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

If A=sin^(28) θ+cos^(36) θ then Ans: 0<A≤1

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

Commented byMJS last updated on 05/Jul/19

0≤cos^(36) θ ≤1 even sin^2 θ +cos^2 θ doesn′t exceed 1 ⇒ 0<sin^(2m) θ +cos^(2n) θ ≤1 with m>1∧n>1

$$\mathrm{0}\leqslant\mathrm{cos}^{\mathrm{36}} \:\theta\:\leqslant\mathrm{1} \\ $$ $$\mathrm{even}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:+\mathrm{cos}^{\mathrm{2}} \:\theta\:\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exceed}\:\mathrm{1} \\ $$ $$\Rightarrow\:\mathrm{0}<\mathrm{sin}^{\mathrm{2}{m}} \:\theta\:+\mathrm{cos}^{\mathrm{2}{n}} \:\theta\:\leqslant\mathrm{1}\:\mathrm{with}\:{m}>\mathrm{1}\wedge{n}>\mathrm{1} \\ $$

Commented byPrithwish sen last updated on 05/Jul/19

−1≤ sinθ≤ 1 and −1≤ cosθ ≤ 1 or, 0≤ sin^(28) θ ≤1 and 0≤ cos^(36) θ ≤ 1 ∵ sin^(28) θ ≠ cos^(36) θ ∀ θ ∈ [−∞,∞] ∴ 0<sin^(28) θ + cos^(36) θ≤ 1 please check.

$$\:−\mathrm{1}\leqslant\:\mathrm{sin}\theta\leqslant\:\mathrm{1}\:\:\:\:\mathrm{and}\:\:−\mathrm{1}\leqslant\:\mathrm{cos}\theta\:\leqslant\:\mathrm{1} \\ $$ $$\:\mathrm{or},\:\mathrm{0}\leqslant\:\mathrm{sin}^{\mathrm{28}} \theta\:\leqslant\mathrm{1}\:\:\:\mathrm{and}\:\mathrm{0}\leqslant\:\mathrm{cos}^{\mathrm{36}} \theta\:\leqslant\:\mathrm{1} \\ $$ $$\because\:\mathrm{sin}^{\mathrm{28}} \theta\:\neq\:\mathrm{cos}^{\mathrm{36}} \theta\:\:\forall\:\theta\:\in\:\left[−\infty,\infty\right] \\ $$ $$\therefore\:\mathrm{0}<\mathrm{sin}^{\mathrm{28}} \theta\:+\:\mathrm{cos}^{\mathrm{36}} \theta\leqslant\:\mathrm{1} \\ $$ $$\mathrm{please}\:\mathrm{check}. \\ $$

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

tnq sir

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

Commented byPrithwish sen last updated on 05/Jul/19

yes sir .I correct it.Is it now ok ?

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

Commented byPrithwish sen last updated on 05/Jul/19

Sir I think ∵ sin^(28) θ and cos^(36) θ are always positive ∴ sin^(28) θ + cos^(36) θ cannot be zero as sin^(28) θ +cos^(36) θ =0 ⇒ sin^(28) θ = cos^(36) θ =0 which is impossible ∴ sin^(28) θ + cos^(36) θ > 0

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

Commented byMJS last updated on 05/Jul/19

you′re right but above there′s still a typo in your first comment: it must be 0<...≤1

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

Commented byPrithwish sen last updated on 05/Jul/19

Thankyou very much sir.

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

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