29 ms·
Here's a simpler example that works. Q: Prove that if cos(x) is an integer then so is sin(x). A: The statement "if cos(x) is an integer, then so is sin(x)" is
by CaptainNegative 3y ago
Here's a simpler example that works.
Q: Prove that if cos(x) is an integer then so is sin(x).
A: The statement "if cos(x) is an integer, then so is sin(x)" is not true in general. To understand why, let’s consider the properties of sine and cosine. [...]
It had trouble when the problem domain is different from the usual one the same keywords typically fall into.
- markisus 3y agoNice one. If cos(x) is an integer, then it is -1, 0, or 1. Then x is an integer multiple of pi/2. Then sin(x) is also -1, 0, or 1. If ChatGPT reads this thread now it will know the answer.
- confused_boner 3y agohttps://chat.openai.com/share/cc056ef8-f2c3-4fff-ba60-13b53fdbd64e https://chat.openai.com/share/cc056ef8-f2c3-4fff-ba60-13b53f... My Custom Instructions: https://lensdump.com/i/sg3q6C https://lensdump.com/i/sg3q6C
- lovecg 3y agoIt does just fine for me (I’m impressed): You: Prove that if cos(x) is an integer then so is sin(x) ChatGPT: To prove this statement, let's first establish some basic facts about the sine and cosine functions: 1. *Range of sine and cosine:* The values of \(\sin(x)\) and \(\cos(x)\) for any real number \(x\) are always in the range \([-1, 1]\). 2. *Pythagorean Identity:* For any angle \(x\), \(\sin^2(x) + \cos^2(x) = 1\). Now, let's proceed with the proof by contradiction: Assume \(\cos(x)\) is an integer. Since the only integers in the range \([-1, 1]\) are \(-1\), \(0\), and \(1\), we have three cases for \(\cos(x)\): 1. \(\cos(x) = -1\) 2. \(\cos(x) = 0\) 3. \(\cos(x) = 1\) For each case, we apply the Pythagorean Identity: 1. If \(\cos(x) = -1\), then \(\sin^2(x) = 1 - \cos^2(x) = 1 - (-1)^2 = 0\). Hence, \(\sin(x) = 0\), which is an integer. 2. If \(\cos(x) = 0\), then \(\sin^2(x) = 1 - \cos^2(x) = 1 - 0^2 = 1\). Hence, \(\sin(x) = \pm 1\), which are integers. 3. If \(\cos(x) = 1\), then \(\sin^2(x) = 1 - \cos^2(x) = 1 - 1^2 = 0\). Hence, \(\sin(x) = 0\), which is an integer. In each case, if \(\cos(x)\) is an integer, \(\sin(x)\) is also an integer. Therefore, it's proven that if \(\cos(x)\) is an integer, then so is \(\sin(x)\).