![\sin A \cdot \cos B = \dfrac{1}{2}\bigg[\sin (A + B) + \sin (A - B)\bigg]](/tpl/images/0735/6017/fbc24.png)
![\text{Product to Sum:}\\ \dfrac{1}{8}\bigg[3\sin \theta -\sin 3\theta +\dfrac{3}{2}\bigg(\sin (\theta +2\theta) +\sin(\theta -2\theta)\bigg)-\dfrac{1}{2}\bigg(\sin (3\theta+2\theta) +\sin (3\theta-2\theta)\bigg)\bigg]](/tpl/images/0735/6017/da418.png)
![\text{Simplify:}\quad \dfrac{1}{8}\bigg[3\sin \theta -\sin 3\theta +\dfrac{3}{2}\bigg(\sin 3\theta +\sin(-\theta)\bigg)-\dfrac{1}{2}\bigg(\sin 5\theta +\sin \theta\bigg)\bigg]](/tpl/images/0735/6017/ba370.png)
![\text{Cofunction:}\quad \dfrac{1}{8}\bigg[3\sin \theta -\sin 3\theta +\dfrac{3}{2}\bigg(\sin 3\theta -\sin(\theta)\bigg)-\dfrac{1}{2}\bigg(\sin 5\theta +\sin \theta\bigg)\bigg]](/tpl/images/0735/6017/fad22.png)
Please someone help me to prove this.
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Answers: 2
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