If cos x * (cos 4x) = sin x * sin (6x), what is x between 0 and 2pi?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Knowing the following formula:

cos a+cos b= 2 cos[(a+b)/2]cos[(a-b)/2] and

cos a - cos b= 2sin [(a+b)/2]sin[(b-a)/2]

Suppose we have:

cos a+cos b= cos x * (cos 4x), but

cos a+cos b= 2 cos[(a+b)/2]cos[(a-b)/2]

From the both identities above, it results that:

cos x * (cos 4x)=2 cos[(a+b)/2]cos[(a-b)/2]

From the principle of an identity it results that:

x=(a+b)/2 => a+b=2x

4x=(a-b)/2 =>a-b=8x

a+b+a-b=2x+8x

2a=10x

a=5x

a+b-a+b=2x-8x

2b=-6x

b=-3x

So, cos 5x+ cos 3x=(1/2)cos x * (cos 4x)

(1/2)(cos 5x+ cos 3x)=cos x * (cos 4x)

sin x * sin (6x)=(1/2)[cos a-cosb]

x=(a+b)/2 => a+b=2x

6x=(b-a)/2 => b-a=12x

a+b+b-a=2x+12x

2b=14x => b=7x

a+b-b+a=2x-12x => 2a=-10x => a=-5x

(1/2)(cos 5x - cos 7x)=(1/2)(sin x * sin (6x))

So, the equivalent of the give identity is:

(1/2)(cos 5x+ cos 3x)=(1/2)(cos 5x - cos 7x)

(cos 5x+ cos 3x)=(cos 5x - cos 7x)

cos 3x=-cos 7x

cos3x+cos7x=0

Now, we'll write the sum as a product, in this way using the properties of a product of 2 factors which is equal to 0.

2cos[(3x+7x)/2]cos[(3x-7x)/2]=0

2cos(5x)cos(2x)=0

cos 5x=0

5x= arccos0+2kpi

5x=pi/2+2kpi

For k=0, x=pi/10, (x=180/10=18 degrees)

For k=1, 5x=2pi-pi/2

5x=3pi/2

x=3pi/10(x=3*18=54 degrees, first quadrant)

cos 2x=0

2x=arccos0 + 2kpi

2x=pi/2+2kpi

For k=0, x=pi/4(x=45 degrees)

For k=1, 2x=2pi-pi/2,

x=pi-pi/4=3pi/4(x=3*45=135, second quadrant)

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neela | High School Teacher | (Level 3) Valedictorian

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cosx*cos4x=sinx*sin6x.

cosA+cosB = 2cos(A+B)/2 * cos[(A-B)/2] and

cosA - cosB = -2sin[(a+B)/2] * sin[(A-B)/2] and

2cosAcosB = (1/2)[ cos(A+B) +cos(A-B)],

2sinA sinB =(1/2)(cos(A-B)-cos(A+B)]

[cos(4x+x) +cos(4x-x)]/2 = [cos(6x-x)-cos (x+6x) ]/2

cos5x+cos3x=cos5x-cos7x

cos7x+cos3x = 0

2cos5x*cos2x=0

cos5x=0  or cos2x = 0

5x=90 degree or 270 degree. So x=90/5= 18 deg or x = 270/5 = 54 deg

2x=0 when 2x=90 or 270. So, x=90/2 = 45 degree or x=270/2 = 135 deg .

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