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November 10th, 2011, 07:56 AM   #1
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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
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November 10th, 2011, 09:39 AM   #2
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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.
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November 10th, 2011, 10:22 AM   #3
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Re: Help with Trignometric Ratios

Can you show me what you mean by solving one of these?
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November 10th, 2011, 10:28 AM   #4
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Re: Help with Trignometric Ratios

2) cos(7?/6) = cos(?)cos(?/6) - sin(?)sin(?/6) = -?(3)/2. The other problems follow a similar pattern.
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November 10th, 2011, 10:49 AM   #5
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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.
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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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