Saurabh Koranglekar
Last Activity: 6 Years ago
To find the maximum value of the expression sin²(2π/3 + x) + sin²(2π/3 - x), we can start by simplifying it using trigonometric identities and properties. Let's break it down step by step.
Understanding the Sine Function
The sine function is periodic and takes values between -1 and 1. For our purposes, we need to focus on the square of the sine function, which will range from 0 to 1. The sine function is also symmetrical, which will help us simplify our expression.
Using Trigonometric Identities
We know that:
- sin²(θ) = (1 - cos(2θ))/2
- sin(a ± b) = sin(a)cos(b) ± cos(a)sin(b)
Let's rewrite our expression using the first identity:
Let A = 2π/3 + x and B = 2π/3 - x. Then, we have:
sin²(A) + sin²(B) = (1 - cos(2A))/2 + (1 - cos(2B))/2
This simplifies to:
sin²(A) + sin²(B) = 1 - (cos(2A) + cos(2B))/2
Finding cos(2A) and cos(2B)
Next, we need to find cos(2A) and cos(2B):
2A = 2(2π/3 + x) = 4π/3 + 2x
2B = 2(2π/3 - x) = 4π/3 - 2x
Thus, we get:
cos(2A) = cos(4π/3 + 2x)
cos(2B) = cos(4π/3 - 2x)
Using the Cosine Addition Formula
Using the cosine addition formula, we can express this as:
cos(4π/3 + 2x) = cos(4π/3)cos(2x) - sin(4π/3)sin(2x
cos(4π/3 - 2x) = cos(4π/3)cos(2x) + sin(4π/3)sin(2x)
Now, since cos(4π/3) = -1/2 and sin(4π/3) = -√3/2, we can substitute these values:
cos(2A) + cos(2B) = -cos(2x) + cos(2x) - √3sin(2x) + √3sin(2x) = -√3sin(2x)
Maximizing the Expression
Thus, we can rewrite our expression as:
sin²(A) + sin²(B) = 1 + (√3/2)sin(2x)
To find the maximum value, we need to maximize sin(2x). The sine function achieves its maximum value of 1. Hence:
Maximum of sin²(A) + sin²(B) = 1 + (√3/2) * 1 = 1 + √3/2.
The Final Result
Therefore, the maximum value of sin²(2π/3 + x) + sin²(2π/3 - x) is:
1 + √3/2.
This value indicates the potential oscillations of the sine function while being influenced by the shifts introduced by the angles involved. Exploring this not only helps in maximizing the function but also deepens our understanding of trigonometric behavior.