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Question Number 137530 by ZiYangLee last updated on 03/Apr/21

$$\mathrm{If}{\sin }^{-1}(\sin \alpha +\sin \beta )+{\sin }^{-1}(\sin \alpha -\sin \beta )=\frac{\pi }{2}$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}{\sin }^2\alpha +{\sin }^2\beta .\left[\frac{1}{2}\right]$$

Answered by mr W last updated on 04/Apr/21

$$\sin \alpha +\sin \beta =u$$$$\sin \alpha -\sin \beta =v$$$${\sin }^2\alpha +{\sin }^2\beta =\frac{u^2+v^2}{2}$$$${\sin }^{-1}u+{\sin }^{-1}v=\frac{\pi }{2}$$$${\sin }^{-1}u=\frac{\pi }{2}-{\sin }^{-1}v$$$$u=\sin (\frac{\pi }{2}-{\sin }^{-1}v)=\cos \left({\sin }^{-1}v\right)$$$$=\sqrt{1-{\sin }^2\left({\sin }^{-1}v\right)}=\sqrt{1-v^2}$$$$\Rightarrow u^2=1-v^2$$$$\Rightarrow u^2+v^2=1$$$$\Rightarrow {\sin }^2\alpha +{\sin }^2\beta =\frac{u^2+v^2}{2}=\frac{1}{2}$$

Commented by otchereabdullai@gmail.com last updated on 04/Apr/21

$$\mathrm{prof}\mathrm{i}\mathrm{want}\mathrm{to}\mathrm{learn}\mathrm{something}\mathrm{from}$$$$\mathrm{you}\mathrm{can}\mathrm{you}\mathrm{please}\mathrm{prove}\mathrm{the}$$$${\sin }^2\mathrm{A}+{\sin }^2\mathrm{B}=1\mathrm{for}\mathrm{me}?$$$$\mathrm{i}\mathrm{only}\mathrm{know}\mathrm{of}{\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}=1$$$$\mathrm{sorry}\mathrm{for}\mathrm{worrying}\mathrm{you}$$$$$$

Commented by otchereabdullai@gmail.com last updated on 04/Apr/21

$$\mathrm{wow}!$$

Commented by otchereabdullai@gmail.com last updated on 04/Apr/21

$$\mathrm{very}\mathrm{nice}\mathrm{solution}\mathrm{but}\mathrm{am}\mathrm{finding}\mathrm{it}$$$$\mathrm{difficult}\mathrm{to}\mathrm{understand}\mathrm{where}\mathrm{the}\cos$$$$\mathrm{and}\mathrm{the}\mathrm{square}\mathrm{root}\mathrm{is}\mathrm{introduce}$$$$\mathrm{please}\mathrm{help}\mathrm{me}$$

Commented by mr W last updated on 04/Apr/21

$$say$$$$A={\sin }^{-1}u\Rightarrow \sin A=u$$$$B={\sin }^{-1}v\Rightarrow \sin B=v$$$$A+B=\frac{\pi }{2}$$$$A=\frac{\pi }{2}-B$$$$\sin A=\sin (\frac{\pi }{2}-B)=\cos B=\sqrt{1-{\sin }^{2}B}$$$${\sin }^{2}A=1-{\sin }^{2}B$$$${\sin }^{2}A+{\sin }^{2}B=1$$$$u^2+v^2=1$$

Commented by otchereabdullai@gmail.com last updated on 04/Apr/21

$$\mathrm{now}\mathrm{is}\mathrm{very}\mathrm{clear}\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{alot}$$$$\mathrm{for}\mathrm{the}\mathrm{kindness}\mathrm{profW}$$

Commented by mr W last updated on 04/Apr/21

$${\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}=1✓$$$$sinceA=\frac{\pi }{2}-B,whichmeans$$$$\cos A=\sin B,soweget$$$${\sin }^2\mathrm{A}+{\sin }^{2}B=1✓$$

Commented by otchereabdullai@gmail.com last updated on 04/Apr/21

$$\mathrm{God}\mathrm{bless}\mathrm{u}\mathrm{profW}$$

Commented by ZiYangLee last updated on 10/Apr/21

$$\mathrm{ProfW},\mathrm{how}\mathrm{can}$$$$\sqrt{1-{\sin }^2\left({\sin }^{-1}b\right)}=\sqrt{1-v^2}$$

Commented by mr W last updated on 16/Apr/21

$$youmean$$$$\sqrt{1-{\sin }^2\left({\sin }^{-1}v\right)}=\sqrt{1-v^2}?$$$$say{\sin }^{-1}v=V,thatmeans$$$$\sin V=v$$$$i.e.\sin \left({\sin }^{-1}v\right)=v$$$$so{\sin }^2\left({\sin }^{-1}v\right)=v^2$$$$\sqrt{1-{\sin }^2\left({\sin }^{-1}v\right)}=\sqrt{1-v^2}$$

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