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In-ABC-if-sin-A-sin-2-B-then-prove-4-cos-2A-4-cos-2B-1-cos-4B-

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Question Number 37914 by gunawan last updated on 19/Jun/18
In △ABC, if sin A=sin^2 B then prove 4 cos 2A−4 cos 2B=1−cos 4B
$$\mathrm{In}\:\bigtriangleup{ABC},\:\mathrm{if}\:\mathrm{sin}\:\mathrm{A}=\mathrm{sin}^{\mathrm{2}} {B}\: \\ $$$$\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{4}\:\mathrm{cos}\:\mathrm{2}{A}−\mathrm{4}\:\mathrm{cos}\:\mathrm{2}{B}=\mathrm{1}−\mathrm{cos}\:\mathrm{4}{B} \\ $$
Answered by $@ty@m last updated on 19/Jun/18
LHS= 4(cos 2A−cos 2B) =4.2sin (A+B)sin (B−A) =8(sin Bcos A+cos Bsin A). (sin Bcos A−cos Bsin A). =8(sin^2 Bcos^2 A−cos^2 Bsin^2 A) =8{sin^2 B(1−sin^2 A)−(1−sin^2 B)sin^2 A} =8{sinA(1−sin^2 A)−(1−sinA)sin^2 A} =8(sin A−sin^3 A−sin^2 A+sin^3 A) =8(sin^2 B−sin^4 B) =8sin^2 Bcos^2 B =2.(2sin Bcos B)^2 =2sin^2 2B =1−cos 4B =RHS
$${LHS}= \\ $$$$\mathrm{4}\left(\mathrm{cos}\:\mathrm{2}{A}−\mathrm{cos}\:\mathrm{2}{B}\right) \\ $$$$=\mathrm{4}.\mathrm{2sin}\:\left({A}+{B}\right)\mathrm{sin}\:\left({B}−{A}\right) \\ $$$$=\mathrm{8}\left(\mathrm{sin}\:{B}\mathrm{cos}\:{A}+\mathrm{cos}\:{B}\mathrm{sin}\:{A}\right). \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{sin}\:{B}\mathrm{cos}\:{A}−\mathrm{cos}\:{B}\mathrm{sin}\:{A}\right). \\ $$$$=\mathrm{8}\left(\mathrm{sin}\:^{\mathrm{2}} {B}\mathrm{cos}^{\mathrm{2}} \:{A}−\mathrm{cos}^{\mathrm{2}} \:{B}\mathrm{sin}^{\mathrm{2}} \:{A}\right) \\ $$$$=\mathrm{8}\left\{\mathrm{sin}\:^{\mathrm{2}} {B}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{A}\right)−\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \:{B}\right)\mathrm{sin}^{\mathrm{2}} \:{A}\right\} \\ $$$$=\mathrm{8}\left\{\mathrm{sin}{A}\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{A}\right)−\left(\mathrm{1}−\mathrm{sin}{A}\right)\mathrm{sin}^{\mathrm{2}} \:{A}\right\} \\ $$$$=\mathrm{8}\left(\mathrm{sin}\:{A}−\mathrm{sin}\:^{\mathrm{3}} {A}−\mathrm{sin}\:^{\mathrm{2}} {A}+\mathrm{sin}\:^{\mathrm{3}} {A}\right) \\ $$$$=\mathrm{8}\left(\mathrm{sin}\:^{\mathrm{2}} {B}−\mathrm{sin}\:^{\mathrm{4}} {B}\right) \\ $$$$=\mathrm{8sin}\:^{\mathrm{2}} {B}\mathrm{cos}\:^{\mathrm{2}} {B} \\ $$$$=\mathrm{2}.\left(\mathrm{2sin}\:{B}\mathrm{cos}\:{B}\right)^{\mathrm{2}} \\ $$$$=\mathrm{2sin}\:^{\mathrm{2}} \mathrm{2}{B} \\ $$$$=\mathrm{1}−\mathrm{cos}\:\mathrm{4}{B} \\ $$$$={RHS} \\ $$

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