| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine if \(\cos(\frac{\pi}{8})\) is equal to \(\frac{\sqrt{2+\sqrt{2}}}{2}\), we can start by using trigonometric identities and properties. First, recall that \(\cos(2\theta) = 2\cos^2(\theta) - 1\). Let's apply this with \(\theta = \frac{\pi}{8}\), which means \(2\theta = \frac{\pi}{4}\). \[ \cos\left(\frac{\pi}{4}\right) = 2\cos^2\left(\frac{\pi}{8}\right) - 1 \] We know that \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), so: \[ \frac{\sqrt{2}}{2} = 2\cos^2\left(\frac{\pi}{8}\right) - 1 \] Solving for \(\cos\left(\frac{\pi}{8}\right)\): \[ \frac{\sqrt{2}}{2} + 1 = 2\cos^2\left(\frac{\pi}{8}\right) \] \[ \frac{\sqrt{2} + 2}{2} = 2\cos^2\left(\frac{\pi}{8}\right) \] \[ \frac{\sqrt{2} + 2}{4} = \cos^2\left(\frac{\pi}{8}\right) \] \[ \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{\sqrt{2} + 2}{4}} \] \[ \cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{\sqrt{2} + 2}}{2} \] Therefore, \(\cos(\frac{\pi}{8})\) is indeed equal to \(\frac{\sqrt{2+\sqrt{2}}}{2}\). |
True |
| deepseekr1 | To determine if \(\cos\left(\frac{\pi}{8}\right)\) equals \(\frac{\sqrt{2 + \sqrt{2}}}{2}\), we use the **half-angle formula** for cosine: \[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos\theta}{2}} \] Let \(\theta = \frac{\pi}{4}\), so \(\frac{\theta}{2} = \frac{\pi}{8}\). Substitute \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\) into the formula: \[ \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} \] Simplify the numerator inside the square root: \[ 1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2} \] Divide by 2: \[ \frac{\frac{2 + \sqrt{2}}{2}}{2} = \frac{2 + \sqrt{2}}{4} \] Take the square root: \[ \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2} \] **Verification via numerical approximation**: - \(\cos\left(\frac{\pi}{8}\right) \approx 0.92388\) - \(\frac{\sqrt{2 + \sqrt{2}}}{2} \approx \frac{\sqrt{3.4142}}{2} \approx 0.9239\) Both values match, confirming the equality. **Answer**: Yes, \(\cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}\). |
True |
| gpto1 | To determine if the identity \[ \cos\left( \frac{\pi}{8} \right) = \frac{\sqrt{2 + \sqrt{2}}}{2} \] is true, we'll use trigonometric identities and known values. **Step 1: Recognize the Half-Angle Relationship** We note that \(\frac{\pi}{8}\) is half of \(\frac{\pi}{4}\), since: \[ \frac{\pi}{8} = \frac{\pi}{4} \times \frac{1}{2} \] This suggests using the half-angle formula for cosine. **Step 2: Recall the Half-Angle Formula for Cosine** The half-angle formula for cosine is: \[ \cos\left( \frac{\theta}{2} \right) = \sqrt{ \frac{1 + \cos\theta}{2} } \] We choose the positive square root because \(\cos\left( \frac{\pi}{8} \right)\) is positive in the first quadrant. **Step 3: Apply the Half-Angle Formula** Let \(\theta = \frac{\pi}{4}\). Then: \[ \cos\left( \frac{\pi}{8} \right) = \sqrt{ \frac{1 + \cos\left( \frac{\pi}{4} \right)}{2} } \] **Step 4: Substitute the Known Value of \(\cos\left( \frac{\pi}{4} \right)\)** We know that: \[ \cos\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \] Substituting this value: \[ \cos\left( \frac{\pi}{8} \right) = \sqrt{ \frac{1 + \frac{\sqrt{2}}{2}}{2} } \] **Step 5: Simplify the Expression Inside the Square Root** First, combine the terms in the numerator: \[ 1 + \frac{\sqrt{2}}{2} = \frac{2}{2} + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2} \] Now, the expression becomes: \[ \cos\left( \frac{\pi}{8} \right) = \sqrt{ \frac{ \frac{2 + \sqrt{2}}{2} }{2} } \] Simplify the denominator: \[ \frac{ \frac{2 + \sqrt{2}}{2} }{2} = \frac{2 + \sqrt{2}}{2 \times 2} = \frac{2 + \sqrt{2}}{4} \] **Step 6: Simplify the Square Root** Now, we have: \[ \cos\left( \frac{\pi}{8} \right) = \sqrt{ \frac{2 + \sqrt{2}}{4} } \] This simplifies to: \[ \cos\left( \frac{\pi}{8} \right) = \frac{ \sqrt{ 2 + \sqrt{2} } }{ \sqrt{4} } = \frac{ \sqrt{ 2 + \sqrt{2} } }{ 2 } \] **Step 7: Conclusion** Therefore, we have shown: \[ \cos\left( \frac{\pi}{8} \right) = \frac{ \sqrt{ 2 + \sqrt{2} } }{ 2 } \] which confirms the identity is true. **Answer:** Yes; using the half-angle identity shows they are equal: cos(π⁄8) = [√(2 + √2)] ⁄ 2. |
True |
| GPT-4o | To determine if \(\cos(\pi/8)\) is equal to \(\frac{\sqrt{2+\sqrt{2}}}{2}\), we can start by using the half-angle identity for cosine. The half-angle identity states that: \[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos(\theta)}{2}} \] In this case, we are interested in \(\cos(\pi/8)\). We can express \(\pi/8\) as \(\pi/4\) divided by 2, so we will use \(\theta = \pi/4\). First, we need to know \(\cos(\pi/4)\). We know that: \[ \cos(\pi/4) = \frac{\sqrt{2}}{2} \] Using the half-angle identity: \[ \cos\left(\frac{\pi/4}{2}\right) = \cos\left(\pi/8\right) = \sqrt{\frac{1 + \cos(\pi/4)}{2}} \] Substitute \(\cos(\pi/4) = \frac{\sqrt{2}}{2}\): \[ \cos(\pi/8) = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} \] Simplify inside the square root: \[ \cos(\pi/8) = \sqrt{\frac{\frac{2}{2} + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{\frac{2 + \sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} \] This simplifies further to: \[ \cos(\pi/8) = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2} \] Thus, we find that: \[ \cos(\pi/8) = \frac{\sqrt{2 + \sqrt{2}}}{2} \] Therefore, \(\cos(\pi/8)\) is indeed equal to \(\frac{\sqrt{2 + \sqrt{2}}}{2}\). |
True |
| llama3.1-70B-Instruct | A nice trigonometry problem! To solve this, we can use the half-angle formula for cosine: $$\cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos\theta}{2}}$$ In this case, we have $\theta = \frac{\pi}{4}$, so: $$\cos\left(\frac{\pi}{8}\right) = \pm\sqrt{\frac{1 + \cos\left(\frac{\pi}{4}\right)}{2}}$$ We know that $\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, so: $$\cos\left(\frac{\pi}{8}\right) = \pm\sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}$$ Simplifying, we get: $$\cos\left(\frac{\pi}{8}\right) = \pm\sqrt{\frac{2 + \sqrt{2}}{4}}$$ $$\cos\left(\frac{\pi}{8}\right) = \pm\frac{\sqrt{2 + \sqrt{2}}}{2}$$ Since $\cos\left(\frac{\pi}{8}\right)$ is positive in the first quadrant, we take the positive value: $$\cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2}$$ Therefore, yes, $\cos\left(\frac{\pi}{8}\right)$ is indeed equal to $\frac{\sqrt{2 + \sqrt{2}}}{2}$. |
True |
| llama3.1-405B-Instruct | To determine if cos(π/8) is equal to (√(2+√2))/2, we can use the half-angle formula for cosine. The half-angle formula for cosine is: cos(x/2) = ±√((1 + cos(x))/2) In this case, we want to find cos(π/8), so we can plug in x = π/4: cos(π/8) = ±√((1 + cos(π/4))/2) We know that cos(π/4) = 1/√2, so: cos(π/8) = ±√((1 + 1/√2)/2) Simplifying, we get: cos(π/8) = ±√((2 + √2)/4) Now, we can take the square root of the numerator and denominator separately: cos(π/8) = ±√(2 + √2)/2 Since cosine is positive in the first quadrant, we can drop the ± sign: cos(π/8) = √(2 + √2)/2 Therefore, the answer is yes, cos(π/8) is equal to (√(2+√2))/2. |
True |
| DeepSeek-Coder-V2 | To determine if cos(π/8) is equal to (√(2+√2))/2, we can use trigonometric identities and algebraic manipulation. Let's go through the steps: 1. **Double Angle Formula**: We start with the double angle formula for cosine: \[ \cos(2\theta) = 2\cos^2(\theta) - 1 \] Here, we set \(\theta = \frac{\pi}{8}\), so: \[ \cos\left(\frac{\pi}{4}\right) = 2\cos^2\left(\frac{\pi}{8}\right) - 1 \] 2. **Known Value**: We know that \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\). Substituting this into the equation, we get: \[ \frac{\sqrt{2}}{2} = 2\cos^2\left(\frac{\pi}{8}\right) - 1 \] 3. **Solve for \(\cos^2\left(\frac{\pi}{8}\right)\)**: \[ \frac{\sqrt{2}}{2} + 1 = 2\cos^2\left(\frac{\pi}{8}\right) \] \[ \frac{\sqrt{2}}{2} + 1 = \frac{\sqrt{2}}{2} + \frac{2}{2} = \frac{\sqrt{2} + 2}{2} \] \[ 2\cos^2\left(\frac{\pi}{8}\right) = \frac{\sqrt{2} + 2}{2} \] \[ \cos^2\left(\frac{\pi}{8}\right) = \frac{\sqrt{2} + 2}{4} \] 4. **Simplify**: To simplify \(\cos^2\left(\frac{\pi}{8}\right)\), we need to check if it matches the given expression \(\left(\frac{\sqrt{2 + \sqrt{2}}}{2}\right)^2\): \[ \left(\frac{\sqrt{2 + \sqrt{2}}}{2}\right)^2 = \frac{2 + \sqrt{2}}{4} \] 5. **Comparison**: We compare the two expressions: \[ \cos^2\left(\frac{\pi}{8}\right) = \frac{\sqrt{2} + 2}{4} \] \[ \left(\frac{\sqrt{2 + \sqrt{2}}}{2}\right)^2 = \frac{2 + \sqrt{2}}{4} \] Since both expressions are equal, we conclude that: \[ \cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2} \] Thus, the answer is yes, \(\cos(\pi/8)\) is equal to \(\frac{\sqrt{2+\sqrt{2}}}{2}\). |
True |
| Qwen2-72B-Instruct | To determine if \(\cos(\frac{\pi}{8})\) is equal to \(\frac{\sqrt{2+\sqrt{2}}}{2}\), we can start by using trigonometric identities and properties. First, recall that \(\cos(2\theta) = 2\cos^2(\theta) - 1\). Let's apply this with \(\theta = \frac{\pi}{8}\), which means \(2\theta = \frac{\pi}{4}\). \[ \cos\left(\frac{\pi}{4}\right) = 2\cos^2\left(\frac{\pi}{8}\right) - 1 \] We know that \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), so: \[ \frac{\sqrt{2}}{2} = 2\cos^2\left(\frac{\pi}{8}\right) - 1 \] Solving for \(\cos\left(\frac{\pi}{8}\right)\): \[ \frac{\sqrt{2}}{2} + 1 = 2\cos^2\left(\frac{\pi}{8}\right) \] \[ \frac{\sqrt{2} + 2}{2} = 2\cos^2\left(\frac{\pi}{8}\right) \] \[ \frac{\sqrt{2} + 2}{4} = \cos^2\left(\frac{\pi}{8}\right) \] \[ \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{\sqrt{2} + 2}{4}} \] \[ \cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{\sqrt{2} + 2}}{2} \] Therefore, \(\cos(\frac{\pi}{8})\) is indeed equal to \(\frac{\sqrt{2+\sqrt{2}}}{2}\). |
True |
| baichuan4 | To determine if $\cos(\pi/8)$ is equal to $(\sqrt{2+\sqrt{2}})/2$, we can use the half-angle formula for cosine. The half-angle formula for cosine of an angle $\theta$ is given by: $$\cos(\theta/2) = \pm \sqrt{\frac{1 + \cos(\theta)}{2}}$$ In our case, we want to find $\cos(\pi/8)$, which is equivalent to finding $\cos((\pi/4)/2)$. Therefore, we can substitute $\theta = \pi/4$ into the formula: $$\cos(\pi/8) = \cos((\pi/4)/2) = \pm \sqrt{\frac{1 + \cos(\pi/4)}{2}}$$ We know that $\cos(\pi/4) = \sqrt{2}/2$, so let's plug that in: $$\cos(\pi/8) = \pm \sqrt{\frac{1 + \sqrt{2}/2}{2}}$$ Simplifying the expression inside the square root: $$\cos(\pi/8) = \pm \sqrt{\frac{2 + \sqrt{2}}{4}}$$ $$\cos(\pi/8) = \pm \sqrt{\frac{2 + \sqrt{2}}{4}} = \pm \frac{\sqrt{2 + \sqrt{2}}}{2}$$ Since the cosine function is positive in the first quadrant and $\pi/8$ lies in the first quadrant, we can discard the negative solution. Therefore, we have: $$\cos(\pi/8) = \frac{\sqrt{2 + \sqrt{2}}}{2}$$ So, yes, $\cos(\pi/8)$ is equal to $(\sqrt{2+\sqrt{2}})/2$. |
True |
| doubao | We can use the half-angle formula for cosine: cos(θ/2) = ±√[(1 + cosθ)/2]. Let θ = π/4, then cos(π/8) = cos(π/4/2) = ±√[(1 + cos(π/4))/2]. cos(π/4) = √2/2, so cos(π/8) = ±√[(1 + √2/2)/2] = ±√[(2 + √2)/4] = ±(√(2 + √2))/2. Since π/8 is in the first quadrant and cosine is positive in the first quadrant, cos(π/8) = (√(2 + √2))/2. So the answer is yes. |
True |
| minimax | To determine if \(\cos(\pi/8)\) is equal to \((\sqrt{2+\sqrt{2}})/2\), we can use trigonometric identities and properties to express \(\cos(\pi/8)\) in a way that allows us to compare it directly to the given expression. First, let's express \(\pi/8\) as a sum of angles for which we know the cosine values. We can use the half-angle formula for cosine, which states that for any angle \(\theta\): \[ \cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos(\theta)}{2}} \] We want to express \(\pi/8\) as \(\pi/4 - \pi/8\), so we can use the cosine of a difference formula: \[ \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) \] Let's set \(\alpha = \pi/4\) and \(\beta = \pi/8\). We know that \(\cos(\pi/4) = \sin(\pi/4) = \frac{\sqrt{2}}{2}\). Now we can write: \[ \cos\left(\frac{\pi}{8}\right) = \cos\left(\frac{\pi}{4} - \frac{\pi}{8}\right) \] Applying the cosine of a difference formula: \[ \cos\left(\frac{\pi}{8}\right) = \cos\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{8}\right) + \sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{8}\right) \] Substituting the known values: \[ \cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2}}{2}\cos\left(\frac{\pi}{8}\right) + \frac{\sqrt{2}}{2}\sin\left(\frac{\pi}{8}\right) \] Now, we can isolate \(\cos(\pi/8)\): \[ \cos\left(\frac{\pi}{8}\right) - \frac{\sqrt{2}}{2}\cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2}}{2}\sin\left(\frac{\pi}{8}\right) \] \[ \left(1 - \frac{\sqrt{2}}{2}\right)\cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2}}{2}\sin\left(\frac{\pi}{8}\right) \] \[ \cos\left(\frac{\pi}{8}\right) = \frac{\frac{\sqrt{2}}{2}\sin\left(\frac{\pi}{8}\right)}{1 - \frac{\sqrt{2}}{2}} \] Now, we can use the Pythagorean identity \(\sin^2(\theta) + \cos^2(\theta) = 1\) to express \(\sin(\pi/8)\) in terms of \(\cos(\pi/8)\): \[ \sin^2\left(\frac{\pi}{8}\right) = 1 - \cos^2\left(\frac{\pi}{8}\right) \] Since we are looking for an expression involving only \(\cos(\pi/8)\), we can't directly use this identity to find \(\sin(\pi/8)\). Instead, we can use the double-angle formula for cosine, which states: \[ \cos(2\theta) = 2\cos^2(\theta) - 1 \] Let \(\theta = \pi/8\), then: \[ \cos\left(\frac{\pi}{4}\right) = 2\cos^2\left(\frac{\pi}{8}\right) - 1 \] \[ \frac{\sqrt{2}}{2} = 2\cos^2\left(\frac{\pi}{8}\right) - 1 \] \[ 2\cos^2\left(\frac{\pi}{8}\right) = \frac{\sqrt{2}}{2} + 1 \] \[ \cos^2\left(\frac{\pi}{8}\right) = \frac{\frac{\sqrt{2}}{2} + 1}{2} \] \[ \cos^2\left(\frac{\pi}{8}\right) = \frac{\sqrt{2} + 2}{4} \] \[ \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{\sqrt{2} + 2}{4}} \] \[ \cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{\sqrt{2} + 2}}{2} \] Now, we can see that: \[ \cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{\sqrt{2} + 2}}{2} \neq \frac{\sqrt{2+\sqrt{2}}}{2} \] Therefore, \(\cos(\pi/8)\) is not equal to \((\sqrt{2+\sqrt{2}})/2\). |
True |
| kimi | To determine if \(\cos(\pi/8)\) is equal to \((\sqrt{2+\sqrt{2}})/2\), we can use the half-angle formula for cosine, which states that: \[ \cos\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 + \cos(\theta)}{2}} \] We want to find \(\cos(\pi/8)\), which is the same as \(\cos\left(\frac{\pi/4}{2}\right)\). Using the half-angle formula with \(\theta = \pi/4\), we get: \[ \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{1 + \cos\left(\frac{\pi}{4}\right)}{2}} \] Since \(\cos(\pi/4) = \frac{\sqrt{2}}{2}\), we can substitute this value into the equation: \[ \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} \] Simplify the expression inside the square root: \[ \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{2 + \sqrt{2}}{4}} \] Now, simplify the square root: \[ \cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2} \] So, yes, \(\cos(\pi/8)\) is indeed equal to \((\sqrt{2+\sqrt{2}})/2\). |
True |
| wenxinyiyan | To determine if cos(π/8) is equal to (√(2+√2))/2, we can use the half-angle formula for cosines. The half-angle formula for cosine is: cos(θ/2) = √((1 + cos(θ)) / 2) In this case, we want to find cos(π/8), which is cos(π/4 / 2). So, θ = π/4. cos(π/8) = √((1 + cos(π/4)) / 2) We know that cos(π/4) = √2/2, so: cos(π/8) = √((1 + √2/2) / 2) = √((2 + √2) / 4) = √(2 + √2) / 2 So, cos(π/8) is indeed equal to (√(2+√2))/2. |
True |
| zhipu | To determine if \(\cos(\pi/8)\) is equal to \((\sqrt{2+\sqrt{2}})/2\), we can use trigonometric identities and some algebraic manipulation. First, let's recall the double-angle formula for cosine: \[ \cos(2\theta) = 2\cos^2(\theta) - 1 \] If we let \(\theta = \pi/8\), then \(2\theta = \pi/4\), and we know that: \[ \cos(\pi/4) = \frac{\sqrt{2}}{2} \] Using the double-angle formula, we have: \[ \cos(\pi/4) = 2\cos^2(\pi/8) - 1 \] Substitute \(\cos(\pi/4) = \frac{\sqrt{2}}{2}\) into the equation: \[ \frac{\sqrt{2}}{2} = 2\cos^2(\pi/8) - 1 \] Now, solve for \(\cos^2(\pi/8)\): \[ \frac{\sqrt{2}}{2} + 1 = 2\cos^2(\pi/8) \] \[ \frac{\sqrt{2}}{2} + \frac{2}{2} = 2\cos^2(\pi/8) \] \[ \frac{\sqrt{2} + 2}{2} = 2\cos^2(\pi/8) \] \[ \frac{\sqrt{2} + 2}{4} = \cos^2(\pi/8) \] Next, take the square root of both sides to solve for \(\cos(\pi/8)\): \[ \cos(\pi/8) = \sqrt{\frac{\sqrt{2} + 2}{4}} \] \[ \cos(\pi/8) = \frac{\sqrt{\sqrt{2} + 2}}{2} \] Now, let's compare this to the given expression \((\sqrt{2+\sqrt{2}})/2\). Notice that: \[ \sqrt{\sqrt{2} + 2} = \sqrt{2 + \sqrt{2}} \] This is because \(\sqrt{2}\) can be written as \(2^{1/2}\), and when squared, it becomes \(2\). Thus, the expression inside the square root matches. Therefore, we have: \[ \cos(\pi/8) = \frac{\sqrt{2 + \sqrt{2}}}{2} \] So, \(\cos(\pi/8)\) is indeed equal to \((\sqrt{2+\sqrt{2}})/2\). ### Conclusion: \[ \boxed{\cos(\pi/8) = \frac{\sqrt{2+\sqrt{2}}}{2}} \] The given statement is true. |
True |
| gpto1 | To determine if the identity \[ \cos\left( \frac{\pi}{8} \right) = \frac{\sqrt{2 + \sqrt{2}}}{2} \] is true, we'll use trigonometric identities and known values. **Step 1: Recognize the Half-Angle Relationship** We note that \(\frac{\pi}{8}\) is half of \(\frac{\pi}{4}\), since: \[ \frac{\pi}{8} = \frac{\pi}{4} \times \frac{1}{2} \] This suggests using the half-angle formula for cosine. **Step 2: Recall the Half-Angle Formula for Cosine** The half-angle formula for cosine is: \[ \cos\left( \frac{\theta}{2} \right) = \sqrt{ \frac{1 + \cos\theta}{2} } \] We choose the positive square root because \(\cos\left( \frac{\pi}{8} \right)\) is positive in the first quadrant. **Step 3: Apply the Half-Angle Formula** Let \(\theta = \frac{\pi}{4}\). Then: \[ \cos\left( \frac{\pi}{8} \right) = \sqrt{ \frac{1 + \cos\left( \frac{\pi}{4} \right)}{2} } \] **Step 4: Substitute the Known Value of \(\cos\left( \frac{\pi}{4} \right)\)** We know that: \[ \cos\left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \] Substituting this value: \[ \cos\left( \frac{\pi}{8} \right) = \sqrt{ \frac{1 + \frac{\sqrt{2}}{2}}{2} } \] **Step 5: Simplify the Expression Inside the Square Root** First, combine the terms in the numerator: \[ 1 + \frac{\sqrt{2}}{2} = \frac{2}{2} + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2} \] Now, the expression becomes: \[ \cos\left( \frac{\pi}{8} \right) = \sqrt{ \frac{ \frac{2 + \sqrt{2}}{2} }{2} } \] Simplify the denominator: \[ \frac{ \frac{2 + \sqrt{2}}{2} }{2} = \frac{2 + \sqrt{2}}{2 \times 2} = \frac{2 + \sqrt{2}}{4} \] **Step 6: Simplify the Square Root** Now, we have: \[ \cos\left( \frac{\pi}{8} \right) = \sqrt{ \frac{2 + \sqrt{2}}{4} } \] This simplifies to: \[ \cos\left( \frac{\pi}{8} \right) = \frac{ \sqrt{ 2 + \sqrt{2} } }{ \sqrt{4} } = \frac{ \sqrt{ 2 + \sqrt{2} } }{ 2 } \] **Step 7: Conclusion** Therefore, we have shown: \[ \cos\left( \frac{\pi}{8} \right) = \frac{ \sqrt{ 2 + \sqrt{2} } }{ 2 } \] which confirms the identity is true. **Answer:** Yes; using the half-angle identity shows they are equal: cos(π⁄8) = [√(2 + √2)] ⁄ 2. |
True |
| deepseekv3 | To determine whether \(\cos\left(\frac{\pi}{8}\right)\) is equal to \(\frac{\sqrt{2 + \sqrt{2}}}{2}\), we can proceed with the following steps: ### Step 1: Express \(\cos\left(\frac{\pi}{8}\right)\) in terms of known angles We know that \(\frac{\pi}{8}\) is half of \(\frac{\pi}{4}\), so we can use the half-angle formula for cosine: \[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos(\theta)}{2}} \] Let \(\theta = \frac{\pi}{4}\), then: \[ \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{1 + \cos\left(\frac{\pi}{4}\right)}{2}} \] ### Step 2: Substitute \(\cos\left(\frac{\pi}{4}\right)\) We know that \(\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), so: \[ \cos\left(\frac{\pi}{8}\right) = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2} \] ### Step 3: Compare the two expressions We have shown that: \[ \cos\left(\frac{\pi}{8}\right) = \frac{\sqrt{2 + \sqrt{2}}}{2} \] Thus, the equality holds true. ### Final Answer: Yes, \(\cos\left(\frac{\pi}{8}\right)\) is equal to \(\frac{\sqrt{2 + \sqrt{2}}}{2}\). |
True |