We are given the trigonometric equation: \(\sin 2\theta = \cos 40^\circ\). We need to find the smallest positive value of \(\theta\) that satisfies this equation.

To solve this equation, we can use the trigonometric identity that relates sine and cosine: \(\cos x = \sin(90^\circ - x)\). Using this identity, we can rewrite the right side of the equation in terms of sine.

Applying the identity to \(\cos 40^\circ\):

\[ \cos 40^\circ = \sin(90^\circ - 40^\circ) = \sin 50^\circ \]

Now, substitute this back into the original equation:

\[ \sin 2\theta = \sin 50^\circ \]

If \(\sin A = \sin B\), the general solution for \(A\) is given by \(A = n \cdot 180^\circ + (-1)^n B\), where \(n\) is an integer.

In our case, \(A = 2\theta\) and \(B = 50^\circ\). So, we have:

\[ 2\theta = n \cdot 180^\circ + (-1)^n 50^\circ \]

To find the possible values of \(\theta\), we can divide by 2:

\[ \theta = n \cdot 90^\circ + (-1)^n 25^\circ \]

We are looking for the smallest positive value of \(\theta\). Let's examine the values of \(\theta\) for different integer values of \(n\):

• For \(n = 0\): \(\theta = 0 \cdot 90^\circ + (-1)^0 25^\circ = 0^\circ + 1 \cdot 25^\circ = 25^\circ\). This is a positive value.
• For \(n = 1\): \(\theta = 1 \cdot 90^\circ + (-1)^1 25^\circ = 90^\circ - 25^\circ = 65^\circ\). This is a positive value.
• For \(n = 2\): \(\theta = 2 \cdot 90^\circ + (-1)^2 25^\circ = 180^\circ + 25^\circ = 205^\circ\). This is a positive value.
• For \(n = -1\): \(\theta = -1 \cdot 90^\circ + (-1)^{-1} 25^\circ = -90^\circ - 25^\circ = -115^\circ\). This is a negative value.

Comparing the positive values we found (\(25^\circ\), \(65^\circ\), \(205^\circ\), ...), the smallest positive value of \(\theta\) is \(25^\circ\).

Alternatively, when \(\sin A = \sin B\), we can also have \(A = 180^\circ - B + n \cdot 360^\circ\). So another possibility for \(2\theta\) is:

\[ 2\theta = 180^\circ - 50^\circ + n \cdot 360^\circ \]

\[ 2\theta = 130^\circ + n \cdot 360^\circ \]

Divide by 2:

\[ \theta = 65^\circ + n \cdot 180^\circ \]

Let's examine values for integer \(n\):

• For \(n = 0\): \(\theta = 65^\circ + 0 \cdot 180^\circ = 65^\circ\). This is a positive value.
• For \(n = 1\): \(\theta = 65^\circ + 1 \cdot 180^\circ = 65^\circ + 180^\circ = 245^\circ\). This is a positive value.
• For \(n = -1\): \(\theta = 65^\circ - 1 \cdot 180^\circ = 65^\circ - 180^\circ = -115^\circ\). This is a negative value.

The positive values obtained from the second case are \(65^\circ\), \(245^\circ\), etc.

Comparing all positive values from both general solutions (\(25^\circ\), \(65^\circ\), \(205^\circ\), \(245^\circ\), ...), the smallest positive value among these is \(25^\circ\).

Trigonometric Identities Used

The core of solving \(\sin 2\theta = \cos 40^\circ\) involves using angle transformation formulas. We used the complementary angle identity:

• \(\cos x = \sin(90^\circ - x)\)

This identity is crucial for converting cosine into sine, allowing us to equate angles based on the property that if \(\sin A = \sin B\), then \(A\) and \(B\) are related in specific ways.

Steps to Solve sin 2θ = cos 40°

1. Start with the given equation: \(\sin 2\theta = \cos 40^\circ\).
2. Convert \(\cos 40^\circ\) to a sine function using the identity \(\cos x = \sin(90^\circ - x)\). This gives \(\cos 40^\circ = \sin(90^\circ - 40^\circ) = \sin 50^\circ\).
3. Rewrite the equation as \(\sin 2\theta = \sin 50^\circ\).
4. Use the general solution for \(\sin A = \sin B\), which states that \(A = n \cdot 180^\circ + (-1)^n B\) or \(A = 180^\circ - B + n \cdot 360^\circ\) for integer \(n\).
5. Apply the first case: \(2\theta = n \cdot 180^\circ + (-1)^n 50^\circ\). Divide by 2 to get \(\theta = n \cdot 90^\circ + (-1)^n 25^\circ\).
6. Find positive values of \(\theta\) by testing integer values for \(n\). For \(n=0\), \(\theta = 25^\circ\). For \(n=1\), \(\theta = 65^\circ\).
7. Apply the second case (alternative form): \(2\theta = 180^\circ - 50^\circ + n \cdot 360^\circ\), which simplifies to \(2\theta = 130^\circ + n \cdot 360^\circ\). Divide by 2 to get \(\theta = 65^\circ + n \cdot 180^\circ\).
8. Find positive values of \(\theta\) by testing integer values for \(n\). For \(n=0\), \(\theta = 65^\circ\). For \(n=1\), \(\theta = 245^\circ\).
9. Compare all positive values found from both cases (\(25^\circ\), \(65^\circ\), \(205^\circ\), \(65^\circ\), \(245^\circ\), ...). The smallest positive value is \(25^\circ\).

Revision Table: Key Trigonometry Concepts

Additional Information: Understanding Trigonometric Solutions

When solving trigonometric equations like \(\sin 2\theta = \sin 50^\circ\), it's important to remember that trigonometric functions are periodic. This means there are infinitely many angles that give the same sine value. The general solution accounts for all these possibilities.

The general solution for \(\sin A = \sin B\), which is \(A = n \cdot 180^\circ + (-1)^n B\), combines two sets of solutions:

• When \(n\) is even (e.g., \(n=2k\)), \((-1)^n = 1\). The solution becomes \(A = 2k \cdot 180^\circ + B = 360^\circ k + B\). This represents angles that are coterminal with \(B\).
• When \(n\) is odd (e.g., \(n=2k+1\)), \((-1)^n = -1\). The solution becomes \(A = (2k+1) \cdot 180^\circ - B = 360^\circ k + 180^\circ - B\). This represents angles that are coterminal with \(180^\circ - B\).

In our case, \(B=50^\circ\). So the two cases essentially cover \(2\theta = 360^\circ k + 50^\circ\) and \(2\theta = 360^\circ k + 180^\circ - 50^\circ = 360^\circ k + 130^\circ\). Dividing by 2 gives \(\theta = 180^\circ k + 25^\circ\) and \(\theta = 180^\circ k + 65^\circ\). The first form of the general solution \( \theta = n \cdot 90^\circ + (-1)^n 25^\circ \) combines these nicely.

• \(n=0\): \(\theta = 0 \cdot 90^\circ + 25^\circ = 25^\circ\) (from \(180^\circ k + 25^\circ\) with \(k=0\))
• \(n=1\): \(\theta = 1 \cdot 90^\circ - 25^\circ = 65^\circ\) (from \(180^\circ k + 65^\circ\) with \(k=0\))
• \(n=2\): \(\theta = 2 \cdot 90^\circ + 25^\circ = 180^\circ + 25^\circ = 205^\circ\) (from \(180^\circ k + 25^\circ\) with \(k=1\))
• \(n=3\): \(\theta = 3 \cdot 90^\circ - 25^\circ = 270^\circ - 25^\circ = 245^\circ\) (from \(180^\circ k + 65^\circ\) with \(k=1\))

And so on. We needed the smallest positive value, which we found to be \(25^\circ\).