# Prove that 2 cos x – 2 cos^{3} x = sin x sin 2x

Trigonometry name itself says that it is a subject that deals with the geometry of triangles and it is very useful for situations when needed to find when there are some sides given and we need the relations between the sides or angles between the sides. In Trigonometry we have different ratios that are sin A, cos A, tan A, cot A, sec A, cosec A with the help of which, the relation between the sides and the angle between the sides of the triangle can be obtained.

### Trigonometric functions

The trigonometric functions define the relation between the sides and angles and the examples are sin A, cos A, tan A, cot A, sec A, cosec A. The relation between different trigonometric functions is a **trigonometric identity**. The identities are very useful to test the inequality in the trigonometric equations.

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Examples are,

- Tan A= sin A/cos A
- sin A = 1/cosec A
- cos A= 1/sec A
- Tan A= 1/cot A

### Prove that 2 cos x – 2 cos^{3} x = sin x sin 2x

There are basic identities that are required in order to solve the above problem statement, lets look at some of the basic identities of the 6 trigonometric functions that are required in this case,

**Prerequisite Identities used in the proof**

- sin 2x = 2 × sin x × cos x
- sin
^{2}x + cos^{2}x = 1 - 1 – cos
^{2}x = sin^{2}x - sin x = 1/cosec x
- cos x – cos y =

**Given Trigonometric equation**

2 cos x – 2 cos^{3}x= sin x sin 2x

LHS = 2 cos x – 2 cos^{3}x

RHS = sin x sin 2x

**Deriving Proof from LHS **

Given LHS2 cos x – 2 cos

^{3}x

Step-1Taking the 2 cos x which is common in both the terms

= 2cos x (1-cos

^{2 }x)

Step-2Substituting 1-cos

^{2}x which is equal to sin^{2}x= 2cos x(sin

^{2 }x)= (sin x) × (2 × sinx × cos x)

Step-3Substituting 2 × sin x × cos x which is sin 2x

= sin x × sin 2x

From step 3, it can be concluded that LHS =sin x × sin 2x which is equal to RHS and thus,

sin x × sin 2x=sin x × sin 2x

LHS = RHS

**Hence Proved.**

**Deriving Proof from the RHS**

Given RHSsin x sin 2x

Step-1Substituting the formula of sin 2x in the given RHS

= sin x × (2 sinx cos x)

= 2 × sin

^{2}x × cosx

Step-2Substituting the formula of sin

^{2}x which is equal to 1-cos^{2}x= 2 × (1-cos

^{2}x) × cos x= 2cos x – cos

^{3}x

From the step 2 it can be concluded that RHS = 2cos x-cos^{3}x which is equal to LHS and thus,

2cos x-cos^{3}x = 2cos x-cos^{3}x

LHS = RHS

**Hence Proved.**

### Sample Problems

**Question 1: Solve the trigonometric identity: 4cos ^{2}x-4cos^{4}x+cos^{2}2x**

**Solution:**

- Taking 2cosx common in the first two terms we get
= 2cosx (2cosx-2cos

^{3}x) + cos^{2}2x

- From the above derivation 2cosx – 2cos
^{3}x = sinxsin2x= 2cosx(sinxsin2x) + cos

^{2}2x

- From the standard identity 2
= sin

^{2}2x + cos^{2}2x= 1

**Question 2: Solve the trigonometric identity: (1/ 8 × cosecx cosec2x cosx – 8 cosecx cosec2x cos ^{3}x)**

**Solution:**

- Taking the 4 cosecx cosec2x common in the denominator
= 1/4 cosecx cosec2x (2cosx -2cos

^{3}x)

- Using identity 4
= (sinx sin 2x)/4 × (2cosx -2cos

^{3}x)

- By the derived identity the equation becomes
= (sinx sin 2x) / (4 × sinx sin2x)

= 1/4 = 0.25

**Question 3: Solve the trigonometric identity: (cos 3x – cosx) / (4cos ^{3}x – 4cosx)**

**Solution:**

- Solving the numerator using the identity 5
=

= -2sinx sin 2x / 4cos

^{3}x – 4cosx

- Solving the denominator by multiplying -1 and dividing by -1 to get the denominator to our derived identity
= 2 sin x sin 2x/2(2cos

^{3}x-2cosx)= sin x sin 2x/(2cos

^{3}x-2cosx)= sin x sin 2x/sin x sin 2x

= 1