Question Number 38151 by ajfour last updated on 22/Jun/18 $${a}\:>\:{b}\:>\:\mathrm{0} \\ $$$${a}^{\mathrm{2}} \mathrm{cos}\:\theta−{b}^{\mathrm{2}} \mathrm{sin}\:\theta=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$${Find}\:\theta\:{in}\:{terms}\:{of}\:{a},\:{b}. \\ $$$$\:\:\theta\:\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right)\:. \\ $$ Commented by behi83417@gmail.com…
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Question Number 103644 by dw last updated on 16/Jul/20 Commented by dw last updated on 16/Jul/20 $$\left[{Trigonometric}\:{Substit}.\right] \\ $$ Answered by 1549442205 last updated on…
Question Number 103643 by bemath last updated on 16/Jul/20 $${if}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\:=\:\frac{\mathrm{5}}{\mathrm{6}} \\ $$$${then}\:\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\:+\:\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\:?\: \\ $$ Answered by Dwaipayan Shikari last updated on 16/Jul/20 $$\left({sinx}+{cosx}\right)^{\mathrm{2}} −\mathrm{2}{sinxcosx}=\mathrm{1} \\…
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Question Number 103459 by bemath last updated on 15/Jul/20 $${what}\:{are}\:{the}\:{complex}\:{solution}\:\mathrm{tan} \\ $$$$\left({z}\right)\:=\:−\mathrm{2}\:? \\ $$ Commented by mr W last updated on 15/Jul/20 $${z}={k}\pi−\mathrm{tan}^{−\mathrm{1}} \mathrm{2} \\…
Question Number 37914 by gunawan last updated on 19/Jun/18 $$\mathrm{In}\:\bigtriangleup{ABC},\:\mathrm{if}\:\mathrm{sin}\:\mathrm{A}=\mathrm{sin}^{\mathrm{2}} {B}\: \\ $$$$\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{4}\:\mathrm{cos}\:\mathrm{2}{A}−\mathrm{4}\:\mathrm{cos}\:\mathrm{2}{B}=\mathrm{1}−\mathrm{cos}\:\mathrm{4}{B} \\ $$ Answered by $@ty@m last updated on 19/Jun/18 $${LHS}=…
Question Number 103314 by Ar Brandon last updated on 14/Jul/20 $$\mathrm{Linearise}\:\mathrm{cos}^{\mathrm{2}\left(\mathrm{p}−\mathrm{1}\right)} \left(\mathrm{t}\right) \\ $$ Answered by OlafThorendsen last updated on 14/Jul/20 $$\mathrm{cos}^{\mathrm{2}\left({p}−\mathrm{1}\right)} \left({t}\right)\:=\:\left[\frac{{e}^{{it}} +{e}^{−{it}} }{\mathrm{2}}\right]^{\mathrm{2}\left({p}−\mathrm{1}\right)}…
Question Number 37745 by kunal1234523 last updated on 17/Jun/18 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{sec}^{\mathrm{2}} \theta\:=\:\frac{\mathrm{4}{xy}}{\left({x}+{y}\right)^{\mathrm{2}} }\:\mathrm{is}\: \\ $$$$\mathrm{only}\:\mathrm{possible}\:\mathrm{when}\:{x}\:=\:{y} \\ $$ Answered by math1967 last updated on 17/Jun/18 $${x},{y}\:{real}\:\therefore\left({x}−{y}\right)^{\mathrm{2}} \geqslant\mathrm{0}…
Question Number 37733 by kunal1234523 last updated on 17/Jun/18 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{sin}\:\theta\:=\:{x}\:+\:\frac{\mathrm{1}}{{x}}\: \\ $$$$\mathrm{is}\:\mathrm{impossible}\:\mathrm{if}\:{x}\:\mathrm{be}\:\mathrm{real}. \\ $$ Answered by MrW3 last updated on 17/Jun/18 $${if}\:{x}>\mathrm{0}: \\ $$$$\mathrm{sin}\:\theta={x}+\frac{\mathrm{1}}{{x}}=\left(\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)^{\mathrm{2}} +\mathrm{2}\geqslant\mathrm{2}…