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Question Number 63466 by vajpaithegrate@gmail.com last updated on 04/Jul/19

$$\mathrm{If}\mathrm{A}={\sin }^{28}\theta +{\cos }^{36}\theta \mathrm{then}$$ $$\mathrm{Ans}:0<\mathrm{A}⩽1$$

Commented byMJS last updated on 05/Jul/19

$$0⩽{\cos }^{36}\theta ⩽1$$ $$\mathrm{even}{\sin }^2\theta +{\cos }^2\theta {\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exceed}1$$ $$\Rightarrow 0<{\sin }^{2m}\theta +{\cos }^{2n}\theta ⩽1\mathrm{with}m>1\wedge n>1$$

Commented byPrithwish sen last updated on 05/Jul/19

$$-1⩽\sin \theta ⩽1\mathrm{and}-1⩽\cos \theta ⩽1$$ $$\mathrm{or},0⩽{\sin }^{28}\theta ⩽1\mathrm{and}0⩽{\cos }^{36}\theta ⩽1$$ $$\because {\sin }^{28}\theta \ne {\cos }^{36}\theta \mathrm{\forall }\theta \in [-\mathrm{\infty },\mathrm{\infty }]$$ $$\therefore 0<{\sin }^{28}\theta +{\cos }^{36}\theta ⩽1$$ $$\mathrm{please}\mathrm{check}.$$

Commented byvajpaithegrate@gmail.com last updated on 05/Jul/19

$$\mathrm{tnq}\mathrm{sir}$$

Commented byPrithwish sen last updated on 05/Jul/19

$$\mathrm{yes}\mathrm{sir}.\mathrm{I}\mathrm{correct}\mathrm{it}.\mathrm{Is}\mathrm{it}\mathrm{now}\mathrm{ok}?$$

Commented byPrithwish sen last updated on 05/Jul/19

$$\mathrm{Sir}\mathrm{I}\mathrm{think}$$ $$\because {\sin }^{28}\theta \mathrm{and}{\cos }^{36}\theta \mathrm{are}\mathrm{always}\mathrm{positive}$$ $$\therefore {\sin }^{28}\theta +{\cos }^{36}\theta \mathrm{cannot}\mathrm{be}\mathrm{zero}$$ $$\mathrm{as}{\sin }^{28}\theta +{\cos }^{36}\theta =0\Rightarrow {\sin }^{28}\theta ={\cos }^{36}\theta =0$$ $$\mathrm{which}\mathrm{is}\mathrm{impossible}$$ $$\therefore {\sin }^{28}\theta +{\cos }^{36}\theta >0$$

Commented byMJS last updated on 05/Jul/19

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{right}$$ $$\mathrm{but}\mathrm{above}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{still}\mathrm{a}\mathrm{typo}\mathrm{in}\mathrm{your}\mathrm{first}$$ $$\mathrm{comment}:\mathrm{it}\mathrm{must}\mathrm{be}0<...⩽1$$

Commented byPrithwish sen last updated on 05/Jul/19

$$\mathrm{Thankyou}\mathrm{very}\mathrm{much}\mathrm{sir}.$$

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