Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 57959 by rahul 19 last updated on 15/Apr/19

Answered by MJS last updated on 16/Apr/19

3sin^(−1)  x =y ⇒ x=sin (y/3)  with x∈[−1; 1] ∧ y∈[−((3π)/2); ((3π)/2)]  3sin^(−1)  x increases for −1≤x≤1    sin^(−1)  (−4x^3 +3x) =y ⇒ −4x^3 +3x=sin y  with (−4x^3 +3x)∈[−1; 1] ∧ y∈[−(π/2); (π/2)]  (−4x^3 +3x)∈[−1; 1] ⇒ x∈[−1; 1]  but −4x^3 +3x has a minimum at  (((−(1/2))),((−1)) )  and a maximum at  (((1/2)),(1) ) ⇒  ⇒ sin^(−1)  (−4x^3 +3x) decreases for −1≤x<−(1/2)  and (1/2)<x≤1 and increases for −(1/2)<x<(1/2)    ⇒ answer is (a)

$$\mathrm{3sin}^{−\mathrm{1}} \:{x}\:={y}\:\Rightarrow\:{x}=\mathrm{sin}\:\frac{{y}}{\mathrm{3}} \\ $$$$\mathrm{with}\:{x}\in\left[−\mathrm{1};\:\mathrm{1}\right]\:\wedge\:{y}\in\left[−\frac{\mathrm{3}\pi}{\mathrm{2}};\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right] \\ $$$$\mathrm{3sin}^{−\mathrm{1}} \:{x}\:\mathrm{increases}\:\mathrm{for}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$$ \\ $$$$\mathrm{sin}^{−\mathrm{1}} \:\left(−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}\right)\:={y}\:\Rightarrow\:−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}=\mathrm{sin}\:{y} \\ $$$$\mathrm{with}\:\left(−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}\right)\in\left[−\mathrm{1};\:\mathrm{1}\right]\:\wedge\:{y}\in\left[−\frac{\pi}{\mathrm{2}};\:\frac{\pi}{\mathrm{2}}\right] \\ $$$$\left(−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}\right)\in\left[−\mathrm{1};\:\mathrm{1}\right]\:\Rightarrow\:{x}\in\left[−\mathrm{1};\:\mathrm{1}\right] \\ $$$$\mathrm{but}\:−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}\:\mathrm{has}\:\mathrm{a}\:\mathrm{minimum}\:\mathrm{at}\:\begin{pmatrix}{−\frac{\mathrm{1}}{\mathrm{2}}}\\{−\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{and}\:\mathrm{a}\:\mathrm{maximum}\:\mathrm{at}\:\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{1}}\end{pmatrix}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{sin}^{−\mathrm{1}} \:\left(−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}\right)\:\mathrm{decreases}\:\mathrm{for}\:−\mathrm{1}\leqslant{x}<−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{2}}<{x}\leqslant\mathrm{1}\:\mathrm{and}\:\mathrm{increases}\:\mathrm{for}\:−\frac{\mathrm{1}}{\mathrm{2}}<{x}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\left(\mathrm{a}\right) \\ $$

Commented by MJS last updated on 16/Apr/19

solving −4x^3 +3x=sin y  x^3 −(3/4)x+(1/4)sin y =0  D=(p^3 /(27))+(q^2 /4)=(((−(3/4))^3 )/(27))+((((1/4)sin y)^2 )/4)=((−1+sin^2  y)/(64))=  =−((cos^2  y)/(64)) ≤0 ⇒ 3 real solutions ⇒ we need  the trigonometric method, which gives  x_1 =sin (y/3)  x_2 =(1/2)((√3)cos (y/3) −sin (y/3))  x_3 =−(1/2)((√3)cos (y/3) +sin (y/3))  ⇒ 3sin^(−1)  x =sin^(−1)  (−4x^3 +3x) in certain cases

$$\mathrm{solving}\:−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}=\mathrm{sin}\:{y} \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{4}}{x}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:{y}\:=\mathrm{0} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}=\frac{\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} }{\mathrm{27}}+\frac{\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:{y}\right)^{\mathrm{2}} }{\mathrm{4}}=\frac{−\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{y}}{\mathrm{64}}= \\ $$$$=−\frac{\mathrm{cos}^{\mathrm{2}} \:{y}}{\mathrm{64}}\:\leqslant\mathrm{0}\:\Rightarrow\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions}\:\Rightarrow\:\mathrm{we}\:\mathrm{need} \\ $$$$\mathrm{the}\:\mathrm{trigonometric}\:\mathrm{method},\:\mathrm{which}\:\mathrm{gives} \\ $$$${x}_{\mathrm{1}} =\mathrm{sin}\:\frac{{y}}{\mathrm{3}} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:\frac{{y}}{\mathrm{3}}\:−\mathrm{sin}\:\frac{{y}}{\mathrm{3}}\right) \\ $$$${x}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{cos}\:\frac{{y}}{\mathrm{3}}\:+\mathrm{sin}\:\frac{{y}}{\mathrm{3}}\right) \\ $$$$\Rightarrow\:\mathrm{3sin}^{−\mathrm{1}} \:{x}\:=\mathrm{sin}^{−\mathrm{1}} \:\left(−\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}{x}\right)\:\mathrm{in}\:\mathrm{certain}\:\mathrm{cases} \\ $$

Commented by rahul 19 last updated on 17/Apr/19

Thank you sir.

$${Thank}\:{you}\:{sir}. \\ $$

Commented by MJS last updated on 17/Apr/19

you′re welcome. I hope my answer is helpful  I′m not 100% sure if this is a legal proof...

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}.\:\mathrm{I}\:\mathrm{hope}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{helpful} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{100\%}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{legal}\:\mathrm{proof}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com