Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 143443 by mnjuly1970 last updated on 14/Jun/21

Commented by amin96 last updated on 14/Jun/21

?

$$? \\ $$$$ \\ $$

Answered by mr W last updated on 14/Jun/21

say R=radius  cos α=(R/b)  (b sin α+a cos α)^2 +(a sin α)^2 =R^2   b^2  sin^2  α+2ab sin αcos α+a^2 =b^2 cos^2  α  a^2 =b(b cos 2α−a sin 2α)  a^2 =b(√(a^2 +b^2 ))(cos ϕ cos 2α−sin ϕ sin 2α)  (a^2 /(b(√(a^2 +b^2 ))))=cos (ϕ+2α)  2α=cos^(−1) (a^2 /(b(√(a^2 +b^2 ))))−cos^(−1) (b^2 /( b(√(a^2 +b^2 ))))  cos 2α=((a(ab+(√(a^2 b^2 +b^4 −a^4 ))))/( b(a^2 +b^2 )))  ⇒tan α=(√((b^3 −a(√(a^2 b^2 +b^4 −a^4 )))/(2a^2 b+b^3 +a(√(a^2 b^2 +b^4 −a^4 )))))

$${say}\:{R}={radius} \\ $$$$\mathrm{cos}\:\alpha=\frac{{R}}{{b}} \\ $$$$\left({b}\:\mathrm{sin}\:\alpha+{a}\:\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +\left({a}\:\mathrm{sin}\:\alpha\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\alpha+\mathrm{2}{ab}\:\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha+{a}^{\mathrm{2}} ={b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\alpha \\ $$$${a}^{\mathrm{2}} ={b}\left({b}\:\mathrm{cos}\:\mathrm{2}\alpha−{a}\:\mathrm{sin}\:\mathrm{2}\alpha\right) \\ $$$${a}^{\mathrm{2}} ={b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left(\mathrm{cos}\:\varphi\:\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{sin}\:\varphi\:\mathrm{sin}\:\mathrm{2}\alpha\right) \\ $$$$\frac{{a}^{\mathrm{2}} }{{b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}=\mathrm{cos}\:\left(\varphi+\mathrm{2}\alpha\right) \\ $$$$\mathrm{2}\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{{a}^{\mathrm{2}} }{{b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}−\mathrm{cos}^{−\mathrm{1}} \frac{{b}^{\mathrm{2}} }{\:{b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\mathrm{2}\alpha=\frac{{a}\left({ab}+\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} −{a}^{\mathrm{4}} }\right)}{\:{b}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\sqrt{\frac{{b}^{\mathrm{3}} −{a}\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} −{a}^{\mathrm{4}} }}{\mathrm{2}{a}^{\mathrm{2}} {b}+{b}^{\mathrm{3}} +{a}\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} −{a}^{\mathrm{4}} }}} \\ $$

Commented by mnjuly1970 last updated on 14/Jun/21

thanks alot Mr  W....

$${thanks}\:{alot}\:{Mr}\:\:{W}.... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com