Question Number 135110 by faysal last updated on 10/Mar/21 Commented by Ñï= last updated on 10/Mar/21 $${I}\:{try}\:\mathrm{2}{csc}\:\mathrm{2}\theta=\mathrm{tan}\:\theta+\mathrm{cot}\:\theta.{It}'{s}\:{too}\:{complicate}. \\ $$$${Maybe}\:{there}\:{have}\:{easy}\:{way}. \\ $$ Terms of Service Privacy…
Question Number 135065 by liberty last updated on 10/Mar/21 $$\mathrm{cosec}\:^{\mathrm{2}} \mathrm{68}°\:+\mathrm{sec}\:^{\mathrm{2}} \mathrm{56}°\:−\mathrm{cot}\:^{\mathrm{2}} \mathrm{34}°\:−\mathrm{tan}\:^{\mathrm{2}} \mathrm{22}°\:=? \\ $$ Commented by som(math1967) last updated on 10/Mar/21 $$\mathrm{2} \\…
Question Number 69500 by TawaTawa last updated on 24/Sep/19 Answered by MJS last updated on 24/Sep/19 $$\mathrm{the}\:\mathrm{shape}\:\mathrm{of}\:\mathrm{the}\:\mathrm{graph}\:\mathrm{looks}\:\mathrm{like}\:\mathrm{a}\:\mathrm{modified} \\ $$$${f}\left({x}\right)={x}^{\mathrm{4}} −{x}^{\mathrm{2}} \\ $$$${f}\left(\mathrm{0}\right)\approx−\mathrm{7}\:\Rightarrow\:{f}\left({x}\right)\approx{x}^{\mathrm{4}} −{x}^{\mathrm{2}} −\mathrm{7} \\…
Question Number 135019 by faysal last updated on 09/Mar/21 Answered by Dwaipayan Shikari last updated on 09/Mar/21 $${sin}\mathrm{6}°{sin}\mathrm{42}°{sin}\mathrm{66}°{sin}\mathrm{78}° \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left({cos}\mathrm{60}°−{cos}\mathrm{72}°\right)\left({cos}\mathrm{36}°−{cos}\mathrm{120}°\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{4}}\right)\left(\frac{\sqrt{\mathrm{5}}+\mathrm{3}}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{16}}=\Lambda \\ $$$${cos}\mathrm{6}°{cos}\mathrm{42}°{cos}\mathrm{66}°{cos}\mathrm{78}° \\…
Question Number 134902 by faysal last updated on 08/Mar/21 Commented by bobhans last updated on 08/Mar/21 $$\mathrm{use}\:\mathrm{that}\:\left(\mathrm{1}+\mathrm{sec}\:\mathrm{2x}\right)\mathrm{cot}\:\mathrm{2x}\:=\:\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2x}}{\mathrm{sin}\:\mathrm{2x}}\:=\mathrm{cot}\:\mathrm{x} \\ $$ Answered by bobhans last updated on…
Question Number 134872 by bemath last updated on 08/Mar/21 $$−−−−−−−−−−−−− \\ $$Trygonometri What is the value of cosA if cosA+3sinA=2? Answered by EDWIN88 last updated…
Question Number 134801 by faysal last updated on 07/Mar/21 Answered by EDWIN88 last updated on 07/Mar/21 $$\mathrm{13}\theta\:=\:\pi\:\Rightarrow\mathrm{cos}\:\mathrm{13}\theta\:=\:−\mathrm{1} \\ $$$$\mathrm{let}\::\:\mathrm{z}\:=\:\mathrm{cos}\:\theta\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{4}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{2z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{cos}\:\mathrm{2}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{4}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{4z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{4}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{4}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta \\ $$$$\mathrm{8z}\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\:\mathrm{8}\theta\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:\mathrm{5}\theta\:\mathrm{cos}\:\mathrm{6}\theta…
Question Number 134788 by bramlexs22 last updated on 07/Mar/21 $$\mathrm{Trigonometry} \\ $$If tan^2(α) and tan^2(β) are the roots of the equation 8x^2−26x+15=0, then what is…
Question Number 134705 by bramlexs22 last updated on 06/Mar/21 $$\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}}…
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