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Question Number 17119 by gourav~ last updated on 01/Jul/17

Commented by prakash jain last updated on 01/Jul/17

((sin (A+3B)+sin (3A+B))/(sin 2A+sin 2B)) =((2sin (((4A+4B)/2))cos (((2B−2A)/2)))/(2sin (A+B)cos (A−B))) =((2sin (2(A+B))cos (A−B))/(2sin (A+B)cos (A−B))) =((sin (2(A+B)))/(sin (A+B))) =((2sin (A+B)cos (A+B))/(sin (A+B))) =2cos (A+B)■

$$\frac{\mathrm{sin}\:\left({A}+\mathrm{3}{B}\right)+\mathrm{sin}\:\left(\mathrm{3}{A}+{B}\right)}{\mathrm{sin}\:\mathrm{2}{A}+\mathrm{sin}\:\mathrm{2}{B}} \\ $$$$=\frac{\mathrm{2sin}\:\left(\frac{\mathrm{4}{A}+\mathrm{4}{B}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{2}{B}−\mathrm{2}{A}}{\mathrm{2}}\right)}{\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right)} \\ $$$$=\frac{\mathrm{2sin}\:\left(\mathrm{2}\left({A}+{B}\right)\right)\mathrm{cos}\:\left({A}−{B}\right)}{\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}−{B}\right)} \\ $$$$=\frac{\mathrm{sin}\:\left(\mathrm{2}\left({A}+{B}\right)\right)}{\mathrm{sin}\:\left({A}+{B}\right)} \\ $$$$=\frac{\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{cos}\:\left({A}+{B}\right)}{\mathrm{sin}\:\left({A}+{B}\right)} \\ $$$$=\mathrm{2cos}\:\left({A}+{B}\right)\blacksquare \\ $$

Commented by gourav~ last updated on 01/Jul/17

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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