How do you find sin2 theta? Where theta is in the fourth quadrant.

Question

#sin2theta if costheta=12/13#

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Answer 1 · 2018-02-18T17:39:45

#sin2θ= -120/169#

Given: #cosθ=12/13#
Therefore draw a triangle in the fourth quadrant and #sinθ= -5/13#

#sin2θ= 2sinθcosθ#

#sin2θ= 2(-5/13)(12/13)#

#sin2θ= -120/169#

Answer 2 · 2018-02-18T17:40:45

#sin2theta=-120/169#

#costheta=12/13#

#costheta="(adj. side)/(hyp)#

#"Comparing"#

#"adj. side=12, hyp=13"#

#"By pythagoras theorem, "#

#"opp=sqrt(hyp^2-adh^2)=sqrt(13^2-12^2)=5#

#sintheta="(opp. side)/(hyp)=5/13#

#sintheta " is negative in fourth quadrant"#

#sintheta=-5/13#

#sin2theta=2sinthetacostheta=2xx12/13xx(-5/13)=-120/169#

#sin2theta=-120/169#