` sin 100` in terms of functions of ` 5 theta ,`

Answer

To express \( \sin 100^\circ \) in terms of functions of \( 5\theta \), we can follow these steps:

### Step 1: Express \( 100^\circ \) in terms of \( 5\theta \)
We know that \( 100^\circ \) can be rewritten as \( 5 \times 20^\circ \). Thus, we can set \( \theta = 20^\circ \).

### Step 2: Use the sine double angle identity
Using the identity for sine, we have:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
In our case, we can express \( \sin(100^\circ) \) as:
\[ \sin(100^\circ) = \sin(5 \times 20^\circ) = \sin(5\theta) \]

### Step 3: Apply the sine multiple angle formula
Using the multiple angle formula for sine, we can express \( \sin(5\theta) \):
\[ \sin(5\theta) = 16 \sin^5(\theta) - 20 \sin^3(\theta) + 5 \sin(\theta) \]
Now substituting \( \theta = 20^\circ \):
\[ \sin(100^\circ) = 16 \sin^5(20^\circ) - 20 \sin^3(20^\circ) + 5 \sin(20^\circ) \]

### Final Expression
Thus, we can express \( \sin(100^\circ) \) in terms of \( \sin(20^\circ) \):
\[ \sin(100^\circ) = 16 \sin^5(20^\circ) - 20 \sin^3(20^\circ) + 5 \sin(20^\circ) \]