The cos(2x) formula is one of the most useful double-angle identities in trigonometry! It helps us find the cosine of a doubled angle using sine, cosine, or tangent. Basically, it lets us express cos(2x) in different ways, making tricky trigonometric problems much easier to handle. This identity is super handy when simplifying complex expressions and even comes in clutch for solving integration problems in calculus.

If you're preparing for standardized tests like the SAT, ACT, PSAT/NMSQT, GED, GRE, GMAT, AP exams, PERT, Accuplacer, or even the MCAT, a strong grasp of trigonometry—including double-angle identities like cos(2x)—can give you a serious advantage. Many of these exams include math sections that test algebra, geometry, and trigonometry, so mastering these concepts can boost your score and improve your college or grad school prospects!

What is the Cos2x Formula in Trigonometry? 

The cos(2x) identity is a key formula in trigonometry that helps us find the cosine of a double angle. Basically, it tells us how to express cos(2x) using different trig functions like sine, cosine, or tangent. Since the angle here is multiplied by 2, it's called a double-angle identity—pretty cool, right? This formula is super useful for simplifying tricky trig problems and shows up a lot in algebra, calculus, and even physics!

The following are the different forms of the identity cos2x.

\( \cos2x=cos^{2}x-\sin^{2}x \)

\( \cos2x= 2\cos^{2}x-1 \)

\( \cos2x= 1- 2 \sin^{2}x \)

\( \cos2x=\frac{1-\tan^{2}x}{1+\tan^{2}x} \)

The formula for cos2x is derived by using the sum angle formula of cosine function.

 We know that, \( \cos(P + Q) =\cos P \cos Q – \sin P \sin Q \).

 To calculate the value of cos2x, we put the value of angle P and angle Q equal to x.

 Putting \( P = Q= x \) we get,

 \(\ cos (x + x) = \cos x \cos x – \sin x \sin x \) 

 \( \cos2x=cos^{2}x-\sin^{2}x \) 

This derives the formula for cos2x.

Cos 2x in Terms of sin x

The formula of Cos2x in terms of Sine function is \( \cos2x= 1- 2 \sin^{2}x \). This can be proved by using the trigonometric identity \( \sin^{2}x +\ cos^{2}x=1 \).

We know that., 

\( \cos2x=cos^{2}x-\sin^{2}x \) 

Using identity \( \sin^{2}x +\ cos^{2}x=1 \) and putting \( \ cos^{2}x=1 - \sin^{2}x\) in above identity, we get.

= \( (1- \sin^{2}x )-\sin^{2}x \)

= \( \cos2x= 1- 2 \sin^{2}x \)

Thus, The formula of Cos2x in terms of Sine function is \( \cos2x= 1- 2 \sin^{2}x \).

Cos 2x in Terms of cos x

The formula of Cos2x in terms of cos function is \( \cos2x= 2 \sin^{2}x -1 \). This can be proved by using the trigonometric identity \( \sin^{2}x +\ cos^{2}x=1 \).

We know that., 

\( \cos2x=cos^{2}x-\sin^{2}x \) 

Using identity \( \sin^{2}x +\ cos^{2}x=1 \) and putting \( \ sin^{2}x=1 - \cos^{2}x\) in above identity, we get.

= \( \ cos^{2}x - (1-\ cos^{2}x ) \)

= \( \ cos^{2}x )- 1 + \ cos^{2}x \)

= \( \cos2x= 2 \cos^{2}x -1 \)

Thus, the formula of Cos2x in terms of cos function is \( \cos2x= 2 \cos^{2}x -1 \) .

Cos2x formula in Terms of tan x

The formula of Cos2x in terms of tan function is \( \cos2x=\frac{1-\tan^{2}x}{1+\tan^{2}x} \). This can be proved by using the trigonometric identities \( \sin^{2}x +\ cos^{2}x=1 \)and \(\tan=\frac{\sin x}{\cos x} \).

We know that., 

\( \cos2x=\cos^{2}x-\sin^{2}x \) 

Dividing \( \cos^{2}x-\sin^{2}x \) by 1 ,we get.

\(=\frac{cos^{2}x-\sin^{2}x}{1}\)

Using identity \( \sin^{2}x +\ cos^{2}x=1 \) and putting \( 1= \sin^{2}x +\ cos^{2}x\).

 \(=\frac{\cos^{2}x-\sin^{2}x}{\sin^{2}x +\ cos^{2}x}\)

 Dividing the numerator and denominator by \( \ cos^{2}x \).

 \(=\frac{\frac{cos^{2}x}{cos^{2}x}-\frac{sin^{2}x}{cos^{2}x}}{\frac{sin^{2}x}{cos^{2}x}+\frac{cos^{2}x}{cos^{2}x}}\)

Using the identity \(\tan=\frac{\sin x}{\cos x} \).

\(=\frac{1-\tan^{2}x}{1+\tan^{2}x}\)

Thus, the formula of Cos2x in terms of tan function is =\(\frac{1-\tan^{2}x}{1+\tan^{2}x} \).

Cos^2x (Cos Square x)

\(\ cos^{2}x \)or Cos Square x is a trigonometric function which represents cos x as a whole square. \(\ cos^{2}x \) is denoted in different forms in terms of different trigonometric functions like cos and sin functions. Formula for \(\ cos^{2}x \) can be derived by using different trigonometric formulas and identities.

sin2x cos2x formula

The following are the formulas for \( \sin2x \cos2x\).

\( \sin2x \cos2x=2\cos x(2\sin x\cos^{2}x -\sin x) \)

Or

\( \sin2x \cos2x=2\cos x(\sin x -2\sin^{3}x) \)

cos2x all formulas

The following are all formulas of \( cos2x \).

\( \cos2x=cos^{2}x-\sin^{2}x \)

\( \cos2x= 2\cos^{2}x-1 \)

\( \cos2x= 1- 2 \sin^{2}x \)

\( \cos2x=\frac{1-\tan^{2}x}{1+\tan^{2}x} \)

Cos^2x Formula

To get various formulas of \(\ cos^{2}x \) we use the identity \( \sin^{2}x +\ cos^{2}x=1 \).

• We subtract \( \sin^{2}x \) from both sides of the identity, we get.

\( \sin^{2}x +\ cos^{2}x -\sin^{2}x =1 -\sin^{2}x \)

\( \ cos^{2}x =1 -\sin^{2}x \). This formula contains \(\ cos^{2}x \).

• The following are \( Cos2x \) formulas that contain \(\ cos^{2}x \).

\( \cos2x=cos^{2}x-\sin^{2}x \) or \( cos^{2}x =\cos2x+\sin^{2}x \).

\( \cos2x= 2\cos^{2}x-1 \) or \( \cos^{2}x= =\frac{1+\cos2x}{2} \) 

Therefore, the following are the formulas of \(\ cos^{2}x \).

\( \ cos^{2}x =1 -\sin^{2}x \).

\( cos^{2}x =\cos2x+\sin^{2}x \).

\( \cos^{2}x= =\frac{1+\cos2x}{2} \).

Different types of math problems can be solved by applying Cos2x Identity.

We can understand the application of Cos2x Identity by the following example.

Example: To find out the value of cos 180° using the Cos2x Identity.

Solution: We know that \( \cos2x=cos^{2}x-\sin^{2}x \) and the value of cos 90° =0 and 

sin 90° =1.

\( \cos 180° =cos^{2} 90° -\sin^{2} 90° \) 

Putting the value of cos 90° =0 and sin 90° =1

\( = 0^{2}-1^{2} \) 

\( = 0 - 1\) =

\( = -1 \) 

Thus, the value of cos 180° is -1.

Solved Examples of Cos2x Formula

Solution: Considering LHS,

 \( \cos3A\) is written as 

 \( \cos3A=\cos(2A+A) \) ….. equation (1)

Using the identity \( \cos(P+Q) = \cos P\cos Q - \sin P \sin Q \) in equation (1), we get.

=\( \cos 2A\cos A - \sin 2A \sin A \) 

\(=(2Cos2x–1) Cos x –(2 \ CosA) A\)

\(=(2\ cos^{2}A –1) A –2 A sin^{2}A)\)

=\( (2\ cos^{2}A –1) \Cos A –2\Cos A ( 1-\ cos ^{2}A)\) 

=\( 2\ cos^{3}A –\Cos A -2\Cos A +2\ cos^{3}A\) 

=\( 4\ cos^{3}A – 3\Cos A \) 

=RHS

Thus, LHS = RHS

Therefore, It is Proved \( \cos3A = 4\cos^{3}A– 3\cos A \).

Solution: We know that

\( \cos2x=\frac{1-\tan^{2}x}{1+\tan^{2}x} \) and \(\tan x=\frac{1}{\cot x} \)

\( \cos2A=\frac{1-\tan^{2}A}{1+\tan^{2}A} \)

=\( \frac{1-\frac{1}{\cot x}}{1+\frac{1}{\cot x}} \)

= \( \frac{1-\frac{1}{\cot^{2}A}}{1+\frac{1}{\cot^{2}A}}\)

= \( \frac{\cot^{2}A -1}{\ cot^{2}A+1} \)

Therefore, the Cos2A Formula in terms of Cot A is expressed as \( \frac{\cot^{2}A -1}{\ cot^{2}A+1} \).

Solution: Given,\(\cos 2p = \sin p\) for −π≤p<π

We know that \( \cos2x= 1- 2 \sin^{2}x \).

\( 1- 2 \sin^{2}x = \sin p \)

\( 2 \sin^{2}x + \sin p -1=0\)

Factoring this quadratic equation considering variable as \sin p

\( (2 \sin p − 1)(\sin p + 1) = 0 \)

\( 2 \sin p − 1 = 0 \)

\(p=\frac{\pi}{6}\)

\( sin p = \frac{1}{2}\)

 \(\sin p + 1 = 0\)

\( sin p = −1\)

\(p= -\frac{\pi}{2}\)

Therefore,the Solution of \( \cos 2p = \sin p\) is \(p=\frac{\pi}{6}\( and\( p= -\frac{\pi}{2}\).

Solution: Considering finding out the formula for \( 1+ \cos2x \).

 We know that, \( \cos(P + Q) =\cos P \cos Q – \sin P \sin Q \).

 Putting \( P = Q= x \) in the above equation. we get,

 \(\ cos (x + x) = \cos x \cos x – \sin x \sin x \) 

 \( \cos2x=cos^{2}x-\sin^{2}x \) 

Adding 1 in the LHS and RHS of the above equation. we get,

 \( 1+ \cos2x= 1+ (cos^{2}x-\sin^{2}x) \) 

 Using the identity \( \sin^{2}x +\ cos^{2}x=1 \).

 \( 1+ \cos2x= \sin^{2}x +\ cos^{2}x+cos^{2}x-\sin^{2}x \) 

= \( 2\ cos^{2}x \)

Thus, the formula for \( 1+ \cos2x \) is \( 2\ cos^{2}x \).

Consider finding out the formula for \( 1- \cos2x \)

 We know that, \( \cos(P + Q) =\cos P \cos Q – \sin P \sin Q \).

 Putting \( P = Q= x \) in the above equation . we get,

 \(\ cos (x + x) = \cos x \cos x – \sin x \sin x \) 

 \( \cos2x=cos^{2}x-\sin^{2}x \) 

Adding 1 in the LHS and RHS of the above equation. We get,

 \( 1- \cos2x= 1-(cos^{2}x-\sin^{2}x )\) 

 Using the identity \( \sin^{2}x +\ cos^{2}x=1 \).

 \( 1- \cos2x= \sin^{2}x +\ cos^{2}x-cos^{2}x+\sin^{2}x \) 

= \(( 2\sin^{2}x )\)

Thus, the formula for \( 1- \cos2x \) is \( 2 \sin^{2}x \).

Solution: We use the identity:

cos⁡2x=12\cos 2x = \frac{1}{2}cos2x=21

We know that cosine is 12\frac{1}{2}21 at angles π3\frac{\pi}{3}3π and −π3-\frac{\pi}{3}−3π in the unit circle.
So,

2x=±π3+2kπ,k∈Z2x = \pm \frac{\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}2x=±3π+2kπ,k∈Z

Solving for xxx,

x=π6+kπ,k∈Zx = \frac{\pi}{6} + k\pi, \quad k \in \mathbb{Z}x=6π+kπ,k∈Z

For 0≤x<2π0 \leq x < 2\pi0≤x<2π:

x=π6,x=7π6x = \frac{\pi}{6}, \quad x = \frac{7\pi}{6}x=6π,x=67π

Final Answer:

x=π6,x=7π6x = \frac{\pi}{6}, \quad x = \frac{7\pi}{6}x=6π,x=67π

Solution: We use the identity:

cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 xcos2x=1−2sin2x

Setting it equal to sin⁡x\sin xsinx:

1−2sin⁡2x=sin⁡x1 - 2\sin^2 x = \sin x1−2sin2x=sinx

Rearranging:

2sin⁡2x+sin⁡x−1=02\sin^2 x + \sin x - 1 = 02sin2x+sinx−1=0

Solving the quadratic equation in sin⁡x\sin xsinx:

(2sin⁡x−1)(sin⁡x+1)=0(2\sin x - 1)(\sin x + 1) = 0(2sinx−1)(sinx+1)=0

Setting each factor to zero:

1. 2sin⁡x−1=02\sin x - 1 = 02sinx−1=0
   sin⁡x=12\sin x = \frac{1}{2}sinx=21
   This occurs at x=π6,5π6+2kπx = \frac{\pi}{6}, \frac{5\pi}{6} + 2k\pix=6π,65π+2kπ.
    
2. sin⁡x+1=0\sin x + 1 = 0sinx+1=0
   sin⁡x=−1\sin x = -1sinx=−1
   This occurs at x=3π2+2kπx = \frac{3\pi}{2} + 2k\pix=23π+2kπ.
    

Final Answer:

x=π6,x=5π6,x=3π2,k∈Zx = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}, \quad x = \frac{3\pi}{2}, \quad k \in \mathbb{Z}x=6π,x=65π,x=23π,k∈Z

Solution: We start with the left-hand side (LHS):

cos⁡4x−sin⁡4x\cos^4 x - \sin^4 xcos4x−sin4x

Using the difference of squares formula:

a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)

We rewrite it as:

(cos⁡2x−sin⁡2x)(cos⁡2x+sin⁡2x)(\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x)(cos2x−sin2x)(cos2x+sin2x)

Using the identity cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1cos2x+sin2x=1, this simplifies to:

cos⁡2x−sin⁡2x\cos^2 x - \sin^2 xcos2x−sin2x

We recognize this as the cos(2x) identity:

cos⁡2x\cos 2xcos2x

Thus,

cos⁡4x−sin⁡4x=cos⁡2x\cos^4x - \sin^4x = \cos 2xcos4x−sin4x=cos2x

which proves the identity.

Summary of Cos2x Formula

The cos(2x) formula is a key identity in trigonometry that helps us find the cosine of a double angle (2x). It’s super useful for solving trig equations, simplifying expressions, and even tackling calculus problems!

• You can express cos(2x) in different ways using cosine, sine, or tangent—whichever makes your math problem easier to solve.
• There are four different forms of the cos(2x) formula:

cos⁡2x=cos⁡2x−sin⁡2x

\cos 2x = \cos^2x - \sin^2x

cos2x=cos2x−sin2x cos⁡2x=2cos⁡2x−1

\cos 2x = 2\cos^2x - 1cos2x=2cos2x−1 

cos⁡2x=1−2sin⁡2x

\cos 2x = 1 - 2\sin^2xcos2x=1−2sin2x 

cos⁡2x=1−tan⁡2x1+tan⁡2x

\cos 2x = \frac{1 - \tan^2x}{1 + \tan^2x}cos2x=1+tan2x1−tan2x

Each version is handy in different situations—whether you're working with cosine, sine, or tangent!

• This formula is a lifesaver when simplifying complex trig expressions and solving tricky integration problems in calculus.
• One identity that pops up a lot in integration is:

cos⁡2x=1+cos⁡2x2\cos^2x = \frac{1 + \cos2x}{2}cos2x=21+cos2x

This makes it way easier to handle squared cosine terms when solving integrals.

Mastering these formulas will make trigonometry so much easier, especially for SAT, ACT, AP Calculus, and other standardized tests!

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