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 May 31st, 2012, 12:16 PM #1 Member   Joined: Dec 2011 Posts: 75 Thanks: 0 trigonometry practice paper Hello all! I always find trigonometric formulas awkward, and I'm stuck on a practice paper; any help? The acute angle (theta) is such that tan (theta) = 3 / 2 a) By considering an appropriate right-angled triangle, find the exact values of sin (theta) and cos (theta) is this simply sin (theta) = 3 and cos (theta) = 2 ? or is it sin (theta) = + or - 15/4 cos (theta) = + or - 13/4 or something else?? I tried to follow some formulas; don't know if I did them right cos (theta) = + or - 1 / square root 1 + tan^2 theta and sin (theta) = + or - tan (theta) / square root 1 + tan^2 theta b) Use the values in part (a), and trigonometric formulas, to find the exact values of cos (2 theta) and sin (2 theta) well I don't have a clue, I'm looking at different double-angle formulas and all sorts, completely lost! They are only given two marks each question, so I assuming they're not that complex; an explanation would be handy, thanks again!
 May 31st, 2012, 12:22 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,168 Thanks: 472 Math Focus: Calculus/ODEs Re: trigonometry practice paper a) Set up a right triangle such that the side opposite ? is 3 and the side adjacent is 2. What is the hypotenuse? b) What are the double-angle identities for sine and cosine?
 May 31st, 2012, 01:25 PM #3 Member   Joined: Dec 2011 Posts: 75 Thanks: 0 Re: trigonometry practice paper ok for a) hypotenuse is square root 13 so sin = 3 / square root 13 and cos = 2 / square root 13 for b) I used the identity sin (2 theta) = 2 sin(theta)cos(theta) so, sin (2 theta) = 2 sin (3 / square root 13) (cos (2 / square root 13) could I put this as 2 ( answer ) or leave it open such as 2 sin answer cos answer? then for cos (2 theta) I used = 2 cos ^2 - 1 cos (2 theta) = 2 (cos (2 / square root 13)^2 - 1 Is this right?
 May 31st, 2012, 01:34 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 You should simplify your answers by, for example, replacing the square of the square root of 13 with 13. What are your simplified results?
 May 31st, 2012, 02:08 PM #5 Member   Joined: Dec 2011 Posts: 75 Thanks: 0 Re: trigonometry practice paper Hey skipjack do you mean, instead of cos (2 theta) = 2 (cos (2 / square root 13)^2) - 1 change to cos (2 theta) = 2 (cos ( 4 / 13 )) - 1 also could other possible simplifying involve, say sin = 3 / square root 13 change to sin = 3 (square root 13) / 13 ? surds?
 May 31st, 2012, 11:48 PM #6 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 Try to be more precise: cos ? = 2/?13, so cos 2? = 2cos²? - 1 = 2(4/13) - 1 = -5/13. Unless you've been taught otherwise, it isn't necessary to write 2/?13 as 2?13 / 13. Using the identity sin 2? ? 2 sin ? cos ?, what simple fraction do you now get for sin 2??
 June 1st, 2012, 06:14 AM #7 Member   Joined: Dec 2011 Posts: 75 Thanks: 0 Re: trigonometry practice paper so would sin 2? ? 2 sin ? cos ? be sin 2? = 2 sin ( 3 / square root 13 ) cos ( 2 / square root 13 ) sin 2? = 2 sin cos ( 6 / square root 13 ) sin 2? = sin cos ( 12 / square root 13 )
 June 1st, 2012, 07:35 AM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,168 Thanks: 472 Math Focus: Calculus/ODEs Re: trigonometry practice paper Not quite. We have $\sin$$2\theta$$\equiv2\sin$$\theta$$\cos$$\theta$$$ so we replace the trigonometric functions with their values: $\sin$$2\theta$$=2$$\frac{3}{\sqrt{13}}$$$$\frac{2} {\sqrt{13}}$$=\frac{12}{13}$
 June 1st, 2012, 09:26 AM #9 Member   Joined: Dec 2011 Posts: 75 Thanks: 0 Re: trigonometry practice paper Ah I see, I don't know why I thought that theta should only be replaced by the values found earlier (like a variable in its own right?!), but we proved that sin (theta) and cos (theta) gave certain values already, so yeah it makes sense just to replace the trig function with the value we obtained like you said! Thanks again MarkFL and skipjack for the help and not giving up on me!

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