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Question Number 24098 by Sudipta Jana last updated on 12/Nov/17

$${\sin }^{-1}\frac{ax}{c}+{\sin }^{-1}\frac{bx}{c}={\sin }^{-1}x[When{a}^2+b^2=c^2]$$

Answered by 951172235v last updated on 05/Feb/19

$$$$$$\frac{\mathrm{ax}\sqrt{{\mathrm{c}}^2-{\mathrm{b}}^2{\mathrm{x}}^2}}{{\mathrm{c}}^2}+\frac{\mathrm{bx}\sqrt{{\mathrm{c}}^2-{\mathrm{a}}^2{\mathrm{x}}^2}}{{\mathrm{c}}^2}=\mathrm{x}$$$$\mathrm{x}=0\mathrm{or}\mathrm{a}\sqrt{{\mathrm{c}}^2-{\mathrm{b}}^2{\mathrm{x}}^2}+\mathrm{b}\sqrt{{\mathrm{c}}^2-{\mathrm{a}}^2{\mathrm{x}}^2}={\mathrm{c}}^2$$$${\mathrm{a}}^2({\mathrm{c}}^2-{\mathrm{b}}^2{\mathrm{x}}^2)+{\mathrm{b}}^2({\mathrm{c}}^2-{\mathrm{a}}^2{\mathrm{x}}^2)+2\mathrm{ab}\sqrt{({\mathrm{c}}^2-{\mathrm{b}}^2{\mathrm{x}}^2)({\mathrm{c}}^2-{\mathrm{a}}^2{\mathrm{x}}^2)}={\mathrm{c}}^4$$$${4\mathrm{a}}^4{\mathrm{b}}^4{\mathrm{x}}^4={4\mathrm{a}}^2{\mathrm{b}}^2({\mathrm{c}}^4-{\mathrm{c}}^4{\mathrm{x}}^2+{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{x}}^4)$$$${\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{x}}^4={\mathrm{c}}^4-{\mathrm{c}}^4{\mathrm{x}}^2+{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{x}}^4$$$$\mathrm{x}=\pm 1$$$$\mathrm{there}\mathrm{are}3\mathrm{solutions}\mathrm{x}=0,1,-1$$$$$$

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