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Category: Trigonometry

Question-134801

Question Number 134801 by faysal last updated on 07/Mar/21 Answered by EDWIN88 last updated on 07/Mar/21 $$\mathrm{13}\theta\:=\:\pi\:\Rightarrow\mathrm{cos}\:\mathrm{13}\theta\:=\:−\mathrm{1} \\ $$$$\mathrm{let}\::\:\mathrm{z}\:=\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{4}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{2z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{4}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{4z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{4}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{4}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{8z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{8}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta…

Find-the-solution-2-3-cos-2x-sin-2x-1-

Question Number 134705 by bramlexs22 last updated on 06/Mar/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{solution}\: \\ $$$$\:\:\:\:\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\:\mathrm{cos}\:\mathrm{2x}\:+\:\mathrm{sin}\:\mathrm{2x}\:=\:\mathrm{1} \\ $$ Answered by john_santu last updated on 06/Mar/21 $$\:{We}\:{can}\:{setting}\:\rightarrow\begin{cases}{{X}=\mathrm{cos}\:\mathrm{2}{x}}\\{{Y}=\mathrm{sin}\:\mathrm{2}{x}}\end{cases} \\ $$$${so}\:{the}\:{equation}\:{becomes}\:\begin{cases}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){X}+{Y}\:=\:\mathrm{1}}\\{{X}^{\mathrm{2}} +{Y}^{\:\mathrm{2}}…

Prove-that-ratio-of-a-regular-pentagon-diagonal-to-its-side-is-5-1-2-

Question Number 3584 by prakash jain last updated on 16/Dec/15 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{pentagon}\:\mathrm{diagonal} \\ $$$$\mathrm{to}\:\mathrm{its}\:\mathrm{side}\:\mathrm{is}\:\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}. \\ $$ Commented by Yozzii last updated on 15/Dec/15 $${Its}\:{side}\:{is}\:{one}\:{of}\:{the}\:\mathrm{5}\:\:{edges}? \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}\:\:\:\:\:\:\:\:{b}…

Given-sin-A-sin-B-0-7-cos-A-cos-B-0-8-find-the-value-of-A-and-B-

Question Number 134581 by EDWIN88 last updated on 05/Mar/21 $$\mathrm{Given}\:\begin{cases}{\mathrm{sin}\:\mathrm{A}+\mathrm{sin}\:\mathrm{B}=\mathrm{0}.\mathrm{7}}\\{\mathrm{cos}\:\mathrm{A}+\mathrm{cos}\:\mathrm{B}=\mathrm{0}.\mathrm{8}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}. \\ $$ Answered by liberty last updated on 05/Mar/21 $$\Leftrightarrow\:\mathrm{2}+\mathrm{2cos}\:\left(\mathrm{A}−\mathrm{B}\right)=\:\frac{\mathrm{49}+\mathrm{64}}{\mathrm{100}} \\ $$$$\Leftrightarrow\mathrm{2cos}\:\left(\mathrm{A}−\mathrm{B}\right)\:=\:−\frac{\mathrm{87}}{\mathrm{100}} \\…

Question-69034

Question Number 69034 by ahmadshah last updated on 18/Sep/19 Answered by $@ty@m123 last updated on 18/Sep/19 $$\mathrm{4cos}\:{x}−\mathrm{3sec}\:{x}−\mathrm{2tan}\:{x} \\ $$$$\mathrm{4cos}\:{x}−\frac{\mathrm{3}}{\mathrm{cos}\:{x}}−\frac{\mathrm{2sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$$=\frac{\mathrm{4cos}\:^{\mathrm{2}} {x}−\mathrm{3}−\mathrm{2sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$$=\frac{\mathrm{4cos}\:^{\mathrm{3}} {x}−\mathrm{3cos}\:{x}−\mathrm{2sin}\:{x}\mathrm{cos}\:{x}}{\mathrm{cos}^{\mathrm{2}}…

if-the-range-of-f-x-y-sec-1-x-1-x-sec-1-y-1-y-xy-lt-0-is-a-b-and-a-b-equals-pi-10-then-is-eqal-to-

Question Number 68991 by pranay02 last updated on 17/Sep/19 $${if}\:{the}\:{range}\:{of}\:{f}\left({x},\:{y}\right)\:=\:{sec}^{−\mathrm{1}} \left({x}+\frac{\mathrm{1}}{{x}}\right)+{sec}^{−\mathrm{1}} \left({y}+\frac{\mathrm{1}}{{y}}\right),\:{xy}<\mathrm{0}\:{is}\:\left({a},\:{b}\right)\:{and}\:\left({a}+{b}\right)\:{equals}\:\frac{\lambda\pi}{\mathrm{10}},\:{then}\:\lambda\:{is}\:{eqal}\:{to}\: \\ $$ Answered by MJS last updated on 18/Sep/19 $${g}\left({t}\right)=\mathrm{sec}^{−\mathrm{1}} \:\left({t}+\frac{\mathrm{1}}{{t}}\right)\:=\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\right)…