Question-120729

Question Number 120729 by help last updated on 02/Nov/20

Answered by TANMAY PANACEA last updated on 02/Nov/20

LHS ((1−cos2θ)/2)+((1−cos(2θ+2α))/2)+((1−cos(2θ+4α))/2) =(3/2)−(1/2){cos2θ+cos(2θ+4α)+cos(2θ+2α)} =(3/2)−(1/2){2cos(((2θ+2θ+4α)/2))cos(((2θ+4α−2θ)/2))+cos(2θ+2α)} =(3/2)−(1/2)cos(2θ+2α){(2cos2α)+1}

$${LHS} \\ $$$$\frac{\mathrm{1}−{cos}\mathrm{2}\theta}{\mathrm{2}}+\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta+\mathrm{2}\alpha\right)}{\mathrm{2}}+\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta+\mathrm{4}\alpha\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\mathrm{2}\theta+{cos}\left(\mathrm{2}\theta+\mathrm{4}\alpha\right)+{cos}\left(\mathrm{2}\theta+\mathrm{2}\alpha\right)\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{2}{cos}\left(\frac{\mathrm{2}\theta+\mathrm{2}\theta+\mathrm{4}\alpha}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{2}\theta+\mathrm{4}\alpha−\mathrm{2}\theta}{\mathrm{2}}\right)+{cos}\left(\mathrm{2}\theta+\mathrm{2}\alpha\right)\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{2}\theta+\mathrm{2}\alpha\right)\left\{\left(\mathrm{2}{cos}\mathrm{2}\alpha\right)+\mathrm{1}\right\} \\ $$

Commented by TANMAY PANACEA last updated on 02/Nov/20

(3/2)−(1/2)cos(2θ+2×((2π)/3)){1+2cos2×((2π)/3)} =(3/2)−(1/2){cos(2θ)cos(240^o )−sin2θsin240^o }{1+2cos240^o } =(3/2)−(1/2){cos2θ×((−1)/2)−sin2θ×((√3)/2)}{1+2×((−1)/2)} =(3/2)−(1/2){((−cos2θ−(√3) sin2θ)/2)}{0} =(3/2)

$$\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{2}\theta+\mathrm{2}×\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\left\{\mathrm{1}+\mathrm{2}{cos}\mathrm{2}×\frac{\mathrm{2}\pi}{\mathrm{3}}\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\mathrm{2}\theta\right){cos}\left(\mathrm{240}^{{o}} \right)−{sin}\mathrm{2}\theta{sin}\mathrm{240}^{{o}} \right\}\left\{\mathrm{1}+\mathrm{2}{cos}\mathrm{240}^{{o}} \right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\mathrm{2}\theta×\frac{−\mathrm{1}}{\mathrm{2}}−{sin}\mathrm{2}\theta×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right\}\left\{\mathrm{1}+\mathrm{2}×\frac{−\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{−{cos}\mathrm{2}\theta−\sqrt{\mathrm{3}}\:{sin}\mathrm{2}\theta}{\mathrm{2}}\right\}\left\{\mathrm{0}\right\} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by help last updated on 02/Nov/20

$${many}\:{thanks} \\ $$

Commented by TANMAY PANACEA last updated on 02/Nov/20

$${most}\:{welcome} \\ $$

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