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Question Number 54563 by 951172235v last updated on 06/Feb/19

If ((sin^3 α)/(sin β)) + ((cos^3 α)/(cos β)) = 1 show that sin 2α+2sin (α+β)=0

$$\mathrm{If}\:\frac{\mathrm{sin}\:^{\mathrm{3}} \alpha}{\mathrm{sin}\:\beta}\:+\:\frac{\mathrm{cos}\:^{\mathrm{3}} \alpha}{\mathrm{cos}\:\beta}\:=\:\mathrm{1}\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{2sin}\:\left(\alpha+\beta\right)=\mathrm{0} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Feb/19

((sin^3 α)/(sinβ))+((cos^3 α)/(cosβ))=((sin^3 α)/(sinα))+((cos^3 α)/(cosα)) sin^3 α(((sinα−sinβ)/(sinαsinβ)))+cos^3 α(((cosα−cosβ)/(cosαcosβ)))=0 ((2sin^3 αcos(((α+β)/2))sin(((α−β)/2)))/(sinαsinβ))−((2cos^3 αsin(((α+β)/2))sin(((α−β)/2)))/(cosαcosβ))=0 take 2sin(((α−β)/2)) common factor if sin(((α−β)/2))=0 so α=β but α=β do not conform sin2α+2sin(α+β)=0 so ((sin^2 αcos(((α+β)/2)))/(sinβ))−((cos^2 αsin(((α+β)/2)))/(cosβ))=0 sin^2 αcos(((α+β)/2))cosβ=cos^2 αsin(((α+β)/2))sinβ tan^2 α=tan(((α+β)/2))tanβ approach 1) a=tan(α/2) b=tan(β/2) (((2a)/(1−a^2 )))^2 =((a+b)/(1−ab))×((2b)/(1−b^2 )) 4a^2 (1−ab)(1−b^2 )=(a+b)2b(1−a^2 )^2 4a^2 (1−b^2 −ab+ab^3 )=(2ab+2b^2 )(1−2a^2 +a^4 ) 4a^2 −4a^2 b^2 −4a^3 b+4a^3 b^3 =2ab−4a^3 b+2a^5 b+2b^2 −4a^2 b^2 +2a^4 b^2 4a^2 +4a^3 b^3 −2ab−2a^4 b−2b^2 −2a^4 b^2 =0 4a^2 +4a^3 b^3 −2ab−4a^4 b^2 −2b^2 =02 2a^2 −ab−b^2 +2a^3 b^2 (b−a)=0 2a^2 −2ab+ab−b^2 −2a^3 b^2 (a−b)=0 2a(a−b)+b(a−b)−2a^3 b^2 (a−b)=9( (a−b)(2a+b−2a^3 b^2 )=0 2a+b=2a^3 b^2 wait pls...

$$\frac{{sin}^{\mathrm{3}} \alpha}{{sin}\beta}+\frac{{cos}^{\mathrm{3}} \alpha}{{cos}\beta}=\frac{{sin}^{\mathrm{3}} \alpha}{{sin}\alpha}+\frac{{cos}^{\mathrm{3}} \alpha}{{cos}\alpha} \\ $$$${sin}^{\mathrm{3}} \alpha\left(\frac{{sin}\alpha−{sin}\beta}{{sin}\alpha{sin}\beta}\right)+{cos}^{\mathrm{3}} \alpha\left(\frac{{cos}\alpha−{cos}\beta}{{cos}\alpha{cos}\beta}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{2}{sin}^{\mathrm{3}} \alpha{cos}\left(\frac{\alpha+\beta}{\mathrm{2}}\right){sin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)}{{sin}\alpha{sin}\beta}−\frac{\mathrm{2}{cos}^{\mathrm{3}} \alpha{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}\right){sin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)}{{cos}\alpha{cos}\beta}=\mathrm{0} \\ $$$${take}\:\mathrm{2}{sin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)\:{common}\:{factor} \\ $$$${if}\:{sin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)=\mathrm{0}\:\:\:{so}\:\alpha=\beta\:\:{but}\:\alpha=\beta\:{do}\:{not}\:{conform} \\ $$$${sin}\mathrm{2}\alpha+\mathrm{2}{sin}\left(\alpha+\beta\right)=\mathrm{0} \\ $$$${so} \\ $$$$\frac{{sin}^{\mathrm{2}} \alpha{cos}\left(\frac{\alpha+\beta}{\mathrm{2}}\right)}{{sin}\beta}−\frac{{cos}^{\mathrm{2}} \alpha{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}\right)}{{cos}\beta}=\mathrm{0} \\ $$$${sin}^{\mathrm{2}} \alpha{cos}\left(\frac{\alpha+\beta}{\mathrm{2}}\right){cos}\beta={cos}^{\mathrm{2}} \alpha{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}\right){sin}\beta \\ $$$${tan}^{\mathrm{2}} \alpha={tan}\left(\frac{\alpha+\beta}{\mathrm{2}}\right){tan}\beta \\ $$$$\left.{approach}\:\mathrm{1}\right)\:{a}={tan}\frac{\alpha}{\mathrm{2}}\:\:{b}={tan}\frac{\beta}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{2}{a}}{\mathrm{1}−{a}^{\mathrm{2}} }\right)^{\mathrm{2}} =\frac{{a}+{b}}{\mathrm{1}−{ab}}×\frac{\mathrm{2}{b}}{\mathrm{1}−{b}^{\mathrm{2}} } \\ $$$$\mathrm{4}{a}^{\mathrm{2}} \left(\mathrm{1}−{ab}\right)\left(\mathrm{1}−{b}^{\mathrm{2}} \right)=\left({a}+{b}\right)\mathrm{2}{b}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} \left(\mathrm{1}−{b}^{\mathrm{2}} −{ab}+{ab}^{\mathrm{3}} \right)=\left(\mathrm{2}{ab}+\mathrm{2}{b}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{2}{a}^{\mathrm{2}} +{a}^{\mathrm{4}} \right) \\ $$$$\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{3}} {b}+\mathrm{4}{a}^{\mathrm{3}} {b}^{\mathrm{3}} =\mathrm{2}{ab}−\mathrm{4}{a}^{\mathrm{3}} {b}+\mathrm{2}{a}^{\mathrm{5}} {b}+\mathrm{2}{b}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{4}} {b}^{\mathrm{2}} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{3}} {b}^{\mathrm{3}} −\mathrm{2}{ab}−\mathrm{2}{a}^{\mathrm{4}} {b}−\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{4}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{3}} {b}^{\mathrm{3}} −\mathrm{2}{ab}−\mathrm{4}{a}^{\mathrm{4}} {b}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} =\mathrm{02} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} −{ab}−{b}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{2}} \left({b}−{a}\right)=\mathrm{0} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{ab}+{ab}−{b}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{2}} \left({a}−{b}\right)=\mathrm{0} \\ $$$$\mathrm{2}{a}\left({a}−{b}\right)+{b}\left({a}−{b}\right)−\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{2}} \left({a}−{b}\right)=\mathrm{9}\left(\right. \\ $$$$\left({a}−{b}\right)\left(\mathrm{2}{a}+{b}−\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{2}{a}+{b}=\mathrm{2}{a}^{\mathrm{3}} {b}^{\mathrm{2}} \\ $$$${wait}\:{pls}... \\ $$$$ \\ $$

Answered by 951172235v last updated on 07/Feb/19

let sin α =xsin β and cosα =ycos β x^2 sin^2 β+y^2 cos^2 β =1 sin^2 β=((1−y^2 )/(x^2 −y^2 )) cos^2 β=((1−x^2 )/(y^2 −x^2 )) x^3 sin^2 β+y^3 cos^2 β=1 x^3 (((1−y^2 )/(x^2 −y^2 )))−y^3 (((1−x^2 )/(x^2 −y^2 )))=1 (((x−y)(x^2 +y^2 +xy)−x^2 y^2 (x−y))/((x−y)(x+y))) =1 x^2 +y^2 +xy−x^2 y^2 =x+y (x+y)^2 −xy−x^2 y^2 =x+y (x+y+xy)(x+y−xy)−(x+y+xy)=0 (x+y+xy)(x+y−xy−1)=0 (x+y+xy)(x−1)(1−y)=0 x,y≠1 x+y = −xy ((sin α)/(sin β)) +((cos α)/(cos β )) =((−(sin αcos α))/(sin βcos β)) sin αcos β+sin βcos α =−sin αcos α sin (α+β)=−sin αcos α sin 2α+2sin (α+β)=0 ans.

$$\mathrm{let}\:\mathrm{sin}\:\alpha\:=\mathrm{xsin}\:\beta\:\:\mathrm{and}\:\mathrm{cos}\alpha\:=\mathrm{ycos}\:\beta \\ $$$$\mathrm{x}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \beta+\mathrm{y}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \beta\:=\mathrm{1}\:\:\:\:\mathrm{sin}\:^{\mathrm{2}} \beta=\frac{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} \beta=\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{x}^{\mathrm{3}} \mathrm{sin}\:^{\mathrm{2}} \beta+\mathrm{y}^{\mathrm{3}} \mathrm{cos}\:^{\mathrm{2}} \beta=\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{3}} \left(\frac{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\right)−\mathrm{y}^{\mathrm{3}} \left(\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$$\frac{\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{xy}\right)−\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \left(\mathrm{x}−\mathrm{y}\right)}{\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{x}+\mathrm{y}\right)}\:=\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{xy}−\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} =\mathrm{x}+\mathrm{y} \\ $$$$\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} −\mathrm{xy}−\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} =\mathrm{x}+\mathrm{y} \\ $$$$\left(\mathrm{x}+\mathrm{y}+\mathrm{xy}\right)\left(\mathrm{x}+\mathrm{y}−\mathrm{xy}\right)−\left(\mathrm{x}+\mathrm{y}+\mathrm{xy}\right)=\mathrm{0} \\ $$$$\left(\mathrm{x}+\mathrm{y}+\mathrm{xy}\right)\left(\mathrm{x}+\mathrm{y}−\mathrm{xy}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\mathrm{x}+\mathrm{y}+\mathrm{xy}\right)\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{1}−\mathrm{y}\right)=\mathrm{0} \\ $$$$\mathrm{x},\mathrm{y}\neq\mathrm{1}\: \\ $$$$\:\mathrm{x}+\mathrm{y}\:=\:−\mathrm{xy} \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{sin}\:\beta}\:+\frac{\mathrm{cos}\:\alpha}{\mathrm{cos}\:\beta\:\:}\:=\frac{−\left(\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha\right)}{\mathrm{sin}\:\beta\mathrm{cos}\:\beta} \\ $$$$\mathrm{sin}\:\alpha\mathrm{cos}\:\beta+\mathrm{sin}\:\beta\mathrm{cos}\:\alpha\:=−\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha \\ $$$$\mathrm{sin}\:\left(\alpha+\beta\right)=−\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{2sin}\:\left(\alpha+\beta\right)=\mathrm{0}\:\:\mathrm{ans}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Feb/19

excellent sir...your every post praiseworthy

$${excellent}\:{sir}...{your}\:{every}\:{post}\:{praiseworthy} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Feb/19

excellent sir ...your every post are praiseworthy

$${excellent}\:{sir}\:...{your}\:{every}\:{post}\:{are}\:{praiseworthy} \\ $$

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