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Math  /  Trigonometry

QuestionEstablish the identity. cos⁡θ+cos⁡(3θ)sin⁡θ−sin⁡(3θ)=−cot⁡θ\frac{\cos \theta+\cos (3 \theta)}{\sin \theta-\sin (3 \theta)}=-\cot \theta
Choose the correct sequence of steps to establish the identity. A. cos⁡θ+cos⁡(3θ)sin⁡θ−sin⁡(3θ)=−2sin⁡(2θ)cos⁡θ2sin⁡(2θ)sin⁡θ=−cot⁡θ\frac{\cos \theta+\cos (3 \theta)}{\sin \theta-\sin (3 \theta)}=\frac{-2 \sin (2 \theta) \cos \theta}{2 \sin (2 \theta) \sin \theta}=-\cot \theta B. cos⁡θ+cos⁡(3θ)sin⁡θ−sin⁡(3θ)=−2cos⁡θcos⁡θ2cos⁡θsin⁡θ=−cot⁡θ\frac{\cos \theta+\cos (3 \theta)}{\sin \theta-\sin (3 \theta)}=\frac{-2 \cos \theta \cos \theta}{2 \cos \theta \sin \theta}=-\cot \theta c. cos⁡θ+cos⁡(3θ)sin⁡θ−sin⁡(3θ)=2cos⁡(2θ)cos⁡θ−2sin⁡θcos⁡(2θ)=−cot⁡θ\frac{\cos \theta+\cos (3 \theta)}{\sin \theta-\sin (3 \theta)}=\frac{2 \cos (2 \theta) \cos \theta}{-2 \sin \theta \cos (2 \theta)}=-\cot \theta D. cos⁡θ+cos⁡(3θ)sin⁡θ−sin⁡(3θ)=−2sin⁡θcos⁡θ2sin⁡θsin⁡θ=−cot⁡θ\frac{\cos \theta+\cos (3 \theta)}{\sin \theta-\sin (3 \theta)}=\frac{-2 \sin \theta \cos \theta}{2 \sin \theta \sin \theta}=-\cot \theta

Studdy Solution
The correct sequence of steps is C, and we've shown that cos⁡θ+cos⁡(3θ)sin⁡θ−sin⁡(3θ)=−cot⁡θ\frac{\cos \theta+\cos (3 \theta)}{\sin \theta-\sin (3 \theta)}=-\cot \theta.

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