If cos theta=0.6 and 270<theta<360, find the exact value of sin 2 theta ?

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Answer 1 · 2018-04-02T02:13:19

#sin2theta=-0.96#

#270^o<theta<360^o# tells us that we're working in the third quadrant, where sine is negative, while cosine is positive. Keep this in mind.

Now, recall the identity

#sin^2theta+cos^2theta=1#

We know #costheta=0.6, cos^2theta=(0.6)^2=0.36#

So,

#0.36+sin^2theta=1#

#sin^2theta=1-0.36#

#sin^2theta=0.64#

#sintheta=+-sqrt(0.64)=+-0.8#

As mentioned earlier, sine in quadrant III is negative, so we want the negative answer.

#sintheta=-0.8#

Now, we want #sin2theta.# Recall the identity #sin2theta=2sinthetacostheta.# We have both sine and cosine, so we can plug them in:

#sin2theta=2(0.6)(-0.8)#

#sin2theta=-0.96#