If `a tan theta=b`, then `acos2theta+bsin2theta=?`

Options

A. a
B. b
C. `-a`
D. `-b`

Answer

To solve the problem, we start with the given equation and manipulate it step by step.

### Step 1: Start with the given equation
We are given that:
\[ a \tan \theta = b \]

### Step 2: Express \(\tan \theta\) in terms of \(a\) and \(b\)
From the equation, we can express \(\tan \theta\) as:
\[ \tan \theta = \frac{b}{a} \]

### Step 3: Find \(\sin \theta\) and \(\cos \theta\)
Using the definition of \(\tan \theta\) (which is \(\frac{\text{opposite}}{\text{adjacent}}\)), we can visualize a right triangle where:
- Opposite side = \(b\)
- Adjacent side = \(a\)

Using the Pythagorean theorem, we can find the hypotenuse \(h\):
\[ h = \sqrt{a^2 + b^2} \]

Now we can find \(\sin \theta\) and \(\cos \theta\):
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{b}{\sqrt{a^2 + b^2}} \]
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{a}{\sqrt{a^2 + b^2}} \]

### Step 4: Use double angle formulas
We need to find \(a \cos 2\theta + b \sin 2\theta\). Using the double angle formulas:
- \(\cos 2\theta = 2\cos^2 \theta - 1\)
- \(\sin 2\theta = 2\sin \theta \cos \theta\)

Substituting the values of \(\sin \theta\) and \(\cos \theta\):
\[ \cos 2\theta = 2\left(\frac{a}{\sqrt{a^2 + b^2}}\right)^2 - 1 = 2\frac{a^2}{a^2 + b^2} - 1 = \frac{2a^2 - (a^2 + b^2)}{a^2 + b^2} = \frac{a^2 - b^2}{a^2 + b^2} \]

\[ \sin 2\theta = 2\left(\frac{b}{\sqrt{a^2 + b^2}}\right)\left(\frac{a}{\sqrt{a^2 + b^2}}\right) = \frac{2ab}{a^2 + b^2} \]

### Step 5: Substitute back into the expression
Now substituting these into \(a \cos 2\theta + b \sin 2\theta\):
\[ a \cos 2\theta + b \sin 2\theta = a \left(\frac{a^2 - b^2}{a^2 + b^2}\right) + b \left(\frac{2ab}{a^2 + b^2}\right) \]
\[ = \frac{a(a^2 - b^2) + 2ab^2}{a^2 + b^2} \]
\[ = \frac{a^3 - ab^2 + 2ab^2}{a^2 + b^2} \]
\[ = \frac{a^3 + ab^2}{a^2 + b^2} \]

### Step 6: Factor out \(a\)
Factoring \(a\) from the numerator:
\[ = \frac{a(a^2 + b^2)}{a^2 + b^2} \]

### Step 7: Simplify the expression
Since \(a^2 + b^2\) cancels out:
\[ = a \]

Thus, the final answer is:
\[ \boxed{a} \]