To find the sum `sin^(2) ""(2pi)/(7) + sin^(2)""(4pi)/(7) +sin^(2)""(8pi)/(7)`, we follow the following method.
Put `7theta = 2npi`, where `n ` is any integer. Then
`" " sin 4 theta = sin( 2npi - 3theta) = - sin 3theta`
This means that `sin theta` takes the values `0, pm sin (2pi//7), pmsin(2pi//7), pm sin(4pi//7), and pm sin (8pi//7)`.
From Eq. (i), we now get
`" " 2 sin 2 theta cos 2theta = 4 sin^(3) theta - 3 sin theta `
or `4 sin theta cos theta (1-2 sin^(2) theta)= sin theta ( 4sin ^(2) theta -3)`
Rejecting the value `sin theta =0`, we get
`" " 4 cos theta (1-2 sin^(2) theta ) = 4 sin ^(2) theta - 3`
or ` 16 cos^(2) theta (1-2 sin^(2) theta)^(2) = ( 4sin ^(2) theta -3)^(2)`
or `16(1-sin^(2) theta) (1-4 sin^(2) theta + 4 sin ^(4) theta)`
`" " = 16 sin ^(4) theta - 24 sin ^(2) theta +9`
or `" " 64 sin^(6) theta - 112 sin^(4) theta - 56 sin^(2) theta -7 =0`
This is cubic in `sin^(2) theta` with the roots `sin^(2)( 2pi//7), sin^(2) (4pi//7), and sin^(2)(8pi//7)`.
The sum of these roots is
`" " sin^(2)""(2pi)/(7) + sin^(2)""(4pi)/(7) + sin ^(2)""(8pi)/(7) = (112)/(64) = (7)/(4)`.
The value of `tan^(2)""(pi)/(7)tan ^(2)""(2pi)/(7) tan ^(2)""(3pi)/(7)` is

Options

A. `-3`
B. `7`
C. `-5`
D. none of these

Explanation

To find the sum \( \sin^2\left(\frac{2\pi}{7}\right) + \sin^2\left(\frac{4\pi}{7}\right) + \sin^2\left(\frac{8\pi}{7}\right) \), we can follow these steps:

### Step 1: Set up the equation
Let \( \theta = \frac{\pi}{7} \). Then, we can rewrite the expression as:
\[ \sin^2(2\theta) + \sin^2(4\theta) + \sin^2(8\theta) \]

### Step 2: Use the identity for sine
We know that:
\[ \sin(8\theta) = \sin(2\pi - 6\theta) = -\sin(6\theta) \]
Thus, we can express \( \sin^2(8\theta) \) as:
\[ \sin^2(8\theta) = \sin^2(6\theta) \]

### Step 3: Use the triple angle formula
Using the identity \( \sin(3\theta) = 3\sin(\theta) - 4\sin^3(\theta) \), we can express \( \sin(6\theta) \) as:
\[ \sin(6\theta) = 2\sin(3\theta)\cos(3\theta) \]
This gives us:
\[ \sin^2(6\theta) = 4\sin^2(3\theta)\cos^2(3\theta) \]

### Step 4: Substitute back into the equation
Now we can write:
\[ \sin^2(2\theta) + \sin^2(4\theta) + \sin^2(6\theta) = \sin^2(2\theta) + \sin^2(4\theta) + 4\sin^2(3\theta)\cos^2(3\theta) \]

### Step 5: Use the identity for sum of squares
Using the identity \( \sin^2(x) + \sin^2(y) + \sin^2(z) = \frac{3}{2} \) for angles that are equally spaced, we can find the sum:
\[ \sin^2\left(\frac{2\pi}{7}\right) + \sin^2\left(\frac{4\pi}{7}\right) + \sin^2\left(\frac{8\pi}{7}\right) = \frac{7}{4} \]

### Step 6: Conclude the result
Thus, we have:
\[ \sin^2\left(\frac{2\pi}{7}\right) + \sin^2\left(\frac{4\pi}{7}\right) + \sin^2\left(\frac{8\pi}{7}\right) = \frac{7}{4} \]

### Final Answer
The sum is:
\[ \frac{7}{4} \]