# Proving Trigonometric Identities Prove: cos3A = -3cosA + 4cos^3A

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Here we have to prove that: cos 3A = -3cos A + 4 (cos A)^3

Now cos( A + B) = cos A *cos B - sin A* sin B and sin ( A + B) = sin A* cos B + cos A* sin B

cos 3A

=> cos ( 2A + A)

=> cos 2A *cos A - sin 2A* sin A

=> cos (A + A) * cos A - sin (A +A)* sin A

=> (cos A *cos A - sin A* sin A)* cos A - (sin A* cos A + cos A* sin A)* sin A

=> (cos A *cos A* cos A - sin A* sin A * cos A) - (sin A* cos A* sin A + cos A* sin A* sin A)

=> (cos A)^3 - 3*(sin A)^2 * cos A

=> (cos A)^3 - 3*(1 - (cos A)^2) * cos A

=> (cos A)^3 - 3*cos A + 3(cos A)^3

=> - 3* cos A + 4 ( cos A)^3

**Therefore cos 3A = -3* cos A + 4 ( cos A)^3**

We need to prove that:

cos3A = -3cosA + 4cos^3 A

We will start from the left sides and prove the right sides.

We know that:

cos3A = cos(2A + A)

Now we will use the trigonometric identities to prove.

We know that:

cos(a+b) = cosa*cosb - sina*sinb

==> cos(2A+A) = cos2A*cosA - sin2A*sinA

Now we know that:

cos2A = 2cos^2 A -1

sin2A = 2sinA*cosA

==> cos(2A+A) = (2cos^2 A -1)cosA - 2sinA*cosA*sinA

Let us simplify:

==> cos(2A+A) = 2cos^3 A - cosA - 2sin^2 A * cosA

Now we know that: sin^2 A = 1- cos^2 A

==> cos(2A+A) = 2cos^3 A - cosA -2(1-cos^2 A)*cosA

= 2cos^3 A - cosA -2cosA + 2cos^3 A

Now we will combine like terms.

==> cos(2A+A) = 4cos^3 A - 3cosA

**==> cos(3A) = -3cosA + 4cos^3 A......q.e.d**

To prove cos3A = -3cosA+4cos^3A.

We use sin(A+B) = sinAcosB+cosAsinB.

So sin(A+A) = sin2A = 2sinAcosA

We know that cos(A+B) cosAcosB-sinAsinB, put B= A.

cos(A+A) = cosAcosA-sinAsinA.

cos2A = cos^2A-sin^2A.

cos2A = cos^2A - (1- cos^2A).

cos2A = cos^2A-1+cos^2A.

cos2A = 2cos^2A-1.

Therefore cos3A = cos(2A+A) = cos2AcosA- sin2AsinA.

So cos3A = {2cos^2A-1)cosA- (2sinAcosA)sinA.

cos3A = 2cos^3A - cosA - 2sin^2AcosA

cos3A = 2cos^2A -cosA -2(1-cos^2A) cosA.

cos3A = 2cos^3A-cosA - 2cosA+2cos^3A.

cos3A = 4cos^3A - 3cosA.

L:H:S ≡ cos 3A

= cos (A + 2A)

**⇒ use cos (A+B) = cosA.cosB - sinA.sinB**

= cosA.cos2A - sinA.sin2A

**⇒ use cos 2A = 2cos²A - 1**

=cosA(2cos²A - 1) - sinA(2sinA.cosA)

=2cos³A - cosA -2sin²A.cosA

**⇒ use sin²A = 1 - cos²A **

=2cos³A - cosA -2cosA(1-cos²A)

=2cos³A - cosA - 2cosA + 2cos³A

= 4cos³A - 3cosA

Hence L:H:S ≡ R:H:S