(a) (cos 3A) / 2 cos 2A - 1) = cos A and hence find cos 15 degree  (b) [(sin^3 (x) + sin 3x) / sin x] + [(cos^3 (x) - cos 3x) /cos x] = 3 (c) [(COS 3A) / (sin A)] + [(sin 3A) / (cos A)] = 2 cot...

(a) (cos 3A) / 2 cos 2A - 1) = cos A and hence find cos 15 degree

 

(b) [(sin^3 (x) + sin 3x) / sin x] + [(cos^3 (x) - cos 3x) /cos x] = 3

(c) [(COS 3A) / (sin A)] + [(sin 3A) / (cos A)] = 2 cot 2A

(d) 4 (cos^3  20 degree + cos^3 40 degree) = 3 (cos 20 degree + cos 40 degree)



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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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b) [(sin^3 (x) + sin 3x) / sin x] + [(cos^3 (x) - cos 3x) /cos x] = 3

We'll write the formula for sin 3x:

sin 3x = sin (2x+x) = sin 2x*cos x + sin x*cos 2x

sin 2x = 2sin x*cos x

cos 2x = 1 - 2(sin x)^2

We'll substitute sin 2x and cos 2x into the formula of sin 3x:

sin 3x = 2 sinx*(cos x)^2 + sin x*[1-2(sin x)^2]

We'll substitute (cos x)^2 = 1 - (sin x)^2 (fundamental formula of trigonometry)

sin 3x = 2sin x*[1 - (sin x)^2] + sin x - 2(sin x)^3

We'll remove the brackets:

sin 3x = 2sin x - 2(sin x)^3 + sin x - 2(sin x)^3

We'll combine like terms:

sin 3x = 3sin x - 4(sin x)^3

We'll calculate the numerator of the first ratio:

(sin x)^3 + sin 3x = (sin x)^3 + 3sin x - 4(sin x)^3

We'll combine like terms:

(sin x)^3 + sin 3x = 3sin x - 3(sin x)^3

We'll factorize by 3 sin x:

3sin x - 3(sin x)^3 = 3 sin x[1 - (sin x)^2]

But 1 - (sin x)^2 = (cos x)^2

3 sin x[1 - (sin x)^2] = 3 sin x (cos x)^2

The first ratio will become:

[(sin^3 (x) + sin 3x) / sin x] = 3 sin x (cos x)^2/sin x

We'll simplify:

 3 sin x (cos x)^2/sin x =  3 (cos x)^2

 

Now we'll calculate the second ratio:

[(cos x)^3 - cos 3x) /cos x]

First, we'll calculate the formula for cos 3x:

cos 3x = cos (2x + x) = cos 2x*cos x - sin 2x*sin x

cos 3x = 4 (cos x)^3 - 3 cos x

We'll substitute cos 3x to the numerator:

(cos x)^3 - cos 3x = (cos x)^3 - 4 (cos x)^3 + 3 cos x

(cos x)^3 - cos 3x = -3  (cos x)^3 + 3 cos x

We'll factorize by -3 cos x

-3  (cos x)^3 - 3 cos x = 3 cos x[1 - (cos x)^2]

But 1 - (cos x)^2 = (sin x)^2

3 cos x[1 - (cos x)^2] = 3 cos x (sin x)^2

The second ratio will become:

[(cos x)^3 - cos 3x) /cos x] = 3 cos x (sin x)^2/cos x

We'll simplify:

3 cos x (sin x)^2/cos x = 3 (sin x)^2

The final expression will be:

[(sin x)^3+ sin 3x) / sin x] + [(cos x)^3 - cos 3x) /cos x] = 3 (cos x)^2 + 3 (sin x)^2

We'll factorize by 3:

3 (cos x)^2 + 3 (sin x)^2 = 3[(cos x)^2 + (sin x)^2]

But (cos x)^2 + (sin x)^2 = 1

3[(cos x)^2 + (sin x)^2] = 3

[(sin x)^3+ sin 3x) / sin x] + [(cos x)^3 - cos 3x) /cos x] = 3

 

a) Now we'll prove the identity:

 (cos 3A) / 2 cos 2A - 1) = cos A

We'll take the formula for cos 3A from the previous point b):

cos 3A = 4 (cos A)^3 - 3 cos A

 cos 2A = 2 (cos A)^2 - 1

So, the difference 2 cos 2A - 1 = 2 (cos A)^2 - 1 - 1

2 (cos A)^2 - 1 - 1 = 2 (cos A)^2 - 2

We'll factorize by 2:

2 (cos A)^2 - 2 = 2[(cos A)^2 - 1]

But (cos A)^2 - 1 = (sin A)^2

2[(cos A)^2 - 1] = 2(sin A)^2

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neela | High School Teacher | (Level 3) Valedictorian

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(a) (cos 3A) / 2 cos 2A - 1) = cos A and hence find cos 15 degree

 cosA =  cos3A/(2cos2a -1)

cos15 = cos(3*15)/ (2cos2*15 -1)

= cos45/{2cos30 -1} = (1/sqrt2)/{2(sqrt3)/2 -1}

= (1/sqr2)/{sqr3 -1} = 1/(sqrt6 -sqrt2) = (sqrt6+srt2)/4.

 

(b) [(sin^3 (x) + sin 3x) / sin x] + [(cos^3 (x) - cos 3x) /cos x] = 3

 we know sin3x = 3sinx -4sin^3x

 Therefore sin^3x+sin3x =3sinx-3sinx^3x. Divide by sinx.

 (sin^3x-sin3x)/sinx = 3-3in^2x...........(1)

 Co3x = 4cos^3-3cosx

-cos3x = 3cosx-4cos^3x

cos^3x = 3cosx-3cos^3x. Divide by cosx:

(cos^3x-cos3x)/cosx = 3-3cos^2x...............(2)

Add (2) and (3) to get:

(sin^3x-sin3x)/sinx + cos^3x-cos3x)/cosx = (6 - 3(sin^2+cos^2x) =6-3 = 3.

(c) [(CoS 3A) / (sin A)] + [(sin 3A) / (cos A)] = 2 cot 2A

Multiply by sinA* cosA:

cos3A*cosA +sin3A*sinA =2 cos2A/sin2A )sinA*cosA = (2cos2A/2sinAcosA) sinAcosA = 2cos2A.

LHS: cos3AcosA - sin3AsinA = cos(3A - A) = cos2A.

 

(d) 4 (cos^3  20 degree + cos^3 40 degree) = 3 (cos 20 degree + cos 40 degree

4cos^3 a = 3cosa+cos3a. Using this we get:

LHS :  3cos20+3cos(3*20) + cos40+ 4cos(3*40)

=  3(cos20 +3cos40) +(cos60 +cos120)

= 3(cos20+cos40) +0, as cos120 = cos(180-60) = -cos60.

= 4(co

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