## Trending News

# Is cos^2 (2x)-sin^2(2x)=0 an identity?

At first glance, I don’t think it is, but I’m not sure. How would I approach this question?

Thank you for your time.

### 8 Answers

- Pramod KumarLv 71 month ago
There is difference in an equation and an identity.

In an Equation all the variables can have a limited number of values. where as in identity. the relation is satisfied by any value given to the variable(s).

For example consider your problem -

cos² 2x - sin² 2x = 0

It has the highest power = 2, Now if the given relation is satisfied by assigning more than two ( say 3) values of x, it is an identity.

If this relation is satisfied with three values of x, then it is an Identity. Lets see -

Put x = 30°, 45°, 60°, 90°

cos²(90°) - sin²(90°) = 0 - 1 = - 1

It means that x = 45° DOES NOT SATISFY the relation.

Similarly => cos²(30°) - sin²(30°) ≠ 0

cos²(45°) - sin²(45°) = 0 .................... Satisfies.

cos²(60°) - sin²(60°) ≠ 0

Hence --

cos² 2x - sin² 2x = 0 is an equation and not an identity.

- Login to reply the answers

- lenpol7Lv 71 month ago
Remember the Trig. Identity

Cos(2z) = Cos^(z0 - Sin^2(Z)

Hence it follows

Cos(4x) = Cos(2x + 2x)

Cos(2x + 2x) = Cos^(2x) - Sin^2(2x)

- Login to reply the answers

- la consoleLv 71 month ago
cos²(2x) - sin²(2x) = 0 ← not an identity but an equation

cos²(2x) = sin²(2x)

sin²(2x)/cos²(2x) = 1

tan²(2x) = 1

tan(2x) = ± 1

First case: tan(2x) = 1

2x = (π/4) + kπ → where k is an integer

x = (π/8) + k.(π/2)

Second case: tan(2x) = - 1

2x = [π - (π/4)] + kπ → where k is an integer

2x = (3π/4) + kπ

x = (3π/8) + k.(π/2)

- Login to reply the answers

- sepiaLv 71 month ago
cos^2 (2x) - sin^2 (2x) = 0

cos(4 x) = 0

Solution:

x = (π n)/4 - π/8, n element Z

- Login to reply the answers

- How do you think about the answers? You can sign in to vote the answer.
- PopeLv 71 month ago
No, it is not an identity. For the simplest of counterexamples, let x = 0.

x = 0

cos²(2x) - sin²(2x)

= cos²(0) - sin²(0)

= 1 + 0

= 1

≠ 0

The equation does have solutions though.

Let cos²(2x) - sin²(2x) = 0.

sin²(2x) = cos²(2x)

sin²(2x) / cos²(2x) = 1

tan²(2x) = 1

tan(2x) = ±1

2x = (1 + 2k)π/4

x = (1 + 2k)π/8, for any integer k

- Login to reply the answers

- TomVLv 71 month ago
Let x = 0 which results in:

cos²(0) - sin²(0) = 0

1 - 0 = 0

1 = 0

This is a false statement. If the original equation is an identity, it must be true for all values of x. It is not. Therefore it is not an identity.

- Login to reply the answers

- Jeff AaronLv 71 month ago
cos^2(2x) - sin^2(2x) = 0

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

cos(2x) = +/- sin(2x)

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

tan(2x) = +/- 1

2x = (pi*n/2) - (pi/4) radians, for any integer n

x = (pi*n/4) - (pi/8) radians, for any integer n

x = 22.5n degrees, for any odd number n

Any other value for x doesn't work, so therefore that's NOT an identity.

- Login to reply the answers

- 1 month ago
It's not an identity because this isn't true for all values of x. However, we can solve for x

cos(2x)^2 - sin(2x)^2 = 0

cos(2 * 2x) = 0

cos(4x) = 0

4x = pi/2 + pi * k

4x = (pi/2) * (1 + 2k)

x = (pi/8) * (1 + 2k)

x = pi/8 , 3pi/8 , 5pi/8 , 7pi/8 , 9pi/8 , 11pi/8 , 13pi/8 , 15pi/8

k is an integer

cos(2x)^2 - sin(2x)2 = 0

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

1 = tan(2x)^2

-1 , 1 = tan(2x)

pi/4 + pi * k , 3pi/4 + pi * k = 2x

pi/4 + (pi/2) * k = 2x

(pi/4) * (1 + 2k) = 2x

(pi/8) * (1 + 2k) = x

Which is exactly what we had before.

- Login to reply the answers