Determine-the-value-of-sin-3-p-cos-6-p-cos-3-p-sin-6-p-

Question Number 6146 by Ninik last updated on 16/Jun/16

Determine the value of sin ((Π/3)+p)cos ((Π/6)+p)−cos ((Π/3)+p)sin ((Π/6)+p)

$${Determine}\:{the}\:{value}\:{of}\:\mathrm{sin}\:\left(\frac{\Pi}{\mathrm{3}}+{p}\right)\mathrm{cos}\:\left(\frac{\Pi}{\mathrm{6}}+{p}\right)−\mathrm{cos}\:\left(\frac{\Pi}{\mathrm{3}}+{p}\right)\mathrm{sin}\:\left(\frac{\Pi}{\mathrm{6}}+{p}\right) \\ $$

Answered by Rasheed Soomro last updated on 16/Jun/16

sin ((π/3)+p)cos ((π/6)+p)−cos ((π/3)+p)sin ((π/6)+p) sin α cos β−cos α sin β=sin (α−β) =sin (((π/3)+p)−((π/6)+p)) =sin ((π/3)−(π/6))=sin ((π/6))=(1/2)

$$\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+{p}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}+{p}\right)−\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}+{p}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}+{p}\right) \\ $$$$\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta−\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta=\mathrm{sin}\:\left(\alpha−\beta\right) \\ $$$$=\mathrm{sin}\:\left(\left(\frac{\pi}{\mathrm{3}}+{p}\right)−\left(\frac{\pi}{\mathrm{6}}+{p}\right)\right) \\ $$$$=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\frac{\pi}{\mathrm{6}}\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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Determine-the-value-of-sin-3-p-cos-6-p-cos-3-p-sin-6-p-

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