Question 820569
Find the exact values of sin2u, cos2u, and tan2u. cotu=-6, 3pi/2
assume given interval: [0,3π/2]
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reference angle in quadrant II where sin>0, cos<0, cot<0, tan<0
cotu=-6=1/tanu
tanu=-1/6
hypotenuse of reference right triangle=&#8730;(1^2+6^2)=&#8730;(1+36)=&#8730;37
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sinu=1/&#8730;37=&#8730;37/37
cosu=-6/&#8730;37=-6&#8730;37/37
tanu=-1/6
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sin2u=2sinucosu=2*&#8730;37/37*-6&#8730;37/37=-444/1369
cos2u=cos^2u-sin^2u=36/37-1/37=35/37
tan2u=(2tanu)/(1-tan^2u)=-2/6/(1-1/36)=(-12/36)/35/36=-12/35
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calculator check:
tanu=-1/6
u&#8776;170.5376&#730;
2u&#8776;341.0754&#730;(In quadrant IV where sin<0, cos>0, tan<0)
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sin2u=sin(341.0754)&#8776;-0.3243..
exact value as calculated=-444/1369&#8776;-0.3243..
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cos2u=cos(341.0754)&#8776;0.9459..
exact value as calculated=35/37&#8776;0.9459..
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tan2u=tan(341.0754)&#8776;-0.3428..
exact value as calculated=-12/35&#8776;-0.3428..