If `sin theta =a and sin 2 theta = b,` find an expression for ` cos theta ` in terms of a and b. Hence find a relation between a and b not involving `theta.`

Answer

To solve the problem, we start with the given information:

1. \( \sin \theta = a \)
2. \( \sin 2\theta = b \)

We need to find an expression for \( \cos \theta \) in terms of \( a \) and \( b \), and then derive a relationship between \( a \) and \( b \) that does not involve \( \theta \).

### Step 1: Use the double angle formula for sine

The double angle formula for sine states that:

\[ \sin 2\theta = 2 \sin \theta \cos \theta \]

Substituting the known values:

\[ b = 2a \cos \theta \]

### Step 2: Solve for \( \cos \theta \)

From the equation \( b = 2a \cos \theta \), we can isolate \( \cos \theta \):

\[ \cos \theta = \frac{b}{2a} \]

### Step 3: Use the Pythagorean identity

We know from trigonometric identities that:

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

Substituting \( \sin \theta = a \) and \( \cos \theta = \frac{b}{2a} \):

\[ a^2 + \left(\frac{b}{2a}\right)^2 = 1 \]

### Step 4: Simplify the equation

Calculating \( \left(\frac{b}{2a}\right)^2 \):

\[ \left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} \]

Now substituting this back into the equation gives:

\[ a^2 + \frac{b^2}{4a^2} = 1 \]

### Step 5: Multiply through by \( 4a^2 \) to eliminate the fraction

\[ 4a^4 + b^2 = 4a^2 \]

### Step 6: Rearranging the equation

Rearranging gives:

\[ 4a^4 - 4a^2 + b^2 = 0 \]

### Step 7: Factor out common terms

We can factor out \( 4a^2 \):

\[ 4a^2(a^2 - 1) + b^2 = 0 \]

### Step 8: Isolate \( b^2 \)

Rearranging gives:

\[ b^2 = 4a^2(1 - a^2) \]

### Step 9: Taking the square root

Taking the square root of both sides gives:

\[ b = 2a \sqrt{1 - a^2} \]

### Final Result

Thus, we have derived the expression for \( \cos \theta \) and the relationship between \( a \) and \( b \):

1. \( \cos \theta = \frac{b}{2a} \)
2. \( b = 2a \sqrt{1 - a^2} \)