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Category: Trigonometry

bemath-1-Given-x-sin-sin-y-cos-cos-maximum-value-of-x-2-y-2-when-2-find-solution-set-the-equation-sin-4-x-sin-4-x-pi-4-1-4-where-x-0-2pi-

Question Number 106958 by bemath last updated on 08/Aug/20 $$\:\:\:@{bemath}@ \\ $$$$\left(\mathrm{1}\right)\:{Given}\:\begin{cases}{{x}=\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta}\\{{y}=\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta}\end{cases} \\ $$$${maximum}\:{value}\:{of}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:{when}\:\alpha=… \\ $$$$\left(\mathrm{2}\right)\:{find}\:{solution}\:{set}\:{the}\:{equation} \\ $$$$\mathrm{sin}\:^{\mathrm{4}} {x}\:+\:\mathrm{sin}\:^{\mathrm{4}} \left({x}+\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\:{where}\:{x}\:\in\:\left[\mathrm{0},\mathrm{2}\pi\right]\: \\ $$ Answered…

bemath-Given-2cos-x-7cos-y-5-2sin-x-7sin-y-6-cos-x-y-

Question Number 106950 by bemath last updated on 08/Aug/20 $$\:\:\:\:\:\:\:\:\:\:^{@{bemath}@} \\ $$$${Given}\:\begin{cases}{\mathrm{2cos}\:{x}+\mathrm{7cos}\:{y}\:=\mathrm{5}}\\{\mathrm{2sin}\:{x}+\mathrm{7sin}\:{y}\:=\:\mathrm{6}}\end{cases} \\ $$$$\mathrm{cos}\:\left({x}−{y}\right)\:=?\: \\ $$ Answered by john santu last updated on 08/Aug/20 $$\:\:\:\:\:\:\:\:\:\blacklozenge\mathrm{JS}\lozenge…

If-i-1-n-sin-i-n-then-cos-1-cos-2-cos-3-cos-n-i-got-0-as-answer-please-who-can-correct-

Question Number 106944 by Rio Michael last updated on 08/Aug/20 $$\mathrm{If}\:\underset{{i}=\:\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{sin}\:\left(\theta_{{i}} \right)\:=\:{n}\:\mathrm{then}\:\:\mathrm{cos}\:\left(\theta_{\mathrm{1}} \right)\:+\:\mathrm{cos}\:\left(\theta_{\mathrm{2}} \right)\:+\:\mathrm{cos}\:\left(\theta_{\mathrm{3}} \right)\:+\:…\:+\:\mathrm{cos}\:\left(\theta_{{n}} \right)\:=\:? \\ $$$$\mathrm{i}\:\mathrm{got}\:\mathrm{0}\:\mathrm{as}\:\mathrm{answer}.\:\mathrm{please}\:\mathrm{who}\:\mathrm{can}\:\mathrm{correct}? \\ $$ Commented by mr…

show-that-a-1-cos2A-1-cos2A-tan-2-A-b-sin2A-1-cos2A-tanA-

Question Number 41394 by Rio Michael last updated on 07/Aug/18 $$\left.{show}\:{that}\:{a}\right)\:\frac{\mathrm{1}−{cos}\mathrm{2}{A}}{\mathrm{1}+{cos}\mathrm{2}{A}}\:\equiv\:{tan}^{\mathrm{2}} {A} \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}\right)\:\frac{{sin}\mathrm{2}{A}}{\mathrm{1}+{cos}\mathrm{2}{A}}\equiv\:{tanA}. \\ $$ Commented by mondodotto@gmail.com last updated on 07/Aug/18 $$\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{part}}\:\boldsymbol{\mathrm{B}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{should}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{proved}} \\…

An-electric-pole-PN-is-such-that-PN-12cm-where-N-is-the-top-of-the-pole-and-P-the-base-At-a-given-moment-of-the-day-the-shadow-of-the-pole-PN-PN-find-a-the-length-NN-b-the-bearing-of-P-from-N

Question Number 41290 by Rio Michael last updated on 04/Aug/18 $${An}\:{electric}\:{pole}\:{PN}\:{is}\:{such}\:{that}\:{PN}=\mathrm{12}{cm}\:{where}\:{N}\:{is}\:{the}\:{top}\:{of}\:{the}\:{pole}\:{and}\:{P}\:{the}\:{base} \\ $$$$.{At}\:{a}\:{given}\:{moment}\:{of}\:{the}\:{day}\:{the}\:\boldsymbol{{shadow}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{pole}}\:{PN}'\:=\:{PN}.\:{find}\: \\ $$$$\left.{a}\right)\:{the}\:{length}\:{NN}' \\ $$$$\left.{b}\right)\:{the}\:{bearing}\:{of}\:{P}\:{from}\:{N}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…

Question-172189

Question Number 172189 by Mikenice last updated on 23/Jun/22 Answered by Mathspace last updated on 24/Jun/22 $${let}\:{u}_{{n}} =\prod_{{k}=\mathrm{1}} ^{{n}} \left(\varphi\left({a}+\frac{{kb}}{{n}}\right)\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow+\infty} {ln}\left({u}_{{n}} \right) \\…

Given-cos-x-cos-y-3-2-cos-x-y-where-x-y-0-2pi-find-x-amp-y-

Question Number 106579 by john santu last updated on 06/Aug/20 $$\mathcal{G}\mathrm{iven}\:\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:\mathrm{y}=\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{cos}\:\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{where}\:\mathrm{x},\mathrm{y}\:\in\:\left[\mathrm{0},\mathrm{2}\pi\:\right].\:\mathrm{find}\:\mathrm{x}\:\&\:\mathrm{y}\: \\ $$ Answered by bobhans last updated on 06/Aug/20 $$\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:\mathrm{y}\:=\:\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{cos}\:\mathrm{xcos}\:\mathrm{y}−\mathrm{sin}\:\mathrm{xsin}\:\mathrm{y} \\ $$$$\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:\mathrm{y}−\mathrm{cos}\:\mathrm{xcos}\:\mathrm{y}+\mathrm{sin}\:\mathrm{xsin}\:\mathrm{y}=\frac{\mathrm{3}}{\mathrm{2}}…