Solving Trigonometric Equations: Finding the Value of Theta

We are given a trigonometric equation involving sine and cosine functions, and we need to find the value of the angle $\theta$. The given equation is:

\(\sin 3\theta = \cos (\theta - 6^\circ)\)

To solve this equation, we can use the relationship between sine and cosine of complementary angles. Two angles are complementary if their sum is $90^\circ$. The key trigonometric identity related to complementary angles is:

\(\sin x = \cos (90^\circ - x)\)

or equivalently,

\(\cos x = \sin (90^\circ - x)\)

Using this identity, we can convert either the sine term to a cosine term or the cosine term to a sine term.

Applying the Complementary Angle Identity

Let's convert the cosine term $\cos (\theta - 6^\circ)$ into a sine term using the identity $\cos x = \sin (90^\circ - x)$. Here, $x = \theta - 6^\circ$.

\(\cos (\theta - 6^\circ) = \sin (90^\circ - (\theta - 6^\circ))\)

Simplifying the angle inside the sine function:

\(90^\circ - (\theta - 6^\circ) = 90^\circ - \theta + 6^\circ = 96^\circ - \theta\)

So, the equation becomes:

\(\sin 3\theta = \sin (96^\circ - \theta)\)

Solving for Theta

If $\sin A = \sin B$, then for angles typically encountered in these types of problems (often within the range $0^\circ$ to $90^\circ$ for $3\theta$ and $\theta - 6^\circ$), the general solution involves \(A = n \cdot 180^\circ + (-1)^n B\). However, for simple cases where angles are expected to be acute, we can often consider the principal solution where the angles are equal or complementary in a specific way within the sine function's periodicity.

A common case when $\sin A = \sin B$ for acute angles is $A = B$. Let's assume this case first:

\(3\theta = 96^\circ - \theta\)

Now, we solve this linear equation for $\theta$:

• Add \(\theta\) to both sides of the equation:
• \(3\theta + \theta = 96^\circ\)
• \(4\theta = 96^\circ\)
• Divide both sides by 4:
• \(\theta = \frac{96^\circ}{4}\)
• \(\theta = 24^\circ\)

Verification

Let's check if $\theta = 24^\circ$ satisfies the original equation $\sin 3\theta = \cos (\theta - 6^\circ)$.

• Left side: \(\sin (3 \times 24^\circ) = \sin 72^\circ\)
• Right side: \(\cos (24^\circ - 6^\circ) = \cos 18^\circ\)

We know that $\sin 72^\circ = \sin (90^\circ - 18^\circ)$, and by the complementary angle identity, $\sin (90^\circ - 18^\circ) = \cos 18^\circ$.

So, $\sin 72^\circ = \cos 18^\circ$, which means the value $\theta = 24^\circ$ satisfies the equation.

Another way to think about $\sin A = \cos B$ for acute angles A and B is that their sum must be $90^\circ$. That is, $A + B = 90^\circ$.

In our equation, $A = 3\theta$ and $B = \theta - 6^\circ$.

So, \(3\theta + (\theta - 6^\circ) = 90^\circ\).

\(4\theta - 6^\circ = 90^\circ\)

\(4\theta = 90^\circ + 6^\circ\)

\(4\theta = 96^\circ\)

\(\theta = \frac{96^\circ}{4}\)

\(\theta = 24^\circ\)

Both methods lead to the same result.

Revision Table: Trigonometric Identities

Additional Information: General Solutions for Trigonometric Equations

While we used the simple case $A=B$ for $\sin A = \sin B$, the general solution is \(A = n \cdot 180^\circ + (-1)^n B\), where \(n\) is an integer. In our case, $A = 3\theta$ and $B = 96^\circ - \theta$.

So, \(3\theta = n \cdot 180^\circ + (-1)^n (96^\circ - \theta)\).

Case 1: \(n\) is even (let \(n = 2k\))

\(3\theta = 2k \cdot 180^\circ + (96^\circ - \theta)\)

\(3\theta + \theta = 360^\circ k + 96^\circ\)

\(4\theta = 360^\circ k + 96^\circ\)

\(\theta = 90^\circ k + 24^\circ\)

If \(k=0\), \(\theta = 24^\circ\). If \(k=1\), \(\theta = 114^\circ\). If \(k=-1\), \(\theta = -66^\circ\). And so on.

Case 2: \(n\) is odd (let \(n = 2k + 1\))

\(3\theta = (2k + 1) \cdot 180^\circ - (96^\circ - \theta)\)

\(3\theta = 360^\circ k + 180^\circ - 96^\circ + \theta\)

\(3\theta - \theta = 360^\circ k + 84^\circ\)

\(2\theta = 360^\circ k + 84^\circ\)

\(\theta = 180^\circ k + 42^\circ\)

If \(k=0\), \(\theta = 42^\circ\). If \(k=1\), \(\theta = 222^\circ\). And so on.

When solving problems from options in tests, the simplest and most common solution for acute angles is usually the one obtained from equating the angles directly or using the \(A+B=90^\circ\) rule for $\sin A = \cos B$. The value $\theta = 24^\circ$ is one of the possible solutions from the general form (when \(k=0\)). The options provided suggest we are looking for a specific principal value, likely acute or positive.