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 November 10th, 2011, 07:56 AM #1 Member   Joined: Nov 2011 Posts: 37 Thanks: 0 Help with Trignometric Ratios Determine the exact value of each Trig ratio: 1) Sin 2?/3 2) Cos 7?/6 3) Sin 5?/4 4) Tan 4?/3 5) Sec 7?/6
 November 10th, 2011, 09:39 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Help with Trignometric Ratios Given cos(?/3) = 1/2, sin(?/3) = ?(3)/2, cos(?/4) = ?(2)/2, sin(?/4) = ?(2)/2, cos(?/6) = ?(3)/2 and sin(?/6) = 1/2 (from 'special' triangles) you may use the angle sum difference formulas for sin, cos and tan: sin(a ± b) = sin(a)cos(b) ± sin(b)cos(a). cos(a ± b) = cos(a)cos(b) ? sin(a)sin(b). tan(a ± b) = (tan(a) ± tan(b))/(1 ? tan(a)tan(b)). Note: sin(?) = 0, cos(?) = -1.
 November 10th, 2011, 10:22 AM #3 Member   Joined: Nov 2011 Posts: 37 Thanks: 0 Re: Help with Trignometric Ratios Can you show me what you mean by solving one of these?
 November 10th, 2011, 10:28 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Help with Trignometric Ratios 2) cos(7?/6) = cos(?)cos(?/6) - sin(?)sin(?/6) = -?(3)/2. The other problems follow a similar pattern.
 November 10th, 2011, 10:49 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Help with Trignometric Ratios Another way of looking at it, you've got the angles shifted by ? so, for example, cos(7?/6) = cos(? + ?/6) = -cos(?/6) = -?(3)/2.
November 10th, 2011, 12:17 PM   #6
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Re: Help with Trignometric Ratios

I recommend studying this too:

[attachment=0:1zds2glw]600px-Unit_circle_angles_color.svg.png[/attachment:1zds2glw]
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