How do you find the value for #sin2theta#, #cos2theta#, and #tan2theta# and the quadrant in which #2theta# lies given #costheta=-3/5# and #theta# is in quadrant III?

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Answer 1 · 2016-09-15T16:31:11

#(1) : sin 2theta=24/25; (2) : cos2theta=-7/25; (3) : tan2theta=-24/7.#

#(4) : 2theta" lies in "Q_(II)#.

# theta in Q_(III) rArr pi lt theta lt 3pi/2... ... (0) rArr sin theta lt 0#

Now, #cos theta =-3/5 rArr sin theta =-sqrt(1-(-3/5)^2)=-4/5#.

#:. sin 2theta=2sinthetacostheta=2(-4/5)(-3/5)=24/25... ...(1)#

#cos2theta=2cos^2theta-1=2(-3/5)^2-1=-7/25... ... (2)#

From #(0), 2pi lt theta lt 3pi rArr 2theta in Q_IuuQ_(II).#

But, #(1) & (2)# confirm that #2theta# lies in #Q_(II)#.

Finally, by #(1) & (2), tan 2theta=sin(2theta)/cos(2theta)=-24/7.#