Question Number 135205 by victoras last updated on 11/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 135206 by victoras last updated on 11/Mar/21 Answered by Olaf last updated on 11/Mar/21 $$\mathrm{81}^{\mathrm{sin}^{\mathrm{2}} {x}} +\mathrm{81}^{\mathrm{cos}^{\mathrm{2}} {x}} \:=\:\mathrm{30} \\ $$$$\mathrm{81}^{\mathrm{sin}^{\mathrm{2}} {x}} +\frac{\mathrm{81}}{\mathrm{81}^{\mathrm{sin}^{\mathrm{2}}…
Question Number 69641 by Henri Boucatchou last updated on 26/Sep/19 $$\boldsymbol{{Solve}}\:\:\boldsymbol{{arctg}}\left(\mathrm{1}−\boldsymbol{{x}}\right)\:+\:\frac{\mathrm{1}}{\boldsymbol{{arcctg}}\left(\mathrm{1}+\boldsymbol{{x}}\right)}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 135111 by faysal last updated on 10/Mar/21 Answered by Ñï= last updated on 10/Mar/21 $$\mathrm{2cos}\:\theta=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}} \\ $$$$\mathrm{4cos}\:^{\mathrm{2}} \theta=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \theta−\mathrm{1}\right)=\mathrm{2cos}\:\mathrm{2}\theta=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{4cos}\:^{\mathrm{2}}…
Question Number 135110 by faysal last updated on 10/Mar/21 Commented by Ñï= last updated on 10/Mar/21 $${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…