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Question Number 93239 by Ar Brandon last updated on 11/May/20

$$\mathrm{Calculate};$$$$i)\cos \left(\arctan x\right)$$$$ii)\cos \left(\arcsin x\right)$$$$iii)\tan \left(\arcsin x\right)$$

Commented by mathmax by abdo last updated on 12/May/20

$$1)vhangementarctanx=tgivex=tant\Rightarrow$$$$cos\left(arctanx\right)=cost=\sqrt{\frac{1}{1+{tan}^{2}t}}=\frac{1}{\sqrt{1+x^2}}$$$$2)letarcsinx=u\Rightarrow x=sinu\Rightarrow$$$$cos\left(arcsinx\right)=cosu=\sqrt{1-{sin}^{2}u}=\sqrt{1-x^2}$$$$3)letarcsinx=u\Rightarrow x=sinu\Rightarrow tan\left(arcsinx\right)=tanu$$$$=\frac{sinu}{cosu}=\frac{sinu}{\sqrt{1-{sin}^{2}u}}=\frac{x}{\sqrt{1-x^2}}$$

Commented by Ar Brandon last updated on 12/May/20

Thanks Mathmax ��

Commented by mathmax by abdo last updated on 12/May/20

$$youarewelcome.$$

Answered by hknkrc46 last updated on 12/May/20

$$i)\arctan x=u\Rightarrow \tan u=x\wedge \cos u=h=?$$$$\tan u=\frac{\sin u}{\cos u}=\frac{\sqrt{1-\cos {}^{2}u}}{\cos u}=\frac{\sqrt{1-h^2}}{h}=x$$$$\Rightarrow \sqrt{1-h^2}=hx\Rightarrow 1-h^2=h^2{x}^2$$$$\Rightarrow h^2{x}^2+h^2=1\Rightarrow h^2(x^2+1)=1$$$$\Rightarrow h^2=\frac{1}{x^2+1}\Rightarrow h=\frac{1}{\sqrt{x^2+1}}$$$$ii)\arcsin x=u\Rightarrow \sin u=x\wedge \cos u=?$$$$\cos u=\sqrt{\cos {}^{2}u}=\sqrt{1-\sin {}^{2}u}=\sqrt{1-x^2}$$$$iii)\arcsin x=u\Rightarrow \sin u=x\wedge \tan u=h=?$$$$h=\tan u=\frac{\sin u}{\cos u}=\frac{\sin u}{\sqrt{1-\sin {}^{2}u}}=\frac{x}{\sqrt{1-x^2}}$$

Commented by Ar Brandon last updated on 12/May/20

hknkrc46 Thank you sir ��

Answered by Rio Michael last updated on 12/May/20

$$\left(i\right)\cos \left(\arctan x\right)$$$$\mathrm{my}\mathrm{approach}$$$$\mathrm{let}\arctan x=u\Rightarrow \tan u=x$$$$\Rightarrow \frac{\sin u}{\cos u}=x$$$$\frac{{\sin }^{2}u}{{\cos }^{2}u}=x^2$$$$\Rightarrow \frac{{\sin }^{2}u}{{\cos }^{2}u}+1=x^2+1$$$$\mathrm{hence}\frac{{\sin }^{2}u+{\cos }^{2}u}{{\cos }^{2}u}=x^2+1$$$$\frac{1}{x^2+1}={\cos }^{2}u$$$$\Rightarrow \cos u=\sqrt{\frac{1}{x^2+1}}\mathrm{for}-\frac{\pi }{2}⩽x⩽\frac{\pi }{2}$$$$\mathrm{since}\mathrm{you}\mathrm{were}\mathrm{not}\mathrm{specific}.$$

Commented by Ar Brandon last updated on 12/May/20

Thanks Mr Rio Michael. Good Day Sir ��

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