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# Show that (cosx)^3 = (1/4)(3cosx + cos3x)?

Thanks heaps.

### 8 Answers

- germanoLv 71 decade agoFavorite Answer
cos³x = (1/4)[3cosx + cos(3x)]

rewrite (3x) as (2x+x):

cos³x = (1/4)[3cosx + cos(2x+ x)] →

recall cosine addiction rule:

cos(a + b) = cos(a) cos(b) - sin(a) sin(b)

thus:

cos³x = (1/4){3cosx + [cos(2x) cosx - sin(2x) sinx]} →

find cos(2x) and sin(2x) using the double-angle identities:

cos(2x) = cos²x - sin²x

sin(2x) = 2sinx cosx

cos³x = (1/4){3cosx + [(cos²x - sin²x)cosx - (2sinx cosx) sinx]} →

cos³x = (1/4)[3cosx + (cos³x - sin²x cosx - 2sin²x cosx)] →

cos³x = (1/4)(3cosx + cos³x - 3sin²x cosx) →

replace sin²x with 1 - cos²x:

cos³x = (1/4)[3cosx + cos³x - 3(1 - cos²x) cosx] →

cos³x = (1/4)[3cosx + cos³x - 3(cosx - cos³x)] →

cos³x = (1/4)(3cosx + cos³x - 3cosx + 3cos³x) →

cos³x = (1/4)(4cos³x) →

cos³x = cos³x

thus your identity is verified

I hope it helps...

Bye!

- Mein Hoon NaLv 71 decade ago
we know

cos(A + B) = cos A cos B - sin A sin B ...1

put B= A to get

cos 2A = cos^2 A - sin ^2 A = 2 cos ^2 A - 1

similarly sin 2A = 2 sin A cos A

now in 1 put B = 2A

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

= 2 cos ^2 A cos A - cos A - 2 cos A sin ^2A

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

= 4 cos^2 A - 3 cos A

cos 3A = 4 cos^3 A - 3 cos A

so cos 3A + 3 cos A = 4 cos^3 A

so cos^3 A= 1/4( cos 3A + 3 cos A)

proved

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- 1 decade ago
cos^3x = (1/4)(3cosx + cos3x)

cos3x = 4cos^3x - 3cosx, so

cos^3x = (1/4)(3cosx + 4cos^3x - 3cosx)

cos^3x = 1/4(4cos^3x)

cos^3x = cos^3x

- Scrander berryLv 71 decade ago
RHS = (1/4)(3cosx + cos(3x))

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

= (1/4)(3cos(x) + cos(x)cos(2x) - sin(x)sin(2x))

= (1/4)(3cos(x) + cos(x)(cos²(x) - sin²(x)) - 2sin²(x)cos(x))

= (1/4)(3cos(x) + cos³(x) - cos(x)sin²(x) - 2sin²(x)cos(x))

= (1/4)(3cos(x) + cos³(x) - 3sin²(x)cos(x))

= (1/4)(3cos(x)(1 - sin²(x))+ cos³(x))

= (1/4)(3cos(x)cos²(x)+ cos³(x))

= (1/4)(4cos³(x))

= cos³(x)

= LHS qed

- Anonymous1 decade ago
u need to use the formula of cos3x and expand it

- 1 decade ago
it doesn't...IDENTIES...(cosx)^2= 1/2(1+cos2x) (ANGLE NEVER CHANGES............so........(cosx)^3=1/3(1+cos2x')