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Category: Trigonometry

nice-calculus-if-a-b-c-0-and-acos-2-x-bsin-2-x-c-then-prove-that-a-cos-2-x-b-sin-2-x-c-

Question Number 133443 by mnjuly1970 last updated on 22/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……{nice}\:\:\:\:\:\:{calculus}……. \\ $$$$\:\:{if}\:\:{a},{b},{c}\:\geqslant\mathrm{0} \\ $$$$\:\:\:\:{and}\:::\:\:\:\:{acos}^{\mathrm{2}} \left({x}\right)+{bsin}^{\mathrm{2}} \left({x}\right)\leqslant{c} \\ $$$$\:\:\:\:{then}\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\:\sqrt{{a}}\:{cos}^{\mathrm{2}} \left({x}\right)+\sqrt{{b}}\:{sin}^{\mathrm{2}} \left({x}\right)\leqslant\sqrt{{c}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…………. \\…

If-sin-23pi-24-2-p-q-1-4-r-then-the-value-of-p-2-q-2-r-2-is-equal-to-

Question Number 133142 by bemath last updated on 19/Feb/21 $$\mathrm{If}\:\mathrm{sin}\:\left(\frac{\mathrm{23}\pi}{\mathrm{24}}\right)\:=\:\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{p}}−\sqrt{\mathrm{q}}−\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{r}}}} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} \right)\:\mathrm{is} \\ $$$$\mathrm{equal}\:\mathrm{to}\: \\ $$ Answered by Ñï= last updated on…

Question-67495

Question Number 67495 by TawaTawa last updated on 28/Aug/19 Commented by Prithwish sen last updated on 28/Aug/19 $$\mathrm{sin}\left(\frac{\mathrm{5}}{\mathrm{2}}\pi+\mathrm{x}\right)=\mathrm{Cos}\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{9}}\right)\Rightarrow\mathrm{Cosx}=\mathrm{Cos}\left(\mathrm{x}+\frac{\mathrm{7}}{\mathrm{18}}\right) \\ $$$$\mathrm{x}=\mathrm{2n}\pi\pm\left(\mathrm{x}+\frac{\mathrm{7}}{\mathrm{18}}\right)\:\:\:\mathrm{n}\in\:\mathbb{Z} \\ $$ Commented by TawaTawa…

prove-Cos-2pi-7-Cos-4pi-7-Cos-8pi-7-1-2-

Question Number 67464 by lalitchand last updated on 27/Aug/19 $$\mathrm{prove}\:\:\:\mathrm{Cos}\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{Cos}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)+\mathrm{Cos}\left(\frac{\mathrm{8}\pi}{\mathrm{7}}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by mind is power last updated on 27/Aug/19 $${Z}^{\mathrm{7}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({z}−\mathrm{1}\right)\left(\mathrm{1}+{z}+{z}^{\mathrm{2}}…