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Question Number 18891 by Tinkutara last updated on 01/Aug/17

If angles A and B satisfy (√2) cos A = cos B + cos^3 B and (√2) sin A = sin B − sin^3 B, then the value of 1620sin^2 (A − B) is

$$\mathrm{If}\:\mathrm{angles}\:{A}\:\mathrm{and}\:{B}\:\mathrm{satisfy}\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:{A}\:= \\ $$$$\mathrm{cos}\:{B}\:+\:\mathrm{cos}^{\mathrm{3}} \:{B}\:\mathrm{and}\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:{A}\:=\:\mathrm{sin}\:{B}\:− \\ $$$$\mathrm{sin}^{\mathrm{3}} \:{B},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{1620sin}^{\mathrm{2}} \left({A}\:−\:{B}\right) \\ $$$$\mathrm{is} \\ $$

Answered by behi.8.3.4.1.7@gmail.com last updated on 02/Aug/17

2cos^2 A=cos^2 B(1+cos^2 B)^2 2sin^2 A=sin^2 B(1−sin^2 B)^2 cosB=t ⇒2=t^2 (1+t^2 )^2 +(1−t^2 ).t^4 =t^2 (1+3t^2 ) ⇒3t^4 +t^2 −2=0⇒t^2 =((−1+(√(1+24)))/6)=(2/3) ⇒cos^2 B=(2/3),sin^2 B=(1/3) ⇒cos^2 A=(1/3).(1+(2/3))^2 =(1/3).((25)/9)=((25)/(27)) ⇒cos^2 A=((25)/(27)),sin^2 A=(2/(27)) sin(A−B)=sinA.cosB−cosA.sinB= =((√2)/(3(√3))).((√2)/(√3))−(5/(3(√3))).(1/(√3))=(2/9)−(5/9)=−(1/3) ⇒1620×sin^2 (A−B)=1620×(1/9)=180 .■

$$\mathrm{2}{cos}^{\mathrm{2}} {A}={cos}^{\mathrm{2}} {B}\left(\mathrm{1}+{cos}^{\mathrm{2}} {B}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{sin}^{\mathrm{2}} {A}={sin}^{\mathrm{2}} {B}\left(\mathrm{1}−{sin}^{\mathrm{2}} {B}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cosB}={t} \\ $$$$\Rightarrow\mathrm{2}={t}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\mathrm{1}−{t}^{\mathrm{2}} \right).{t}^{\mathrm{4}} ={t}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{3}{t}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{3}{t}^{\mathrm{4}} +{t}^{\mathrm{2}} −\mathrm{2}=\mathrm{0}\Rightarrow{t}^{\mathrm{2}} =\frac{−\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{24}}}{\mathrm{6}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{cos}^{\mathrm{2}} {B}=\frac{\mathrm{2}}{\mathrm{3}},{sin}^{\mathrm{2}} {B}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{cos}^{\mathrm{2}} {A}=\frac{\mathrm{1}}{\mathrm{3}}.\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{25}}{\mathrm{9}}=\frac{\mathrm{25}}{\mathrm{27}} \\ $$$$\Rightarrow{cos}^{\mathrm{2}} {A}=\frac{\mathrm{25}}{\mathrm{27}},{sin}^{\mathrm{2}} {A}=\frac{\mathrm{2}}{\mathrm{27}} \\ $$$${sin}\left({A}−{B}\right)={sinA}.{cosB}−{cosA}.{sinB}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}\sqrt{\mathrm{3}}}.\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}}−\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{3}}}.\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}=\frac{\mathrm{2}}{\mathrm{9}}−\frac{\mathrm{5}}{\mathrm{9}}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{1620}×{sin}^{\mathrm{2}} \left({A}−{B}\right)=\mathrm{1620}×\frac{\mathrm{1}}{\mathrm{9}}=\mathrm{180}\:.\blacksquare \\ $$

Answered by ajfour last updated on 01/Aug/17

(√2)sin (A−B)=(√2)sin Acos B−(√2)cos Asin B =(sin B−sin^3 B)cos B −(cos B+cos^3 B)sin B =−(sin Bcos B)(1) ⇒2sin^2 (A−B)=sin^2 Bcos^2 B ..(i) As (√2)cos A=cos B(1+cos^2 B) and (√2)sin A=sin B(1−sin^2 B) squaring and adding above two eqns: 2=cos^2 B(1+cos^2 B)^2 +(1−cos^2 B)(cos^2 B)^2 2=cos^2 B+2cos^4 B+cos^6 B +cos^4 B−cos^6 B ⇒ 3cos^4 B+cos^2 B−2=0 3cos^4 B+3cos^2 B−2cos^2 B−2=0 ⇒ (3cos^2 B−2)(cos^2 B+1)=0 ⇒ cos^2 B=(2/3) and sin^2 B=(1/3) Using in (i): 2sin^2 (A−B)=sin^2 Bcos^2 B =(1/3)×(2/3) = (2/9) So 1620sin^2 (A−B)=810×(2/9) 1620sin^2 (A−B)=180 .

$$\sqrt{\mathrm{2}}\mathrm{sin}\:\left(\mathrm{A}−\mathrm{B}\right)=\sqrt{\mathrm{2}}\mathrm{sin}\:\mathrm{Acos}\:\mathrm{B}−\sqrt{\mathrm{2}}\mathrm{cos}\:\mathrm{Asin}\:\mathrm{B} \\ $$$$\:\:=\left(\mathrm{sin}\:\mathrm{B}−\mathrm{sin}\:^{\mathrm{3}} \mathrm{B}\right)\mathrm{cos}\:\mathrm{B} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(\mathrm{cos}\:\mathrm{B}+\mathrm{cos}\:^{\mathrm{3}} \mathrm{B}\right)\mathrm{sin}\:\mathrm{B} \\ $$$$\:\:=−\left(\mathrm{sin}\:\mathrm{Bcos}\:\mathrm{B}\right)\left(\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{2sin}\:^{\mathrm{2}} \left(\mathrm{A}−\mathrm{B}\right)=\mathrm{sin}\:^{\mathrm{2}} \mathrm{Bcos}\:^{\mathrm{2}} \mathrm{B}\:\:..\left(\mathrm{i}\right) \\ $$$$\:\:\:\mathrm{As}\:\:\sqrt{\mathrm{2}}\mathrm{cos}\:\mathrm{A}=\mathrm{cos}\:\mathrm{B}\left(\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{B}\right) \\ $$$$\mathrm{and}\:\:\:\sqrt{\mathrm{2}}\mathrm{sin}\:\mathrm{A}=\mathrm{sin}\:\mathrm{B}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{B}\right) \\ $$$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{adding}\:\mathrm{above}\:\mathrm{two}\:\mathrm{eqns}: \\ $$$$\:\:\mathrm{2}=\mathrm{cos}\:^{\mathrm{2}} \mathrm{B}\left(\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{B}\right)^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{B}\right)\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{B}\right)^{\mathrm{2}} \\ $$$$\:\:\mathrm{2}=\mathrm{cos}\:^{\mathrm{2}} \mathrm{B}+\mathrm{2cos}\:^{\mathrm{4}} \mathrm{B}+\mathrm{cos}\:^{\mathrm{6}} \mathrm{B} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{cos}\:^{\mathrm{4}} \mathrm{B}−\mathrm{cos}\:^{\mathrm{6}} \mathrm{B} \\ $$$$\Rightarrow\:\:\:\mathrm{3cos}\:^{\mathrm{4}} \mathrm{B}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{B}−\mathrm{2}=\mathrm{0} \\ $$$$\:\mathrm{3cos}\:^{\mathrm{4}} \mathrm{B}+\mathrm{3cos}\:^{\mathrm{2}} \mathrm{B}−\mathrm{2cos}\:^{\mathrm{2}} \mathrm{B}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\:\:\left(\mathrm{3cos}\:^{\mathrm{2}} \mathrm{B}−\mathrm{2}\right)\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{B}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{B}=\frac{\mathrm{2}}{\mathrm{3}}\:\:\:\mathrm{and}\:\:\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{B}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{Using}\:\mathrm{in}\:\left(\mathrm{i}\right): \\ $$$$\:\:\:\:\mathrm{2sin}\:^{\mathrm{2}} \left(\mathrm{A}−\mathrm{B}\right)=\mathrm{sin}\:^{\mathrm{2}} \mathrm{Bcos}\:^{\mathrm{2}} \mathrm{B} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{2}}{\mathrm{3}}\:=\:\frac{\mathrm{2}}{\mathrm{9}} \\ $$$$\:\:\mathrm{So}\:\:\mathrm{1620sin}\:^{\mathrm{2}} \left(\mathrm{A}−\mathrm{B}\right)=\mathrm{810}×\frac{\mathrm{2}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{1620sin}\:^{\mathrm{2}} \left(\mathrm{A}−\mathrm{B}\right)=\mathrm{180}\:. \\ $$

Commented by behi.8.3.4.1.7@gmail.com last updated on 02/Aug/17

it is perfect mr Ajfour.

$${it}\:{is}\:{perfect}\:{mr}\:{Ajfour}. \\ $$

Commented by ajfour last updated on 01/Aug/17

Is this correct?

$$\mathrm{Is}\:\mathrm{this}\:\mathrm{correct}? \\ $$

Commented by Tinkutara last updated on 01/Aug/17

Thank you very much ajfour Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{ajfour}\:\mathrm{Sir}! \\ $$

Commented by ajfour last updated on 01/Aug/17

thanks for confirming.

$$\mathrm{thanks}\:\mathrm{for}\:\mathrm{confirming}. \\ $$

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