Use the given information to find the values of sin 2theta. cos2theta. and tan 2theta. Sin theta =
-15/17; The terminal side of theta lies in quadrant III.sin 2theta = (Simplify sour answer,
including any radicals. Use integers or fractions for any numbers in the expression) cos2theta =
Square (Simplify sour answer, including any radicals. Use integers or fractions for any numbers
in the expression) tan 2 theta =Square (Simplify sour answer, including any radicals. Use
integers or fractions for any numbers in the expression)
Solution
sin = -15/17 is in QIII
now by sin^2 +cos^2 =1
cos^2 = 1- (-15/17)^2
cos = -8/17
so now sin2 =2sin.cos
= 2(-15/17) .(-8/17)
= 240/289
cos2 =cos^2 -sin^2
= (-8/17)^2 - (-15/17)^2
=64/289 - 225/289
= -161/289
tan(2) = sin2/cos = 240/-161
= -240/161.
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Use the given information to find the values of sin 2theta. cos2theta.pdf
1. Use the given information to find the values of sin 2theta. cos2theta. and tan 2theta. Sin theta =
-15/17; The terminal side of theta lies in quadrant III.sin 2theta = (Simplify sour answer,
including any radicals. Use integers or fractions for any numbers in the expression) cos2theta =
Square (Simplify sour answer, including any radicals. Use integers or fractions for any numbers
in the expression) tan 2 theta =Square (Simplify sour answer, including any radicals. Use
integers or fractions for any numbers in the expression)
Solution
sin = -15/17 is in QIII
now by sin^2 +cos^2 =1
cos^2 = 1- (-15/17)^2
cos = -8/17
so now sin2 =2sin.cos
= 2(-15/17) .(-8/17)
= 240/289
cos2 =cos^2 -sin^2
= (-8/17)^2 - (-15/17)^2
=64/289 - 225/289
= -161/289
tan(2) = sin2/cos = 240/-161
= -240/161
