Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

Question Number 103155 by 9027201563 last updated on 13/Jul/20

If cos^(−1) x+cos^(−1) y+cos^(−1) z+cos^(−1) u=2π,  then  x^(1999) +y^(2000) +z^(2001) +u^(2002) =

$$\mathrm{If}\:\mathrm{cos}^{−\mathrm{1}} {x}+\mathrm{cos}^{−\mathrm{1}} {y}+\mathrm{cos}^{−\mathrm{1}} {z}+\mathrm{cos}^{−\mathrm{1}} {u}=\mathrm{2}\pi, \\ $$$$\mathrm{then}\:\:{x}^{\mathrm{1999}} +{y}^{\mathrm{2000}} +{z}^{\mathrm{2001}} +{u}^{\mathrm{2002}} = \\ $$

Commented by 9027201563 last updated on 13/Jul/20

solution needed urgently

$$\mathrm{solution}\:\mathrm{needed}\:\mathrm{urgently} \\ $$

Commented by prakash jain last updated on 13/Jul/20

I dont think you can get a unique  answer  assuming  cos^(−1) x=cos^(−1) y=π  cos^(−1) z=cos^(−1) u=0  x=−1,y=−1,z=1,u=1  sum=2  cos^(−1) x=cos^(−1) y=cos^(−1) z=cos^(−1) u=(π/2)  x=y=z=u=0  sum=0

$$\mathrm{I}\:\mathrm{dont}\:\mathrm{think}\:\mathrm{you}\:\mathrm{can}\:\mathrm{get}\:\mathrm{a}\:\mathrm{unique} \\ $$$$\mathrm{answer} \\ $$$$\mathrm{assuming} \\ $$$$\mathrm{cos}^{−\mathrm{1}} {x}=\mathrm{cos}^{−\mathrm{1}} {y}=\pi \\ $$$$\mathrm{cos}^{−\mathrm{1}} {z}=\mathrm{cos}^{−\mathrm{1}} {u}=\mathrm{0} \\ $$$${x}=−\mathrm{1},{y}=−\mathrm{1},{z}=\mathrm{1},{u}=\mathrm{1} \\ $$$${sum}=\mathrm{2} \\ $$$$\mathrm{cos}^{−\mathrm{1}} {x}=\mathrm{cos}^{−\mathrm{1}} {y}=\mathrm{cos}^{−\mathrm{1}} {z}=\mathrm{cos}^{−\mathrm{1}} {u}=\frac{\pi}{\mathrm{2}} \\ $$$${x}={y}={z}={u}=\mathrm{0} \\ $$$${sum}=\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com