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Question Number 114577 by Khalmohmmad last updated on 19/Sep/20

x=(π/(30)) ((cos14x×cos6x)/(cos4x))=?

$$\mathrm{x}=\frac{\pi}{\mathrm{30}} \\ $$$$\frac{\mathrm{cos14x}×\mathrm{cos6x}}{\mathrm{cos4x}}=? \\ $$

Commented by bemath last updated on 19/Sep/20

((cos 20x+cos 8x)/(2cos 4x)) = ((cos ((2π)/3)+cos ((4π)/(15)))/(2cos ((2π)/(15)))) = ((−(1/2)+2cos^2 (((2π)/(15)))−1)/(2cos ((2π)/(15))))

$$\frac{\mathrm{cos}\:\mathrm{20}{x}+\mathrm{cos}\:\mathrm{8}{x}}{\mathrm{2cos}\:\mathrm{4}{x}}\:=\:\frac{\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}}{\mathrm{2cos}\:\frac{\mathrm{2}\pi}{\mathrm{15}}} \\ $$$$=\:\frac{−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{15}}\right)−\mathrm{1}}{\mathrm{2cos}\:\frac{\mathrm{2}\pi}{\mathrm{15}}} \\ $$

Commented by MJS_new last updated on 19/Sep/20

we can express this with roots but I haven′t got the time right now

$$\mathrm{we}\:\mathrm{can}\:\mathrm{express}\:\mathrm{this}\:\mathrm{with}\:\mathrm{roots}\:\mathrm{but}\:\mathrm{I}\:\mathrm{haven}'\mathrm{t} \\ $$$$\mathrm{got}\:\mathrm{the}\:\mathrm{time}\:\mathrm{right}\:\mathrm{now} \\ $$

Commented by MJS_new last updated on 20/Sep/20

I get ((13+7(√5)−(√(390+174(√5))))/8)

$$\mathrm{I}\:\mathrm{get}\:\frac{\mathrm{13}+\mathrm{7}\sqrt{\mathrm{5}}−\sqrt{\mathrm{390}+\mathrm{174}\sqrt{\mathrm{5}}}}{\mathrm{8}} \\ $$

Commented by bobhans last updated on 20/Sep/20

how sir?

$${how}\:{sir}? \\ $$

Commented by MJS_new last updated on 20/Sep/20

(π/(30))=6° cos 84° = sin 6° =sin (36°−30°) cos 36° =((1+(√5))/4) cos 24° =cos (60°−36°) (or use other sums/differences) now apply formulas for sin(a±b), cos (a±b)

$$\frac{\pi}{\mathrm{30}}=\mathrm{6}° \\ $$$$\mathrm{cos}\:\mathrm{84}°\:=\:\mathrm{sin}\:\mathrm{6}°\:=\mathrm{sin}\:\left(\mathrm{36}°−\mathrm{30}°\right) \\ $$$$\mathrm{cos}\:\mathrm{36}°\:=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\mathrm{cos}\:\mathrm{24}°\:=\mathrm{cos}\:\left(\mathrm{60}°−\mathrm{36}°\right) \\ $$$$\left(\mathrm{or}\:\mathrm{use}\:\mathrm{other}\:\mathrm{sums}/\mathrm{differences}\right) \\ $$$$\mathrm{now}\:\mathrm{apply}\:\mathrm{formulas}\:\mathrm{for}\:\mathrm{sin}\left({a}\pm{b}\right),\:\mathrm{cos}\:\left({a}\pm{b}\right) \\ $$

Answered by 1549442205PVT last updated on 20/Sep/20

f(x)=((cos14x×cos6x)/(cos4x))=((cos20x+cos8x)/(2cos4x)) f((π/(30)))=((cos20x+2cos^2 4x−1)/(2cos4x)) =((cos((2π)/3)+2cos^2 ((2π)/(15))−1)/(2cos((2π)/(15))))=((−(1/2)+2cos^2 24°−1)/(2cos24°))

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{cos14x}×\mathrm{cos6x}}{\mathrm{cos4x}}=\frac{\mathrm{cos20x}+\mathrm{cos8x}}{\mathrm{2cos4x}}\: \\ $$$$\mathrm{f}\left(\frac{\pi}{\mathrm{30}}\right)=\frac{\mathrm{cos20x}+\mathrm{2cos}^{\mathrm{2}} \mathrm{4x}−\mathrm{1}}{\mathrm{2cos4x}} \\ $$$$=\frac{\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{2cos}^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{15}}−\mathrm{1}}{\mathrm{2cos}\frac{\mathrm{2}\pi}{\mathrm{15}}}=\frac{−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2cos}^{\mathrm{2}} \mathrm{24}°−\mathrm{1}}{\mathrm{2cos24}°} \\ $$

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