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Question-112466




Question Number 112466 by Aina Samuel Temidayo last updated on 08/Sep/20
Answered by 1549442205PVT last updated on 08/Sep/20
∣cosx∣≥1−sin^2 x⇔∣cosx∣≥cos^2 x=∣cosx∣^2   ⇔∣cosx∣≤1.The last inequality is   always true ∀x.Hence,∣cosx∣≥1−sin^2 x  is true (q.e.d)
$$\mid\mathrm{cosx}\mid\geqslant\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}\Leftrightarrow\mid\mathrm{cosx}\mid\geqslant\mathrm{cos}^{\mathrm{2}} \mathrm{x}=\mid\mathrm{cosx}\mid^{\mathrm{2}} \\ $$$$\Leftrightarrow\mid\mathrm{cosx}\mid\leqslant\mathrm{1}.\mathrm{The}\:\mathrm{last}\:\mathrm{inequality}\:\mathrm{is}\: \\ $$$$\mathrm{always}\:\mathrm{true}\:\forall\mathrm{x}.\mathrm{Hence},\mid\mathrm{cosx}\mid\geqslant\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{is}\:\mathrm{true}\:\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$
Commented by Aina Samuel Temidayo last updated on 08/Sep/20
Ok. Thanks.
$$\mathrm{Ok}.\:\mathrm{Thanks}. \\ $$

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