It is given that
(sinθ+6cosθ)2+(3sinθ−2cosθ)2=28(\sin\theta + 6\cos\theta)^2 + (3\sin\theta - 2\cos\theta)^2 = 28(sinθ+6cosθ)2+(3sinθ−2cosθ)2=28
and that θ \theta\,θ is obtuse.
Find the exact value of sinθ\sin\thetasinθ. Fully justify your answer.
Find the exact value of cos2θ\cos 2\thetacos2θ.
Find the exact value of tanθ\tan\thetatanθ, giving your answer in surd form.