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 116768 by bemath last updated on 06/Oct/20

sin (4sin^(−1) (x)) = sin (2sin^(−1) (x))

$$\mathrm{sin}\:\left(\mathrm{4sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)\:=\:\mathrm{sin}\:\left(\mathrm{2sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\right) \\ $$

Answered by bobhans last updated on 06/Oct/20

letting sin^(−1) (x) = z⇒sin z = x  ⇒ sin (4z) = sin (2z)   ⇒2sin (2z)cos (2z)−sin (2z) = 0  ⇒ { ((sin (2z)=0 or)),((cos (2z) = (1/2))) :}  case(1) ⇒sin 2z = 0  ⇒ 2sin z.cos z = 0 ; 2x(√(1−x^2 )) = 0  we get x = 0 or x = ± 1  case(2) ⇒cos (2z) = (1/2)  ⇒ 1−2sin^2  z = (1/2) ; (1/2)−2x^2  = 0   ⇒((1/2))^2 −x^2  = 0 ⇒x = ± (1/2)  therefore we get solution set is   x∈ { 0, ±(1/2) , ±1 }

$$\mathrm{letting}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\:=\:\mathrm{z}\Rightarrow\mathrm{sin}\:\mathrm{z}\:=\:\mathrm{x} \\ $$$$\Rightarrow\:\mathrm{sin}\:\left(\mathrm{4z}\right)\:=\:\mathrm{sin}\:\left(\mathrm{2z}\right)\: \\ $$$$\Rightarrow\mathrm{2sin}\:\left(\mathrm{2z}\right)\mathrm{cos}\:\left(\mathrm{2z}\right)−\mathrm{sin}\:\left(\mathrm{2z}\right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\mathrm{sin}\:\left(\mathrm{2z}\right)=\mathrm{0}\:\mathrm{or}}\\{\mathrm{cos}\:\left(\mathrm{2z}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:\Rightarrow\mathrm{sin}\:\mathrm{2z}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{2sin}\:\mathrm{z}.\mathrm{cos}\:\mathrm{z}\:=\:\mathrm{0}\:;\:\mathrm{2x}\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{x}\:=\:\mathrm{0}\:\mathrm{or}\:\mathrm{x}\:=\:\pm\:\mathrm{1} \\ $$$$\mathrm{case}\left(\mathrm{2}\right)\:\Rightarrow\mathrm{cos}\:\left(\mathrm{2z}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:\mathrm{z}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:;\:\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2x}^{\mathrm{2}} \:=\:\mathrm{0}\: \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{0}\:\Rightarrow\mathrm{x}\:=\:\pm\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{therefore}\:\mathrm{we}\:\mathrm{get}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{is}\: \\ $$$$\mathrm{x}\in\:\left\{\:\mathrm{0},\:\pm\frac{\mathrm{1}}{\mathrm{2}}\:,\:\pm\mathrm{1}\:\right\} \\ $$

Answered by MJS_new last updated on 06/Oct/20

A=sin (4arcsin x) =4x(1−2x^2 )(√(1−x^2 ))  B=sin (2arcsin x) =2x(√(1−x^2 ))  A−B=0  2x(1−4x^2 )(√(1−x^2 ))=0  ⇒ x=0∨x=±(1/2)∨x=±1

$${A}=\mathrm{sin}\:\left(\mathrm{4arcsin}\:{x}\right)\:=\mathrm{4}{x}\left(\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${B}=\mathrm{sin}\:\left(\mathrm{2arcsin}\:{x}\right)\:=\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${A}−{B}=\mathrm{0} \\ $$$$\mathrm{2}{x}\left(\mathrm{1}−\mathrm{4}{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\mathrm{0}\vee{x}=\pm\frac{\mathrm{1}}{\mathrm{2}}\vee{x}=\pm\mathrm{1} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com