Statement Given that sinθ=513\sin\theta=\dfrac{5}{13}sinθ=135 with θ\thetaθ in the second quadrant (90∘<θ<180∘)(90^\circ<\theta<180^\circ)(90∘<θ<180∘), find sin2θ\sin 2\thetasin2θ, cos2θ\cos 2\thetacos2θ and tan2θ\tan 2\thetatan2θ. Solution In the second quadrant cosθ<0\cos\theta<0cosθ<0: cosθ=−1−25169=−1213\cos\theta=-\sqrt{1-\tfrac{25}{169}}=-\dfrac{12}{13}cosθ=−1−16925=−1312. sin2θ=2sinθcosθ=2⋅513⋅(−1213)=−120169≈−0.7101\sin 2\theta=2\sin\theta\cos\theta=2\cdot\tfrac{5}{13}\cdot\big(-\tfrac{12}{13}\big)=-\dfrac{120}{169}\approx -0{.}7101sin2θ=2sinθcosθ=2⋅135⋅(−1312)=−169120≈−0.7101. cos2θ=1−2sin2θ=1−50169=119169≈0.7041\cos 2\theta=1-2\sin^2\theta=1-\dfrac{50}{169}=\dfrac{119}{169}\approx 0{.}7041cos2θ=1−2sin2θ=1−16950=169119≈0.7041. tan2θ=sin2θcos2θ=−120119≈−1.0084\tan 2\theta=\dfrac{\sin 2\theta}{\cos 2\theta}=-\dfrac{120}{119}\approx -1{.}0084tan2θ=cos2θsin2θ=−119120≈−1.0084. sin2θ=−120169; cos2θ=119169; tan2θ=−120119\boxed{\sin 2\theta=-\tfrac{120}{169};\ \cos 2\theta=\tfrac{119}{169};\ \tan 2\theta=-\tfrac{120}{119}}sin2θ=−169120; cos2θ=169119; tan2θ=−119120