a. Given that sin2 𝑥, cos2 𝑥 𝑎𝑛𝑑 5cos2𝑥 − 3sin2 𝑥 are in arithmetic progression, find the value of

Question

i. cos2𝑥
ii. The common difference

Answer 1

We know that the common difference for an arithmetic progression is equal to the difference between any two consecutive terms. So we can start by finding the difference between sin2 𝑥 and cos2 𝑥:

cos2 𝑥 - sin2 𝑥

Using the identity sin2 𝑥 + cos2 𝑥 = 1, we can write cos2 𝑥 = 1 - sin2 𝑥. Substituting this into the equation above, we get:

(1 - sin2 𝑥) - sin2 𝑥
= 1 - 2sin2 𝑥

Now we can find the common difference by subtracting 5cos2𝑥 − 3sin2 𝑥 from 1 - 2sin2 𝑥:

(1 - 2sin2 𝑥) - (5cos2𝑥 − 3sin2 𝑥)
= 1 - 2sin2 𝑥 - 5cos2𝑥 + 3sin2 𝑥
= 1 - 5cos2𝑥 + sin2 𝑥

We want this expression to be equal to the common difference, so:

1 - 5cos2𝑥 + sin2 𝑥 = d

where d is the common difference.

Now we have two equations:

cos2 𝑥 = 1 - sin2 𝑥
1 - 5cos2𝑥 + sin2 𝑥 = d

We can substitute the first equation into the second to get an equation in terms of sin2 𝑥:

1 - 5(1 - cos2 𝑥) + sin2 𝑥 = d
1 - 5(1 - (1 - sin2 𝑥)) + sin2 𝑥 = d
6sin2 𝑥 - 4 = d

So the common difference is 6sin2 𝑥 - 4. We can now use this to find cos2 𝑥:

cos2 𝑥 = sin2 𝑥 + (6sin2 𝑥 - 4)
cos2 𝑥 = 7sin2 𝑥 - 4

We don't have enough information to solve for sin2 𝑥, but this is the value of cos2 𝑥 in terms of sin2 𝑥.

Answer 2

cos^2x - sin^2x = 5cos2x - 3sin^2x
cos2x = 5cos2x - 3(1 - cos2x)/2
cos2x = 3/11
d = 5(3/11) - 3(1 - 3/11)/2 = 3/11

Answer 3

Your value for cos2x is incorrect. Let's start with the equation:

cos^2x - sin^2x = 5cos2x - 3sin^2x

We can use the identity cos^2x + sin^2x = 1 to substitute for sin^2x:

cos^2x - (1 - cos^2x) = 5cos2x - 3(1 - cos^2x)

Simplifying, we get:

2cos^2x - 1 = 8cos^2x - 3

6cos^2x = 4

cos^2x = 2/3

Now we can find cos2x:

cos2x = 2cos^2x - 1

cos2x = 2(2/3) - 1

cos2x = 1/3

To find the common difference, we use the formula:

d = (5cos2x - 3sin^2x) - (cos^2x - sin^2x)

Substituting our values, we get:

d = (5/3) - (2/3)

d = 1

So the value of cos2x is 1/3 and the common difference is 1.