SOLUTION: Find the exact values of sin2u, cos2u, and tan2u. cotu=-6, 3pi/2<u<2pi

Question 820569: Find the exact values of sin2u, cos2u, and tan2u. cotu=-6, 3pi/2
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Find the exact values of sin2u, cos2u, and tan2u. cotu=-6, 3pi/2
assume given interval: [0,3π/2]
***
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=√(1^2+6^2)=√(1+36)=√37
..
sinu=1/√37=√37/37
cosu=-6/√37=-6√37/37
tanu=-1/6
..
sin2u=2sinucosu=2*√37/37*-6√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
..
calculator check:
tanu=-1/6
u≈170.5376˚
2u≈341.0754˚(In quadrant IV where sin<0, cos>0, tan<0)
..
sin2u=sin(341.0754)≈-0.3243..
exact value as calculated=-444/1369≈-0.3243..
..
cos2u=cos(341.0754)≈0.9459..
exact value as calculated=35/37≈0.9459..
..
tan2u=tan(341.0754)≈-0.3428..
exact value as calculated=-12/35≈-0.3428..

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