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) I f \cos (α + β) = \frac{4}{5} a n d \sin (α - β) = \frac{5}{13}

w h e r e 0 < α < \frac{π}{4} . F i n d \tan 2 α .

(2) \cos^{4} (\frac{π}{8}) + \cos^{4} (\frac{3 π}{8}) + \cos^{4} (\frac{5 π}{8}) + \cos^{4} (\frac{7 π}{8}) ?

Commented by Dwaipayan Shikari last updated on 31/Jul/20

2) \frac{17}{16}

Answered by bemath last updated on 01/Aug/20

(1) \tan 2 α = \tan ((α + β) + (α - β))

= \frac{\tan (α + β) + \tan (α - β)}{1 - \tan (α + β) \tan (α - β)}

= \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{15}{48}} = \frac{36 + 20}{33} = \frac{56}{33} . ⊸

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

\tan (α + β) = \frac{\sin (α + β)}{\cos (α + β)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}

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

(2) \cos (\frac{5 π}{8}) = \cos (\frac{3 π}{8}) a n d \cos (\frac{7 π}{8}) = \cos (\frac{π}{8})

\Leftrightarrow 2 {\cos^{4} (\frac{π}{8}) + \cos^{4} (\frac{3 π}{8})} = 2 {{(\cos^{2} (\frac{π}{8}) + \cos^{2} (\frac{3 π}{8}))}^{2}

- 2 \cos^{2} (\frac{π}{8}) \cos^{2} (\frac{3 π}{8})}

\Leftrightarrow l e t A = {[\cos^{2} (\frac{π}{8}) + \cos^{2} (\frac{3 π}{8})]}^{2} = [{(\cos \frac{π}{8} + \cos \frac{3 π}{8})}^{2} - 2

\cos \frac{π}{8} . \cos \frac{3 π}{8}]^{2} =

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

2) \cos^{4} \frac{5 π}{8} = \cos^{4} (π - \frac{3 π}{8}) = \cos^{4} \frac{3 π}{8}

\cos^{4} \frac{7 π}{8} = \cos^{4} (π - \frac{π}{8}) = \cos^{4} \frac{π}{8}

\cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8} + \cos^{4} \frac{5 π}{8} + \cos^{4} \frac{7 π}{8}

2 (\cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8})

2 (\cos^{4} \frac{π}{8} + \sin^{4} \frac{π}{8}) ★

{(\cos^{2} \frac{π}{8} + \sin^{2} \frac{π}{8})}^{2}

+ {(\cos^{2} \frac{π}{8} - \sin^{2} \frac{π}{8})}^{2}

= 1^{2} + \cos^{2} \frac{π}{4}

= 1 + \frac{1}{2} = \frac{3}{2} ans

★ 2 (a^{2} + b^{2}) = {(a + b)}^{2} + {(a - b)}^{2}

\cos^{4} \frac{3 π}{8} = \cos^{4} (\frac{π}{2} - \frac{π}{8}) = \sin^{4} \frac{π}{8}

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

There is a mistake ocurring at fourth

line up to below \cos^{4} \frac{3 π}{8} \neq \sin^{4} \frac{3 π}{4} .

Commented by bobhans last updated on 01/Aug/20

t y p o s i r \cos^{4} (\frac{3 π}{4}) i t s h o u l d b e \cos^{4} (\frac{3 π}{8})

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

yes sir I fix it

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

2) A = \cos^{4} (\frac{π}{8}) + \cos^{2} (\frac{7 π}{8}) + \cos^{4} (\frac{3 π}{8}) + \cos^{4} (\frac{5 π}{8})

= \cos^{2} (\frac{π}{8}) + \cos^{4} (π - \frac{π}{8}) + \cos^{4} (\frac{3 π}{8}) + \cos^{4} (π - \frac{3 π}{8})

= {2 \cos}^{4} (\frac{π}{8}) + {2 \cos}^{4} (\frac{3 π}{8})

= 2 {\cos^{4} (\frac{π}{8}) + \cos^{4} (\frac{π}{2} - \frac{π}{8})} = 2 {\cos^{4} (\frac{π}{8}) + \sin^{4} (\frac{π}{8})}

= 2 {{(\cos^{2} (\frac{π}{8}) + \sin^{2} (\frac{π}{8}))}^{2} - {2 \cos}^{2} (\frac{π}{8}) \sin^{2} (\frac{π}{8})}

= 2 {1 - 2 {(\frac{1}{2} \sin (\frac{π}{4}))}^{2} = 2 {1 - \frac{1}{2} {(\frac{1}{\sqrt{2}})}^{2}} = 2 {1 - \frac{1}{4}} = 2 \times \frac{3}{4} = \frac{3}{2}

\Rightarrow A = \frac{3}{2}