Trigonometry Questions

Question Number 105764 by bobhans last updated on 31/Jul/20

$$\left(\mathrm{1}\right){If}\:\mathrm{cos}\:\left(\alpha+\beta\right)\:=\:\frac{\mathrm{4}}{\mathrm{5}}\:{and}\:\mathrm{sin}\:\left(\alpha−\beta\right)\:=\:\frac{\mathrm{5}}{\mathrm{13}} \\$$ $${where}\:\mathrm{0}\:<\:\alpha<\:\frac{\pi}{\mathrm{4}}.\:{Find}\:\mathrm{tan}\:\mathrm{2}\alpha\:. \\$$ $$\left(\mathrm{2}\right)\:\:\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right)? \\$$

Commented byDwaipayan Shikari last updated on 31/Jul/20

$$\left.\mathrm{2}\right)\frac{\mathrm{17}}{\mathrm{16}} \\$$

Answered by bemath last updated on 01/Aug/20

$$\left(\mathrm{1}\right)\mathrm{tan}\:\mathrm{2}\alpha\:=\:\mathrm{tan}\:\left(\left(\alpha+\beta\right)+\left(\alpha−\beta\right)\right) \\$$ $$\:\:=\:\frac{\mathrm{tan}\:\left(\alpha+\beta\right)+\mathrm{tan}\:\left(\alpha−\beta\right)}{\mathrm{1}−\mathrm{tan}\:\left(\alpha+\beta\right)\mathrm{tan}\:\left(\alpha−\beta\right)} \\$$ $$=\:\frac{\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{12}}}{\mathrm{1}−\frac{\mathrm{15}}{\mathrm{48}}}\:=\:\frac{\mathrm{36}+\mathrm{20}}{\mathrm{33}}\:=\:\frac{\mathrm{56}}{\mathrm{33}}\:.\multimap \\$$

Commented byPRITHWISH SEN 2 last updated on 31/Jul/20

$$\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\mathrm{sin}\:\left(\alpha+\beta\right)}{\mathrm{cos}\:\left(\alpha+\beta\right)}\:=\:\frac{\frac{\mathrm{3}}{\mathrm{5}}}{\frac{\mathrm{4}}{\mathrm{5}}}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\$$

Answered by john santu last updated on 31/Jul/20

$$\left(\mathrm{2}\right)\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)=\:\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\:{and}\:\mathrm{cos}\:\left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right)=\mathrm{cos}\:\left(\frac{\pi}{\mathrm{8}}\right) \\$$ $$\Leftrightarrow\:\mathrm{2}\left\{\:\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\right\}\:=\:\mathrm{2}\left\{\left(\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\right)^{\mathrm{2}} \right. \\$$ $$\left.−\:\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\right\} \\$$ $$\Leftrightarrow{let}\:{A}=\:\:\left[\:\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{cos}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\:\right]^{\mathrm{2}} =\:\left[\left(\mathrm{cos}\:\frac{\pi}{\mathrm{8}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)^{\mathrm{2}} −\mathrm{2}\right. \\$$ $$\left.\mathrm{cos}\:\frac{\pi}{\mathrm{8}}.\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\:\right]\:^{\mathrm{2}} \:=\: \\$$

Answered by som(math1967) last updated on 01/Aug/20

$$\left.\mathrm{2}\right)\mathrm{cos}^{\mathrm{4}} \frac{\mathrm{5}\pi}{\mathrm{8}}=\mathrm{cos}^{\mathrm{4}} \left(\pi−\frac{\mathrm{3}\pi}{\mathrm{8}}\right)=\mathrm{cos}^{\mathrm{4}} \frac{\mathrm{3}\pi}{\mathrm{8}} \\$$ $$\mathrm{cos}^{\mathrm{4}} \frac{\mathrm{7}\pi}{\mathrm{8}}=\mathrm{cos}^{\mathrm{4}} \left(\pi−\frac{\pi}{\mathrm{8}}\right)=\mathrm{cos}^{\mathrm{4}} \frac{\pi}{\mathrm{8}} \\$$ $$\mathrm{cos}^{\mathrm{4}} \frac{\pi}{\mathrm{8}}+\mathrm{cos}^{\mathrm{4}} \frac{\mathrm{3}\pi}{\mathrm{8}}+\mathrm{cos}^{\mathrm{4}} \frac{\mathrm{5}\pi}{\mathrm{8}}+\mathrm{cos}^{\mathrm{4}} \frac{\mathrm{7}\pi}{\mathrm{8}} \\$$ $$\mathrm{2}\left(\mathrm{cos}^{\mathrm{4}} \frac{\pi}{\mathrm{8}}+\mathrm{cos}^{\mathrm{4}} \frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\$$ $$\mathrm{2}\left(\mathrm{cos}^{\mathrm{4}} \frac{\pi}{\mathrm{8}}\:+\mathrm{sin}^{\mathrm{4}} \frac{\pi}{\mathrm{8}}\right)\:\bigstar \\$$ $$\left(\mathrm{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\:+\mathrm{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\right)^{\mathrm{2}} \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\mathrm{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\:−\mathrm{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\right)^{\mathrm{2}} \\$$ $$=\mathrm{1}^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} \frac{\pi}{\mathrm{4}} \\$$ $$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{ans} \\$$ $$\bigstar\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)=\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{2}} +\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} \\$$ $$\mathrm{cos}^{\mathrm{4}} \frac{\mathrm{3}\pi}{\mathrm{8}}=\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}}\:−\frac{\pi}{\mathrm{8}}\right)=\mathrm{sin}^{\mathrm{4}} \frac{\pi}{\mathrm{8}} \\$$

Commented by1549442205PVT last updated on 01/Aug/20

$$\boldsymbol{\mathrm{There}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{mistake}}\:\boldsymbol{\mathrm{ocurring}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{fourth}} \\$$ $$\boldsymbol{\mathrm{line}}\:\boldsymbol{\mathrm{up}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{below}}\:\boldsymbol{\mathrm{cos}}^{\mathrm{4}} \frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{8}}\neq\boldsymbol{\mathrm{sin}}^{\mathrm{4}} \frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{4}}. \\$$

Commented bybobhans last updated on 01/Aug/20

$${typo}\:{sir}\:\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\:{it}\:{should}\:{be}\:\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\$$ $$\\$$

Commented bysom(math1967) last updated on 01/Aug/20

$$\mathrm{yes}\:\mathrm{sir}\:\mathrm{I}\:\mathrm{fix}\:\mathrm{it} \\$$

Answered by mathmax by abdo last updated on 03/Aug/20

$$\left.\mathrm{2}\right)\mathrm{A}\:=\mathrm{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right)+\mathrm{cos}^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+\mathrm{cos}^{\mathrm{4}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right) \\$$ $$=\mathrm{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{cos}^{\mathrm{4}} \left(\pi−\frac{\pi}{\mathrm{8}}\right)+\mathrm{cos}^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+\mathrm{cos}^{\mathrm{4}} \left(\pi−\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\$$ $$=\mathrm{2cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)\:+\mathrm{2cos}^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\$$ $$=\mathrm{2}\left\{\mathrm{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{8}}\right)\right\}\:=\mathrm{2}\left\{\mathrm{cos}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{sin}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{8}}\right)\right\} \\$$ $$=\mathrm{2}\left\{\:\left(\mathrm{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)+\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\right)^{\mathrm{2}} −\mathrm{2cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\right\} \\$$ $$=\mathrm{2}\left\{\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}\right)\right)^{\mathrm{2}} \:=\mathrm{2}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \right\}\:=\mathrm{2}\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right\}=\mathrm{2}×\frac{\mathrm{3}}{\mathrm{4}}\:=\frac{\mathrm{3}}{\mathrm{2}}\right. \\$$ $$\Rightarrow\mathrm{A}\:=\frac{\mathrm{3}}{\mathrm{2}} \\$$