Question

If $3\sin\theta + 4\cos\theta = 3$ and $\theta \neq (2n+1)\frac{\pi}{2}$, then $\sin 2\theta =$

• A. $\dfrac{336}{625}$
• B. $\dfrac{-7}{25}$
• C. $\dfrac{24}{25}$
• D. $\dfrac{-336}{625}$

Hint:

The tangent half-angle substitution is very powerful for equations of the form $a\sin\theta + b\cos\theta = c$. However, remember that this substitution is not defined for $\theta = (2n+1)\pi$, as $\tan(\theta/2)$ would be undefined. Also, be sure to check for extraneous solutions introduced by the algebraic manipulation, as seen here with the $\theta \neq (2n+1)\pi/2$ condition.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
This problem involves solving a trigonometric equation of the form $a\sin\theta + b\cos\theta = c$. A standard method is to use the tangent half-angle substitution, which converts the trigonometric equation into an algebraic equation.
Step 2: Key Formula or Approach
We use the Weierstrass substitution (tangent half-angle formulas): Let $t = \tan(\theta/2)$. Then: \[ \sin\theta = \frac{2t}{1+t^2} \quad \text{and} \quad \cos\theta = \frac{1-t^2}{1+t^2} \] We substitute these into the given equation to solve for $t$. Once we find $t$, we can find $\sin\theta$ and $\cos\theta$, and finally $\sin(2\theta) = 2\sin\theta\cos\theta$.
Step 3: Detailed Explanation
The given equation is $3\sin\theta + 4\cos\theta = 3$. Substitute the half-angle formulas: \[ 3\left(\frac{2t}{1+t^2}\right) + 4\left(\frac{1-t^2}{1+t^2}\right) = 3 \] Multiply the entire equation by $(1+t^2)$ to clear the denominator: \[ 6t + 4(1-t^2) = 3(1+t^2) \] \[ 6t + 4 - 4t^2 = 3 + 3t^2 \] Rearrange the terms to form a quadratic equation in $t$: \[ 7t^2 - 6t - 1 = 0 \] Factor the quadratic equation: \[ 7t^2 - 7t + t - 1 = 0 \] \[ 7t(t-1) + 1(t-1) = 0 \] \[ (7t+1)(t-1) = 0 \] This gives two possible values for $t = \tan(\theta/2)$: Case 1: $t = 1$. If $\tan(\theta/2) = 1$, then $\theta/2 = k\pi + \pi/4 \implies \theta = 2k\pi + \pi/2$. This is of the form $(4m+1)\pi/2$, which falls under the restricted values $\theta \neq (2n+1)\frac{\pi}{2}$. Therefore, we must discard this solution. Case 2: $t = -1/7$. This is the valid solution. Now we find $\sin\theta$ and $\cos\theta$ using this value of $t$. \[ \sin\theta = \frac{2t}{1+t^2} = \frac{2(-1/7)}{1+(-1/7)^2} = \frac{-2/7}{1+1/49} = \frac{-2/7}{50/49} = \frac{-2}{7} \times \frac{49}{50} = -\frac{7}{25} \] \[ \cos\theta = \frac{1-t^2}{1+t^2} = \frac{1-(-1/7)^2}{1+(-1/7)^2} = \frac{1-1/49}{1+1/49} = \frac{48/49}{50/49} = \frac{48}{50} = \frac{24}{25} \] Now, we calculate $\sin(2\theta)$: \[ \sin(2\theta) = 2\sin\theta\cos\theta = 2 \left(-\frac{7}{25}\right) \left(\frac{24}{25}\right) = -\frac{2 \times 7 \times 24}{625} = -\frac{336}{625} \] Step 4: Final Answer
The value of $\sin(2\theta)$ is $-\frac{336}{625}$. (Note: Options were not provided in the source).