If F(X) = Cos X Cos 2X Cos 4X Cos 8X Cos 16X, Then F "(Π/4) Is Equal To


I can express it for $x$ only with $cos(2x)=cos^2(x)-sin^2(x)$ và $cos(4x)=cos(2x+2x)$, but it only seems to become a really big expression and I have no clue how to lớn proceed after... Any suggestions?

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You can use the double angle identity$$ sin 2x = 2sin x cos x$$By multiplying $sin x$,$$eginalign*cos(x)cos(2x)cos(4x) &= frac18 \frac12 sin (2x) cos(2x)cos(4x) &= frac18sin x \frac14 sin (4x)cos(4x) &= frac18sin x \frac18 sin (8x) &= frac18sin x \sin (8x) &= sin xendalign*$$The last equation will have $16$ roots in $<0,2pi)$, but $0$ and $pi$ vày not solve the original equation since they are introduced by $sin x$.


For $x e npi$ (it"s true, because $ cosxcdotcos2xcdotcos4x,Big|_x=npi efrac18$)$$ cosxcdotcos2xcdotcos4x=fracsinxcdotcosxcdotcos2xcdotcos4xsinx = \=frac12cdot fracsin2xcdotcos2xcdotcos4xsinx=frac14cdot fracsin4xcdotcos4xsinx=frac18cdot fracsin8xsinx,$$therefore,$$sin8x=sinx.$$


Let $z=e^ix$. Then your relation says $$left(z+z^-1 ight)left(z^2+z^-2 ight)left(z^4+z^-4 ight)=1$$ That is: $$eginalignz^7+z^5+z^3+z+z^-1+z^-3+z^-5+z^-7&=1\z^-7fracz^16-1z^2-1&=1& ext(for $z^2 eq1$)\z^16-1&=z^9-z^7\z^16-z^9+z^7-1&=0\left(z^7-1 ight)left(z^9+1 ight)&=0endalign$$

So $z$ is either a $7$th root of $1$ or a $9$th root of $-1$. Which means your $x$ is among $$left\frac2pi7, frac4pi7,frac6pi7,frac8pi7,frac10pi7,frac12pi7,fracpi9,frac3pi9,frac5pi9,frac7pi9,frac11pi9,frac13pi9,frac15pi9,frac17pi9 ight$$ or translates by $2pi k$. We"ve left out $0pi/7$ and $9pi/9$ since they cause $z^2-1$ to equal $0$, invalidating an earlier step here. You can directly check that these angles do not satisfy the original equation.

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edited Mar 14, năm 2016 at 22:19
answered Mar 14, năm nhâm thìn at 19:37

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Not the easiest way, but still a valid method to tackle problems lượt thích this. Use the tan-half-angle substitution

$$ eginalign x & = 2 an^-1(t) \cos(x) và = frac1-t^21+t^2 \sin(x) & = frac2 t1+t^2 endalign $$

Also we need to expand the $cos(n x)$ terms to make

$$ cos(x) left( 2 cos^2(x)-1 ight) left(1-8 sin^2(x) cos^2(x) ight) = frac18 $$

Now change the variables $x ightarrow t$ for

$$ left( frac1-t^21+t^2 ight) left( fract^4-6 t^2+1(1+t^2)^2 ight) left( fract^8-38 t^6+70 t^4-28 t^2+1(1+t^2)^4 ight) = frac18 $$

All this can be factored as

$$ frac(1-3 t^2)(3 t^12-90t^10+705 t^8-1660 t^6+1365 t^4-266 t^2+7)8(1+t^2)^7 = 0 $$

(thank you CAS)

This is an even function, are we are only concerned about positive roots. The first immediate root is $t=frac1sqrt3$, which is

$$ x_1 = 2 an^-1 left( frac1sqrt3 ight) = 2 fracpi6 =fracpi3$$

The other roots I got where

$$ eginalign t_2 và = 0.176326980708464 và x_2 = 0.349065850398863 \ t_3 & = 0.481574618807528 và x_3 = 0.897597901025654 \ t_4 và = 1.1917535925942 & x_4 = 1.74532925199432\ t_5 và = 1.2539603376627 & x_5 = 1.79519580205130 \ t_6 và =2.74747741945462 và x_6 = 2.44346095279206 \ t_7 & = 4.38128626753482 & x_7 = 2.69279370307696endalign$$