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Question Number 45829 by jashim last updated on 17/Oct/18
Given A= sin^2 θ + cos^4 θ, then for all  real θ
$$\mathrm{Given}\:{A}=\:\mathrm{sin}^{\mathrm{2}} \theta\:+\:\mathrm{cos}^{\mathrm{4}} \theta,\:\mathrm{then}\:\mathrm{for}\:\mathrm{all} \\ $$$$\mathrm{real}\:\theta \\ $$
Commented by MJS last updated on 17/Oct/18
this isn′t a clear question.  another possible answer is  A=sin^2  θ +cos^4  θ ∧ θ∈R ⇒ A≥0
$$\mathrm{this}\:\mathrm{isn}'\mathrm{t}\:\mathrm{a}\:\mathrm{clear}\:\mathrm{question}. \\ $$$$\mathrm{another}\:\mathrm{possible}\:\mathrm{answer}\:\mathrm{is} \\ $$$${A}=\mathrm{sin}^{\mathrm{2}} \:\theta\:+\mathrm{cos}^{\mathrm{4}} \:\theta\:\wedge\:\theta\in\mathbb{R}\:\Rightarrow\:{A}\geqslant\mathrm{0} \\ $$
Commented by MJS last updated on 17/Oct/18
A=(7/8)+(1/8)cos 4θ
$${A}=\frac{\mathrm{7}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{4}\theta \\ $$

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