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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: 7,935 Thanks: 1129 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: 7,935 Thanks: 1129 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: 7,935 Thanks: 1129 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 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Help with Trignometric Ratios
I recommend studying this too: [attachment=0:1zds2glw]600pxUnit_circle_angles_color.svg.png[/attachment:1zds2glw] 

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