1-If-cos-4-5-and-sin-5-13-where-0-lt-lt-pi-4-Find-tan-2-2-cos-4-pi-8-cos-4-3pi-8-cos-4-5pi-8-cos-4-7pi-8-

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

(1)If cos (α+β) = (4/5) and sin (α−β) = (5/(13)) where 0 < α< (π/4). Find tan 2α . (2) cos^4 ((π/8))+cos^4 (((3π)/8))+cos^4 (((5π)/8))+cos^4 (((7π)/8))?

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

Commented by Dwaipayan 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

(1)tan 2α = tan ((α+β)+(α−β)) = ((tan (α+β)+tan (α−β))/(1−tan (α+β)tan (α−β))) = (((3/4)+(5/(12)))/(1−((15)/(48)))) = ((36+20)/(33)) = ((56)/(33)) .⊸

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

Commented by PRITHWISH SEN 2 last updated on 31/Jul/20

tan (α+β)=((sin (α+β))/(cos (α+β))) = ((3/5)/(4/5)) = (3/4)

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

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

(2)cos (((5π)/8))= cos (((3π)/8)) and cos (((7π)/8))=cos ((π/8)) ⇔ 2{ cos^4 ((π/8))+cos^4 (((3π)/8))} = 2{(cos^2 ((π/8))+cos^2 (((3π)/8)))^2 − 2cos^2 ((π/8))cos^2 (((3π)/8))} ⇔let A= [ cos^2 ((π/8))+cos^2 (((3π)/8)) ]^2 = [(cos (π/8)+cos ((3π)/8))^2 −2 cos (π/8).cos ((3π)/8) ]^2 =

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

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

2)cos^4 ((5π)/8)=cos^4 (π−((3π)/8))=cos^4 ((3π)/8) cos^4 ((7π)/8)=cos^4 (π−(π/8))=cos^4 (π/8) cos^4 (π/8)+cos^4 ((3π)/8)+cos^4 ((5π)/8)+cos^4 ((7π)/8) 2(cos^4 (π/8)+cos^4 ((3π)/8)) 2(cos^4 (π/8) +sin^4 (π/8)) ★ (cos^2 (π/8) +sin^2 (π/8))^2 +(cos^2 (π/8) −sin^2 (π/8))^2 =1^2 +cos^2 (π/4) =1+(1/2)=(3/2) ans ★2(a^2 +b^2 )=(a+b)^2 +(a−b)^2 cos^4 ((3π)/8)=cos^4 ((π/2) −(π/8))=sin^4 (π/8)

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

Commented by 1549442205PVT last updated on 01/Aug/20

There is a mistake ocurring at fourth line up to below cos^4 ((3𝛑)/8)≠sin^4 ((3𝛑)/4).

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

Commented by bobhans last updated on 01/Aug/20

typo sir cos^4 (((3π)/4)) it should be cos^4 (((3π)/8))

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

Commented by som(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

2)A =cos^4 ((π/8))+cos^2 (((7π)/8))+cos^4 (((3π)/8))+cos^4 (((5π)/8)) =cos^2 ((π/8))+cos^4 (π−(π/8))+cos^4 (((3π)/8))+cos^4 (π−((3π)/8)) =2cos^4 ((π/8)) +2cos^4 (((3π)/8)) =2{cos^4 ((π/8))+cos^4 ((π/2)−(π/8))} =2{cos^4 ((π/8))+sin^4 ((π/8))} =2{ (cos^2 ((π/8))+sin^2 ((π/8)))^2 −2cos^2 ((π/8))sin^2 ((π/8))} =2{1−2((1/2)sin((π/4)))^2 =2{1−(1/2)((1/( (√2))))^2 } =2{1−(1/4)}=2×(3/4) =(3/2) ⇒A =(3/2)

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

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