Question Number 74130 by mathocean1 last updated on 19/Nov/19 $$\left.\mathrm{h}\left.\mathrm{e}\left.\mathrm{l}\left.\mathrm{l}\left.\mathrm{o}\right]\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{in}\:\right]−\Pi;\Pi\right]×\right]−\Pi;\Pi\right]\:\mathrm{please} \\ $$$$\begin{cases}{\mathrm{x}−\mathrm{y}=\frac{\Pi}{\mathrm{6}}}\\{\mathrm{cosx}−\sqrt{\mathrm{3}}\mathrm{cosy}=−\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$ Answered by Tanmay chaudhury last updated on 19/Nov/19 $${cos}\left(\frac{\pi}{\mathrm{6}}+{y}\right)−\sqrt{\mathrm{3}}\:{cosy}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cosy}−\frac{\mathrm{1}}{\mathrm{2}}{siny}−\sqrt{\mathrm{3}}\:{cosy}=\frac{−\mathrm{1}}{\mathrm{2}}…
Question Number 139609 by bemath last updated on 29/Apr/21 $$\mathrm{If}\:\mathrm{tan}\:\mathrm{14}°\:=\:\mathrm{x}\:\mathrm{then}\:\mathrm{tan}\:\mathrm{18}°\:=? \\ $$ Answered by qaz last updated on 29/Apr/21 $$\mathrm{18}°=\frac{\mathrm{18}×\mathrm{14}°}{\mathrm{14}}=\frac{\mathrm{9}}{\mathrm{7}}\mathrm{tan}^{−\mathrm{1}} {x} \\ $$$$\Rightarrow\mathrm{tan}\:\mathrm{18}°=\mathrm{tan}\:\left(\frac{\mathrm{9}}{\mathrm{7}}\mathrm{tan}^{−\mathrm{1}} {x}\right) \\…
Question Number 8503 by MNG last updated on 13/Oct/16 $${Q}.\:\mathrm{1}\:\:{sinA}+{sinB}+{sinC}\:=\:\mathrm{4}{cos}\frac{{A}}{\mathrm{2}}\:{cos} \\ $$$$\frac{{B}}{\mathrm{2}}\:{cos}\:\frac{{C}}{\mathrm{2}}\:. \\ $$$${Q}.\mathrm{2}\:\:{cosA}\:{cosB}\:−\:{cosC}\:=\:\mathrm{4}{cos}\frac{{A}}{\mathrm{2}}\:{cos}\frac{{B}}{\mathrm{2}} \\ $$$${cos}\frac{{C}}{\mathrm{2}}\:−\mathrm{1} \\ $$$$ \\ $$$${Q}.\mathrm{3}\:\:\frac{{sin}\mathrm{2}{A}+{sin}\mathrm{2}{B}+{sin}\mathrm{2}{C}}{{sinA}+{sinB}+{sinC}}\:=\mathrm{8}{sin}\:\frac{{A}}{\mathrm{2}} \\ $$$${sin}\frac{{B}}{\mathrm{2}}\:{sin}\frac{{C}}{\mathrm{2}} \\ $$$$ \\…
Question Number 8452 by tawakalitu last updated on 11/Oct/16 Commented by 123456 last updated on 12/Oct/16 $$\mathrm{sin}\:{x}+\mathrm{cos}\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}=\mathrm{sin}\:{x},{b}=\mathrm{cos}\:{x} \\ $$$${a}+{b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1}…
Question Number 139521 by mnjuly1970 last updated on 28/Apr/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:#\:\:\:{nice}\:…\:{calculus}# \\ $$$$\:\:\:\:\:\:\:\:{if}\:\:\:\:\left({pq}\right)^{{sin}^{\mathrm{2}} \left({x}\right)} +\left({pq}\right)^{{cos}^{\mathrm{2}} \left({x}\right)} ={p}+{q} \\ $$$$\:\:\:\:\:\:\:\:\:\:{then}\:\:\:\:{tan}\left({x}\right)=? \\ $$ Answered by qaz last updated…
Question Number 8402 by rhm last updated on 10/Oct/16 $${Q}.\:\left({cos}^{\mathrm{2}} \mathrm{66}^{\mathrm{0}} −{sin}^{\mathrm{2}} \mathrm{6}^{\mathrm{0}} \right)\left({cos}^{\mathrm{2}} \mathrm{48}^{\mathrm{0}} −{sin}^{\mathrm{2}} \mathrm{12}^{\mathrm{0}} \right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$ Answered by sandy_suhendra…
Question Number 8403 by rhm last updated on 10/Oct/16 $$\left({Q}.\mathrm{1}\right)\:{cos}\:\mathrm{4}{A}=\mathrm{1}−\mathrm{8}{cos}^{\mathrm{2}} {A}\:+\:\mathrm{8}{cos}^{\mathrm{2}} \:{A} \\ $$$$\left({Q}.\mathrm{2}\right)\:\:\frac{{sec}\mathrm{8}\:{A}\:−\mathrm{1}}{{sec}\mathrm{4}\:{A}\:−\mathrm{1}}\:=\:\frac{{tan}\:\mathrm{8}{A}}{{tan}\:\mathrm{2}{A}} \\ $$$$\left({Q}.\mathrm{3}\right)\:\:\:{tanA}+{tan}\left(\mathrm{60}^{\mathrm{0}} +{A}\right)+{tan}\left(\mathrm{120}^{\mathrm{0}} \right. \\ $$$$\left.+\mathrm{4}\right)\:=\:\mathrm{3}{tan}\:\mathrm{3}{A}\: \\ $$$$\left({Q}.\mathrm{4}\right)\:\:\:{sinA}\:{sin}\:\left(\mathrm{60}^{\mathrm{0}} −{A}\right)\:\:{sin}\left(\mathrm{60}^{\mathrm{0}} +{A}\right)\: \\…
Question Number 8401 by rhm last updated on 10/Oct/16 $${Q}\:.\mathrm{1}\:\:\:\mathrm{1}+{cos}^{\mathrm{2}} \mathrm{2}{A}=\mathrm{2}\left({cos}^{\mathrm{4}} \:{A}+{sin}^{\mathrm{4}} \:{A}\right) \\ $$$$\mathrm{2}.\:\:{sin}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} \left(\mathrm{120}^{\mathrm{0}} +{A}\right)+{sin}^{\mathrm{2}} \left(\mathrm{120}^{\mathrm{0}} −{A}\right) \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\…
Question Number 139409 by henderson last updated on 26/Apr/21 $$\mathrm{hi}\:! \\ $$$$\mathrm{prove}\:\mathrm{this}\::\: \\ $$$${cos}\:\frac{\pi}{\mathrm{10}}\:+\:{cos}\:\frac{\mathrm{4}\pi}{\mathrm{10}}\:+\:{cos}\:\frac{\mathrm{6}\pi}{\mathrm{10}}\:+\:{cos}\:\frac{\mathrm{9}\pi}{\mathrm{10}}\:=\:\mathrm{0}. \\ $$$$\left(\mathrm{by}\:\mathrm{the}\:\mathrm{easiest}\:\mathrm{possible}\:\mathrm{way}…!\right) \\ $$ Answered by mr W last updated on…
Question Number 8311 by lepan last updated on 07/Oct/16 $${Solve}\:{the}\:{equation}\:\mathrm{6}{cos}\mathrm{2}{a}−\mathrm{5}{sin}\mathrm{2}{a}=\mathrm{1}.\mathrm{8} \\ $$$${for}\mathrm{0}°\leqslant{a}\leqslant\mathrm{180}°. \\ $$ Commented by 123456 last updated on 07/Oct/16 $$\mathrm{6cos2}{a}−\mathrm{5sin2}{a}=\mathrm{1}.\mathrm{8} \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{2}{a}+\mathrm{sin}^{\mathrm{2}}…