Question Number 38288 by ajfour last updated on 23/Jun/18 $$\mathrm{tan}\:\alpha−\mathrm{tan}\:\beta\:=\:\mathrm{2tan}\:\theta \\ $$$${a}\mathrm{sin}\:\alpha−{b}\mathrm{sin}\:\beta\:=\:{l}\mathrm{sin}\:\theta \\ $$$${express}\:\mathrm{sin}\:\alpha,\:\mathrm{sin}\:\beta\:\:{in}\:{terms}\:{of}\:\boldsymbol{\theta}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18 $${t}_{\mathrm{1}} ={tan}\frac{\alpha}{\mathrm{2}}\:\:\:{t}_{\mathrm{2}}…
Question Number 103823 by bobhans last updated on 17/Jul/20 $$\mathrm{tan}\:\left({x}\right)\:=\:\mathrm{4}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{cot}\:\left(\mathrm{2}{x}\right) \\ $$ Answered by Dwaipayan Shikari last updated on 17/Jul/20 $$\frac{{sinx}}{{cosx}}={cos}\mathrm{2}{x}\left(\mathrm{4}−\frac{\mathrm{1}}{{sin}\mathrm{2}{x}}\right) \\ $$$$\frac{{sinx}}{{cosx}}=\frac{\mathrm{2}{sin}\mathrm{4}{x}−{cos}\mathrm{2}{x}}{\mathrm{2}{sinxcosx}} \\ $$$$\mathrm{2}{sin}^{\mathrm{2}}…
Question Number 169295 by thfchristopher last updated on 28/Apr/22 $$\mathrm{Prove}\:\mathrm{without}\:\mathrm{using}\:\mathrm{Mathematical}\: \\ $$$$\mathrm{Induction}\:\mathrm{that}: \\ $$$$\mathrm{cos}^{{n}} {x}=\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{cos}\:\left({n}−\mathrm{2}{k}\right){x}\: \\ $$$$\mathrm{where}\:\:{x}\:\mathrm{is}\:\mathrm{any}\:\mathrm{real}\:\mathrm{number}\:\mathrm{and}\:{n}\:\mathrm{is}\:\mathrm{any} \\ $$$$\mathrm{positive}\:\mathrm{odd}\:\mathrm{integer}. \\…
Question Number 38190 by ajfour last updated on 22/Jun/18 Commented by ajfour last updated on 22/Jun/18 $$\:\:\:\:\:\:\:\boldsymbol{{What}}\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{time}}\:\boldsymbol{{in}}\:\boldsymbol{{the}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{outer}}\:\boldsymbol{{clock}}\:? \\ $$$$\:\:{Ignore}\:{the}\:{minute}\:{hand}\:{of}\:{the} \\ $$$${bigger}\:{clock}.\:{Then}\:{it}\:{will}\:{be}\:{a} \\ $$$${correct}\:{question},\:{i}\:{think}\:{now}.…
Question Number 38151 by ajfour last updated on 22/Jun/18 $${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} \\…