  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. 
October 17th, 2013, 10:34 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond  Re: Stumped simplifying this trig expression
I get 2?(6)/25. Construct a right triangle with hypotenuse 5, adjacent side 1 and opposite side ?(5²  1²) = 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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