Question

$\frac{sin3\theta-cos^3\theta}{sin\theta - cos\theta}-\frac{cos\theta}{\sqrt{1+cot^2\theta}}-2tan\theta cot\theta = -1$

• A. $\theta \in (0,\frac{\pi}{2})$
• B. $\theta \in (\frac{\pi}{2},\pi)$
• C. $\theta \in (\pi,\frac{3\pi}{2})$
• D. $\theta \in (\frac{3\pi}{2},2\pi)$

Answer 1

Answer: (A), (B)

①, ③

Answer 2

The given equation is: $\frac{\sin^3\theta-\cos^3\theta}{\sin\theta - \cos\theta}-\frac{\cos\theta}{\sqrt{1+\cot^2\theta}}-2\tan\theta \cot\theta = -1$

Let's simplify each term:

Term 1: $\frac{\sin^3\theta-\cos^3\theta}{\sin\theta - \cos\theta}$ Using the identity $a^3-b^3 = (a-b)(a^2+ab+b^2)$, we get: $\frac{(\sin\theta - \cos\theta)(\sin^2\theta + \sin\theta\cos\theta + \cos^2\theta)}{\sin\theta - \cos\theta}$ Provided $\sin\theta - \cos\theta \neq 0$ (i.e., $\theta \neq n\pi + \frac{\pi}{4}$), this simplifies to: $\sin^2\theta + \cos^2\theta + \sin\theta\cos\theta = 1 + \sin\theta\cos\theta$

Term 2: $\frac{\cos\theta}{\sqrt{1+\cot^2\theta}}$ Using the identity $1+\cot^2\theta = \csc^2\theta$: $\frac{\cos\theta}{\sqrt{\csc^2\theta}} = \frac{\cos\theta}{|\csc\theta|}$ Since $\csc\theta = \frac{1}{\sin\theta}$, we have $|\csc\theta| = \frac{1}{|\sin\theta|}$. So, the term becomes $\cos\theta |\sin\theta|$.

Term 3: $2\tan\theta \cot\theta$ Using the identity $\tan\theta \cot\theta = 1$: $2 \tan\theta \cot\theta = 2(1) = 2$ This is valid provided $\tan\theta$ and $\cot\theta$ are defined, which means $\sin\theta \neq 0$ and $\cos\theta \neq 0$. So $\theta \neq n\frac{\pi}{2}$.

Now, substitute the simplified terms back into the equation: $(1 + \sin\theta\cos\theta) - \cos\theta |\sin\theta| - 2 = -1$ $\sin\theta\cos\theta - \cos\theta |\sin\theta| - 1 = -1$ $\sin\theta\cos\theta - \cos\theta |\sin\theta| = 0$ Factor out $\cos\theta$: $\cos\theta (\sin\theta - |\sin\theta|) = 0$ This equation holds if either $\cos\theta = 0$ or $\sin\theta - |\sin\theta| = 0$.

Case 1: $\cos\theta = 0$ This implies $\theta = (2n+1)\frac{\pi}{2}$ for integer $n$. For $\theta \in (0, 2\pi)$, possible values are $\theta = \frac{\pi}{2}$ and $\theta = \frac{3\pi}{2}$. However, the original expression is undefined when $\cos\theta = 0$ (because $\tan\theta$ is undefined). Therefore, these values are not solutions.

Case 2: $\sin\theta - |\sin\theta| = 0$ This implies $\sin\theta = |\sin\theta|$. This condition is true if and only if $\sin\theta \ge 0$. For $\theta \in (0, 2\pi)$, $\sin\theta \ge 0$ when $\theta \in (0, \pi]$.

Now, we must consider the domain restrictions for the original expression:

1. $\sin\theta - \cos\theta \neq 0 \implies \tan\theta \neq 1 \implies \theta \neq n\pi + \frac{\pi}{4}$. For $\theta \in (0, 2\pi)$, this means $\theta \neq \frac{\pi}{4}$ and $\theta \neq \frac{5\pi}{4}$.
2. $\cot\theta$ is defined $\implies \sin\theta \neq 0 \implies \theta \neq n\pi$. For $\theta \in (0, 2\pi)$, this means $\theta \neq \pi$. (Note: $\theta \neq 0, 2\pi$ are already excluded by the open interval).
3. $\tan\theta$ is defined $\implies \cos\theta \neq 0 \implies \theta \neq (2n+1)\frac{\pi}{2}$. For $\theta \in (0, 2\pi)$, this means $\theta \neq \frac{\pi}{2}$ and $\theta \neq \frac{3\pi}{2}$.

Combining the condition $\sin\theta \ge 0$ (i.e., $\theta \in (0, \pi]$) with the domain restrictions: We must exclude $\frac{\pi}{4}$, $\frac{\pi}{2}$, and $\pi$ from the interval $(0, \pi]$. So, the solution set for $\theta$ is $\theta \in (0, \frac{\pi}{4}) \cup (\frac{\pi}{4}, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi)$.

Let's check the given options: ① $\theta \in (0,\frac{\pi}{2})$: This interval is $(0, \frac{\pi}{2})$. The solution set within this interval is $(0, \frac{\pi}{4}) \cup (\frac{\pi}{4}, \frac{\pi}{2})$. This option contains the values that are solutions, but also includes the excluded point $\frac{\pi}{4}$. However, if the question asks for an interval where the equation holds except for specific points, this is a common way to represent it.

③ $\theta \in (\frac{\pi}{2},\pi)$: This interval is $(\frac{\pi}{2}, \pi)$. All values in this interval satisfy $\sin\theta > 0$. The excluded points $\frac{\pi}{4}, \frac{5\pi}{4}, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$ are not in this open interval. So, for all $\theta \in (\frac{\pi}{2}, \pi)$, the equation holds.

② $\theta \in (\pi,\frac{3\pi}{2})$: In this interval, $\sin\theta < 0$. So $\sin\theta - |\sin\theta| = \sin\theta - (-\sin\theta) = 2\sin\theta$. The equation becomes $\cos\theta (2\sin\theta) = 0$. This implies either $\cos\theta = 0$ or $\sin\theta = 0$. For $\theta \in (\pi, \frac{3\pi}{2})$, $\cos\theta < 0$ and $\sin\theta < 0$, so neither is zero. Thus, no solutions in this interval.

④ $\theta \in (\frac{3\pi}{2},2\pi)$: In this interval, $\sin\theta < 0$. Similar to option ②, the equation becomes $2\sin\theta\cos\theta = 0$, which has no solutions in this interval.

Therefore, the intervals where the equation holds are $(0, \frac{\pi}{4}) \cup (\frac{\pi}{4}, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi)$. Both option ① and option ③ represent parts of this solution set. Option ① is $(0, \frac{\pi}{2})$, which contains the solutions $(0, \frac{\pi}{4}) \cup (\frac{\pi}{4}, \frac{\pi}{2})$. Option ③ is $(\frac{\pi}{2}, \pi)$, which contains the solutions $(\frac{\pi}{2}, \pi)$.

Since the question asks to select the correct options, and both intervals are part of the solution set where the equation is true (excluding the points where the expression is undefined), both ① and ③ are correct.