Solving Trigonometric Equation $\cos 2\theta = \sin \theta$

We are given the trigonometric equation $\cos 2\theta = \sin \theta$, and we need to find the value of $\theta$ such that $0^\circ < \theta < 90^\circ$.

There are several ways to solve this trigonometric equation. Let's explore two common methods using trigonometric identities.

Method 1: Using Complementary Angle Identity

We know that $\sin \theta = \cos(90^\circ - \theta)$. Using this identity, we can rewrite the given equation:

$\cos 2\theta = \cos(90^\circ - \theta)$

For the cosine of two angles to be equal, the angles must be related in the form $A = 360^\circ n \pm B$, where $n$ is an integer. So, we have two cases:

Case 1: $2\theta = 360^\circ n + (90^\circ - \theta)$

• $2\theta + \theta = 360^\circ n + 90^\circ$
• $3\theta = 360^\circ n + 90^\circ$
• $\theta = \frac{360^\circ n + 90^\circ}{3}$
• $\theta = 120^\circ n + 30^\circ$

We are given that $0^\circ < \theta < 90^\circ$. Let's find values of $n$ that satisfy this condition:

• If $n = 0$, $\theta = 120^\circ(0) + 30^\circ = 30^\circ$. This value is within the range $0^\circ < \theta < 90^\circ$.
• If $n = 1$, $\theta = 120^\circ(1) + 30^\circ = 150^\circ$. This value is not within the range.
• If $n = -1$, $\theta = 120^\circ(-1) + 30^\circ = -120^\circ + 30^\circ = -90^\circ$. This value is not within the range.

Case 2: $2\theta = 360^\circ n - (90^\circ - \theta)$

• $2\theta = 360^\circ n - 90^\circ + \theta$
• $2\theta - \theta = 360^\circ n - 90^\circ$
• $\theta = 360^\circ n - 90^\circ$

Again, considering the range $0^\circ < \theta < 90^\circ$:

• If $n = 0$, $\theta = 360^\circ(0) - 90^\circ = -90^\circ$. Not in the range.
• If $n = 1$, $\theta = 360^\circ(1) - 90^\circ = 270^\circ$. Not in the range.

From Method 1, the only solution in the given range is $\theta = 30^\circ$.

Method 2: Using Double Angle Identity for Cosine

We know that $\cos 2\theta = 1 - 2\sin^2 \theta$. Substitute this into the given equation:

$1 - 2\sin^2 \theta = \sin \theta$

Rearrange this into a quadratic equation in terms of $\sin \theta$:

$2\sin^2 \theta + \sin \theta - 1 = 0$

Let $x = \sin \theta$. The equation becomes:

$2x^2 + x - 1 = 0$

Factor the quadratic equation:

$(2x - 1)(x + 1) = 0$

This gives two possible solutions for $x$:

• $2x - 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}$
• $x + 1 = 0 \Rightarrow x = -1$

Now substitute back $\sin \theta = x$:

• $\sin \theta = \frac{1}{2}$
• $\sin \theta = -1$

We are given that $0^\circ < \theta < 90^\circ$. In this range, the sine function is positive.

So, $\sin \theta = \frac{1}{2}$ is the relevant solution. For $\sin \theta = \frac{1}{2}$ in the first quadrant ($0^\circ$ to $90^\circ$), the angle $\theta$ is $30^\circ$.

The solution $\sin \theta = -1$ gives $\theta = 270^\circ + 360^\circ n$, which is not in the specified range $0^\circ < \theta < 90^\circ$.

Both methods confirm that the solution to the equation $\cos 2\theta = \sin \theta$ in the range $0^\circ < \theta < 90^\circ$ is $\theta = 30^\circ$.

Final Answer Determination

Based on the calculations using trigonometric identities, the value of $\theta$ that satisfies the given conditions is $30^\circ$. Let's check this value:

• Left side: $\cos(2 \times 30^\circ) = \cos(60^\circ) = \frac{1}{2}$
• Right side: $\sin(30^\circ) = \frac{1}{2}$

Since the left side equals the right side, $\theta = 30^\circ$ is the correct solution. Also, $30^\circ$ lies between $0^\circ$ and $90^\circ$.

Revision Table: Key Trigonometric Identities

Additional Information: Solving Trigonometric Equations

Solving trigonometric equations often involves using identities to simplify the equation or express it in terms of a single trigonometric function. Once simplified, the equation might become a linear or quadratic equation in terms of the trigonometric function (like $\sin \theta$, $\cos \theta$, or $\tan \theta$).

It is crucial to consider the specified range for the angle $\theta$. Trigonometric functions are periodic, meaning they repeat their values at regular intervals. Without a range restriction, there would be infinitely many solutions. The given range helps narrow down the possible values of $\theta$.

When using inverse trigonometric functions (like $\arcsin$, $\arccos$, $\arctan$), remember that they provide the principal value. You often need to consider the periodicity and the quadrant(s) based on the sign of the trigonometric value and the given range to find all solutions within that range.

In this problem, the range $0^\circ < \theta < 90^\circ$ limited the solutions significantly, allowing us to easily pick the correct value from the possible results obtained after solving the equation.