| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To solve this problem, we need to simplify the given expression and then substitute \( x = \pi / 60 \). Let's start by simplifying the expression inside the parentheses: \[ \left( \frac{1 + \cos(2x) + i \sin(2x)}{1 + \cos(2x) - i \sin(2x)} \right) \] We can use the Euler's formula \( e^{ix} = \cos(x) + i \sin(x) \) to simplify the expression. Let's rewrite the expression using Euler's formula: \[ \left( \frac{1 + \cos(2x) + i \sin(2x)}{1 + \cos(2x) - i \sin(2x)} \right) = \left( \frac{1 + e^{i2x}}{1 + e^{-i2x}} \right) \] Now, let's simplify the expression further: \[ \left( \frac{1 + e^{i2x}}{1 + e^{-i2x}} \right) = \left( \frac{1 + e^{i2x}}{1 + \frac{1}{e^{i2x}}} \right) = \left( \frac{1 + e^{i2x}}{\frac{e^{i2x} + 1}{e^{i2x}}} \right) \] Simplifying the expression, we get: \[ \left( \frac{1 + e^{i2x}}{\frac{e^{i2x} + 1}{e^{i2x}}} \right) = (e^{i2x}) \] Now, we need to raise this expression to the power of 30: \[ (e^{i2x})^{30} = e^{i60x} \] Finally, we substitute \( x = \pi / 60 \): \[ e^{i60(\pi / 60)} = e^{i\pi} = \cos(\pi) + i \sin(\pi) = -1 + 0i = \boxed{-1} \] Therefore, the solution to the given problem is \( -1 \). |
-1.0 |
| deepseekr1 | To find the value of \(\left(\frac{1 + \cos(2x) + i \sin(2x)}{1 + \cos(2x) - i \sin(2x)}\right)^{30}\) when \(x = \frac{\pi}{60}\), we start by simplifying the expression using trigonometric identities and properties of complex numbers. 1. **Trigonometric Identities**: - Use the double angle identities: \(1 + \cos(2x) = 2\cos^2(x)\) and \(\sin(2x) = 2\sin(x)\cos(x)\). - Substitute these into the numerator and denominator: \[ \text{Numerator: } 2\cos^2(x) + i \cdot 2\sin(x)\cos(x) = 2\cos(x)(\cos(x) + i\sin(x)) \] \[ \text{Denominator: } 2\cos^2(x) - i \cdot 2\sin(x)\cos(x) = 2\cos(x)(\cos(x) - i\sin(x)) \] 2. **Simplify the Fraction**: - Cancel the common factor \(2\cos(x)\) in numerator and denominator: \[ \frac{\cos(x) + i\sin(x)}{\cos(x) - i\sin(x)} \] 3. **Complex Conjugate**: - Recognize that \(\cos(x) + i\sin(x) = e^{ix}\) and \(\cos(x) - i\sin(x) = e^{-ix}\). - The fraction simplifies to: \[ \frac{e^{ix}}{e^{-ix}} = e^{i2x} \] 4. **Exponentiation**: - Raise the result to the 30th power: \[ \left(e^{i2x}\right)^{30} = e^{i60x} \] 5. **Substitute \(x = \frac{\pi}{60}\)**: - Substitute \(x = \frac{\pi}{60}\) into the exponent: \[ e^{i60 \cdot \frac{\pi}{60}} = e^{i\pi} \] - Using Euler's formula, \(e^{i\pi} = -1\). 6. **Final Calculation**: - Raise \(-1\) to the 30th power: \[ (-1)^{30} = 1 \] Thus, the value of the given expression is \(\boxed{1}\). |
-1.0 |
| gpto1 | To simplify the given expression and find its value, let's start by expressing the numerator and denominator in exponential form using Euler's formula. **Given:** \[ \left( \frac{1 + \cos(2x) + i\sin(2x)}{1 + \cos(2x) - i\sin(2x)} \right)^{30}, \text{ where } x = \frac{\pi}{60} \] **Step 1: Express numerator and denominator using trigonometric identities.** Recall the double-angle identities: \[ \cos(2x) = 2\cos^2 x - 1 \] \[ \sin(2x) = 2\sin x \cos x \] Therefore, \[ 1 + \cos(2x) = 2\cos^2 x \] \[ \sin(2x) = 2\sin x \cos x \] **Step 2: Rewrite the numerator and denominator.** Let’s write the numerator \(A\) and denominator \(B\) as: \[ A = 1 + \cos(2x) + i\sin(2x) = 2\cos^2 x + i(2\sin x \cos x) = 2\cos x (\cos x + i\sin x) \] \[ B = 1 + \cos(2x) - i\sin(2x) = 2\cos^2 x - i(2\sin x \cos x) = 2\cos x (\cos x - i\sin x) \] **Step 3: Use Euler's formula.** Euler's formula states: \[ e^{i\theta} = \cos \theta + i\sin \theta \] \[ e^{-i\theta} = \cos \theta - i\sin \theta \] So we can rewrite \(A\) and \(B\) using exponentials: \[ A = 2\cos x \cdot e^{i x} \] \[ B = 2\cos x \cdot e^{-i x} \] **Step 4: Simplify the fraction.** Compute \(R\): \[ R = \frac{A}{B} = \frac{2\cos x \cdot e^{i x}}{2\cos x \cdot e^{-i x}} = e^{i x} \cdot e^{i x} = e^{2i x} \] **Step 5: Compute \(R^{30}\).** \[ R^{30} = (e^{2i x})^{30} = e^{60i x} \] **Step 6: Substitute \(x = \frac{\pi}{60}\).** \[ R^{30} = e^{60i \cdot \frac{\pi}{60}} = e^{i \pi} = \cos \pi + i\sin \pi = -1 + i(0) = -1 \] **Final Answer:** \[ \left( \frac{1 + \cos(2x) + i\sin(2x)}{1 + \cos(2x) - i\sin(2x)} \right)^{30} = -1 \] **Answer: -1** |
-1.0 |
| GPT-4o | To solve the given problem, we need to simplify the expression \((\frac{1 + \cos(2x) + i \sin(2x)}{1 + \cos(2x) - i \sin(2x)})^{30}\) with \(x = \pi/60\). First, let's examine the expression inside the parentheses. We have: \[ \frac{1 + \cos(2x) + i \sin(2x)}{1 + \cos(2x) - i \sin(2x)} \] Recall Euler's formula, which states: \[ e^{i\theta} = \cos(\theta) + i \sin(\theta) \] If we let \(\theta = 2x\), then: \[ e^{i(2x)} = \cos(2x) + i \sin(2x) \] Now consider the numerator and the denominator of our fraction: \[ 1 + \cos(2x) + i \sin(2x) = 1 + e^{i(2x)} - 1 = e^{i(2x)} \] \[ 1 + \cos(2x) - i \sin(2x) = 1 + e^{-i(2x)} - 1 = e^{-i(2x)} \] Hence, the original expression simplifies to: \[ \frac{e^{i(2x)}}{e^{-i(2x)}} = e^{i(2x) - (-i(2x))} = e^{i(4x)} \] Next, we need to raise this result to the 30th power: \[ \left( e^{i(4x)} \right)^{30} = e^{i(4x \cdot 30)} = e^{i(120x)} \] Substitute \(x = \pi/60\): \[ 120x = 120 \cdot \frac{\pi}{60} = 2\pi \] Thus: \[ e^{i(120x)} = e^{i(2\pi)} \] Since \(e^{i(2\pi)} = \cos(2\pi) + i \sin(2\pi) = 1\): \[ e^{i(2\pi)} = 1 \] Therefore, the final result is: \[ \left( \frac{1 + \cos(2x) + i \sin(2x)}{1 + \cos(2x) - i \sin(2x)} \right)^{30} = 1 \] So, the solution is: \[ 1 \] |
-1.0 |
| llama3.1-70B-Instruct | ## Step 1: Identify the given values and the expression to simplify We are given the expression $(\frac{1 + cos(2x) + i*sin(2x)}{1 + cos(2x) - i*sin(2x)})^30$ and the value of $x = \pi / 60$. ## Step 2: Simplify the expression using Euler's formula Euler's formula states that $e^{i\theta} = cos(\theta) + i*sin(\theta)$. We can rewrite the given expression using this formula. ## Step 3: Rewrite the numerator and denominator using Euler's formula The numerator can be rewritten as $1 + e^{i2x}$ and the denominator can be rewritten as $1 + e^{-i2x}$. ## Step 4: Simplify the expression further The expression can be simplified as $(\frac{1 + e^{i2x}}{1 + e^{-i2x}})^30$. ## Step 5: Use the property of exponents to simplify the expression Using the property of exponents, we can rewrite the expression as $(\frac{e^{i2x} + 1}{e^{-i2x} + 1})^30$. ## Step 6: Simplify the expression using the property of complex conjugates The expression can be simplified as $(\frac{e^{i2x} + 1}{e^{i2x} + 1})^30$. ## Step 7: Cancel out the common terms The expression can be simplified as $1^{30}$. ## Step 8: Evaluate the expression The expression evaluates to $1$. The final answer is: $\boxed{1}$ |
-1.0 |
| llama3.1-405B-Instruct | ## Step 1: Simplify the expression using Euler's formula We can simplify the expression $(\frac{1 + cos(2x) + i*sin(2x)}{1 + cos(2x) - i*sin(2x)})^30$ using Euler's formula, which states that $e^{ix} = cos(x) + i*sin(x)$. ## Step 2: Apply Euler's formula to the numerator and denominator Applying Euler's formula to the numerator and denominator, we get $(\frac{1 + e^{2ix}}{1 + e^{-2ix}})^30$. ## Step 3: Simplify the expression further We can simplify the expression further by multiplying the numerator and denominator by $e^{2ix}$, which gives us $(\frac{e^{2ix} + e^{4ix}}{e^{2ix} + 1})^30$. ## Step 4: Factor out the common term Factoring out the common term $e^{2ix}$ from the numerator and denominator, we get $(\frac{1 + e^{2ix}}{1 + e^{-2ix}})^30$. ## Step 5: Simplify the expression using the given value of x Substituting $x = \pi / 60$ into the expression, we get $(\frac{1 + e^{i\pi/30}}{1 + e^{-i\pi/30}})^30$. ## Step 6: Evaluate the expression Evaluating the expression, we get $(\frac{1 + cos(\pi/30) + i*sin(\pi/30)}{1 + cos(\pi/30) - i*sin(\pi/30)})^30$. ## Step 7: Simplify the expression using trigonometric identities Using the trigonometric identity $cos(\pi/30) = \frac{\sqrt{3}}{2}$ and $sin(\pi/30) = \frac{1}{2}$, we can simplify the expression further. ## Step 8: Evaluate the expression Evaluating the expression, we get $(\frac{1 + \frac{\sqrt{3}}{2} + i*\frac{1}{2}}{1 + \frac{\sqrt{3}}{2} - i*\frac{1}{2}})^30$. ## Step 9: Simplify the expression Simplifying the expression, we get $(\frac{2 + \sqrt{3} + i}{2 + \sqrt{3} - i})^30$. ## Step 10: Rationalize the denominator Rationalizing the denominator, we get $(\frac{(2 + \sqrt{3} + i)(2 + \sqrt{3} + i)}{(2 + \sqrt{3} - i)(2 + \sqrt{3} + i)})^30$. ## Step 11: Simplify the expression Simplifying the expression, we get $(\frac{7 + 4\sqrt{3} + 4i + 2i\sqrt{3}}{7 + 4\sqrt{3}})^30$. ## Step 12: Simplify the expression further Simplifying the expression further, we get $(1 + \frac{4i + 2i\sqrt{3}}{7 + 4\sqrt{3}})^30$. ## Step 13: Evaluate the expression Evaluating the expression, we get $(1 + i*\frac{4 + 2\sqrt{3}}{7 + 4\sqrt{3}})^30$. ## Step 14: Simplify the expression Simplifying the expression, we get $(1 + i*\frac{2(2 + \sqrt{3})}{7 + 4\sqrt{3}})^30$. ## Step 15: Evaluate the expression Evaluating the expression, we get $(1 + i*\frac{2(2 + \sqrt{3})}{(2 + \sqrt{3})(2 + \sqrt{3} - 1)})^30$. ## Step 16: Simplify the expression Simplifying the expression, we get $(1 + i*\frac{2}{2 + \sqrt{3} - 1})^30$. ## Step 17: Evaluate the expression Evaluating the expression, we get $(1 + i*\frac{2})^30$. ## Step 18: Simplify the expression Simplifying the expression, we get $(1 + i)^30$. ## Step 19: Evaluate the expression Evaluating the expression, we get $2^{15}i^{15}$. ## Step 20: Simplify the expression Simplifying the expression, we get $32768i^{15}$. ## Step 21: Evaluate the expression Evaluating the expression, we get $32768i^{4*3 + 3}$. ## Step 22: Simplify the expression Simplifying the expression, we get $32768(i^4)^3i^3$. ## Step 23: Evaluate the expression Evaluating the expression, we get $32768(1)^3i^3$. ## Step 24: Simplify the expression Simplifying the expression, we get $32768i^3$. ## Step 25: Evaluate the expression Evaluating the expression, we get $32768i^2i$. ## Step 26: Simplify the expression Simplifying the expression, we get $32768(-1)i$. ## Step 27: Evaluate the expression Evaluating the expression, we get $-32768i$. The final answer is: $\boxed{-32768i}$ |
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| DeepSeek-Coder-V2 | To solve the given problem, we need to evaluate the expression \((\frac{1 + \cos(2x) + i \sin(2x)}{1 + \cos(2x) - i \sin(2x)})^{30}\) with \(x = \pi / 60\). First, let's simplify the expression inside the parentheses. We can use Euler's formula, which states that \(e^{i\theta} = \cos(\theta) + i \sin(\theta)\). Therefore, we can rewrite the numerator and the denominator using Euler's formula: \[ 1 + \cos(2x) + i \sin(2x) = 1 + e^{i \cdot 2x} \] \[ 1 + \cos(2x) - i \sin(2x) = 1 + e^{-i \cdot 2x} \] Thus, the expression becomes: \[ \frac{1 + e^{i \cdot 2x}}{1 + e^{-i \cdot 2x}} \] Next, we can simplify this fraction by multiplying the numerator and the denominator by \(e^{-i \cdot 2x}\): \[ \frac{(1 + e^{i \cdot 2x}) \cdot e^{-i \cdot 2x}}{(1 + e^{-i \cdot 2x}) \cdot e^{-i \cdot 2x}} = \frac{e^{-i \cdot 2x} + 1}{e^{-i \cdot 2x} + 1} = 1 \] However, this simplification seems incorrect. Let's try another approach by recognizing that the expression can be simplified using complex conjugate properties. We can rewrite the expression as: \[ \frac{1 + e^{i \cdot 2x}}{1 + e^{-i \cdot 2x}} = \frac{1 + e^{i \cdot 2x}}{1 + \overline{e^{i \cdot 2x}}} \] Since \(e^{i \cdot 2x}\) is a complex number, we can denote it as \(z = e^{i \cdot 2x}\). The expression then becomes: \[ \frac{1 + z}{1 + \overline{z}} \] To simplify this, we can multiply the numerator and the denominator by the conjugate of the denominator: \[ \frac{(1 + z)(1 + \overline{z})}{(1 + \overline{z})(1 + \overline{z})} = \frac{1 + z + \overline{z} + z \overline{z}}{1 + 2\overline{z} + |\overline{z}|^2} \] Since \(z = e^{i \cdot 2x}\), we have \(z \overline{z} = |z|^2 = 1\) and \(\overline{z} = e^{-i \cdot 2x}\). Therefore: \[ \frac{1 + e^{i \cdot 2x} + e^{-i \cdot 2x} + 1}{1 + 2e^{-i \cdot 2x} + 1} = \frac{2 + 2\cos(2x)}{2 + 2\cos(2x)} = 1 \] This simplification also seems incorrect. Let's reconsider the original approach and use the polar form of complex numbers. We can rewrite the expression as: \[ \frac{1 + e^{i \cdot 2x}}{1 + e^{-i \cdot 2x}} = \frac{1 + \cos(2x) + i \sin(2x)}{1 + \cos(2x) - i \sin(2x)} \] Multiplying the numerator and the denominator by the conjugate of the denominator: \[ \frac{(1 + \cos(2x) + i \sin(2x))(1 + \cos(2x) + i \sin(2x))}{(1 + \cos(2x) - i \sin(2x))(1 + \cos(2x) + i \sin(2x))} \] The denominator simplifies to: \[ (1 + \cos(2x))^2 + (\sin(2x))^2 = 1 + 2\cos(2x) + \cos^2(2x) + \sin^2(2x) = 2 + 2\cos(2x) \] The numerator simplifies to: \[ (1 + \cos(2x))^2 + 2i \sin(2x)(1 + \cos(2x)) - (\sin(2x))^2 = 1 + 2\cos(2x) + \cos^2(2x) + 2i \sin(2x) + 2i \sin(2x) \cos(2x) - \sin^2(2x) \] This approach seems overly complex. Let's use the polar form directly: \[ \frac{1 + e^{i \cdot 2x}}{1 + e^{-i \cdot 2x}} = \frac{1 + e^{i \cdot 2x}}{1 + e^{-i \cdot 2x}} = e^{i \cdot 2x} \] Thus, the expression simplifies to: \[ (e^{i \cdot 2x})^{30} = e^{i \cdot 60 \cdot 2x} = e^{i \cdot 120x} \] With \(x = \pi / 60\), we have: \[ e^{i \cdot 120 \cdot \pi / 60} = e^{i \cdot 2\pi} = 1 \] Therefore, the final answer is: \[ \boxed{1} \] |
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| Qwen2-72B-Instruct | To solve this problem, we need to simplify the given expression and then substitute $x = \pi / 60$. Let's start by simplifying the expression inside the parentheses: $(\frac{1 + cos(2x) + i*sin(2x)}{1 + cos(2x) - i*sin(2x)})$ We can use the Euler's formula $e^{ix} = cos(x) + i*sin(x)$ to simplify the expression. Let's rewrite the expression using Euler's formula: $(\frac{1 + cos(2x) + i*sin(2x)}{1 + cos(2x) - i*sin(2x)}) = (\frac{1 + e^{i2x}}{1 + e^{-i2x}})$ Now, let's simplify the expression further: $(\frac{1 + e^{i2x}}{1 + e^{-i2x}}) = (\frac{1 + e^{i2x}}{1 + \frac{1}{e^{i2x}}}) = (\frac{1 + e^{i2x}}{\frac{e^{i2x} + 1}{e^{i2x}}})$ Simplifying the expression, we get: $(\frac{1 + e^{i2x}}{\frac{e^{i2x} + 1}{e^{i2x}}}) = (e^{i2x})$ Now, we need to raise this expression to the power of 30: $(e^{i2x})^{30} = e^{i60x}$ Finally, we substitute $x = \pi / 60$: $e^{i60(\pi / 60)} = e^{i\pi} = cos(\pi) + i*sin(\pi) = -1 + 0i = \boxed{-1}$ Therefore, the solution to the given problem is $-1$. |
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| baichuan4 | To solve this problem, we'll first simplify the complex fraction and then raise it to the power of 30. Finally, we'll substitute $x = \pi / 60$ to get the final answer. Let's start by simplifying the complex fraction: $$ \left(\frac{1 + \cos(2x) + i\sin(2x)}{1 + \cos(2x) - i\sin(2x)}\right)^{30} $$ We can multiply the numerator and denominator by the conjugate of the denominator to eliminate the imaginary part: $$ \left(\frac{(1 + \cos(2x) + i\sin(2x))(1 + \cos(2x) + i\sin(2x))}{(1 + \cos(2x) - i\sin(2x))(1 + \cos(2x) + i\sin(2x))}\right)^{30} $$ This gives us: $$ \left(\frac{1 + 2\cos(2x) + \cos^2(2x) + i(2\sin(2x) + \sin(2x)\cos(2x))}{1 + 2\cos(2x) + \cos^2(2x) - i(2\sin(2x) + \sin(2x)\cos(2x))}\right)^{30} $$ Now, we can simplify the fraction by canceling out the real and imaginary parts: $$ \left(\frac{1 + 2\cos(2x) + \cos^2(2x)}{1 + 2\cos(2x) + \cos^2(2x)}\right)^{30} \cdot \left(\frac{i(2\sin(2x) + \sin(2x)\cos(2x))}{-i(2\sin(2x) + \sin(2x)\cos(2x))}\right)^{30} $$ This simplifies to: $$ \left(1\right)^{30} \cdot \left(-1\right)^{30} = 1 $$ Now, let's substitute $x = \pi / 60$: $$ 1^{30} = 1 $$ So, the final answer is $1$. |
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| doubao | Let $z=\frac{1+\cos(2x)+i\sin(2x)}{1+\cos(2x)-i\sin(2x)}$. Multiply the numerator and denominator by the conjugate of the denominator: \[ \begin{align*} z&=\frac{(1+\cos(2x)+i\sin(2x))(1+\cos(2x)+i\sin(2x))}{(1+\cos(2x)-i\sin(2x))(1+\cos(2x)+i\sin(2x))}\\ &=\frac{(1+\cos(2x))^2-(i\sin(2x))^2}{(1+\cos(2x))^2+(i\sin(2x))^2}\\ &=\frac{1+2\cos(2x)+\cos^2(2x)-i^2\sin^2(2x)}{1+2\cos(2x)+\cos^2(2x)+\sin^2(2x)}\\ &=\frac{1+2\cos(2x)+\cos^2(2x)+\sin^2(2x)}{1+2\cos(2x)+\cos^2(2x)+\sin^2(2x)}\\ &=\frac{2+2\cos(2x)}{2+2\cos(2x)}=1 \end{align*} \] So $(\frac{1+\cos(2x)+i\sin(2x)}{1+\cos(2x)-i\sin(2x)})^30=1^{30}=1$. When $x=\frac{\pi}{60}$, the result is still 1. |
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| minimax | To solve the given problem, we can simplify the expression using Euler's formula, which states that for any real number $x$: $$ e^{ix} = \cos(x) + i\sin(x) $$ The given expression is: $$ \left(\frac{1 + \cos(2x) + i\sin(2x)}{1 + \cos(2x) - i\sin(2x)}\right)^{30} $$ First, let's simplify the numerator and the denominator using Euler's formula for $e^{i2x}$: $$ \cos(2x) + i\sin(2x) = e^{i2x} $$ $$ \cos(2x) - i\sin(2x) = e^{-i2x} $$ Now, we can rewrite the expression as: $$ \left(\frac{e^{i2x}}{e^{-i2x}}\right)^{30} $$ Since $e^{i2x}$ and $e^{-i2x}$ are complex conjugates, their magnitudes are the same, and dividing them results in a complex number with magnitude 1. Specifically, dividing $e^{i2x}$ by $e^{-i2x}$ gives us: $$ \frac{e^{i2x}}{e^{-i2x}} = e^{i2x} \cdot e^{i2x} = e^{i4x} $$ Now, we raise this to the 30th power: $$ (e^{i4x})^{30} = e^{i120x} $$ Using Euler's formula again, we can convert this back to trigonometric form: $$ e^{i120x} = \cos(120x) + i\sin(120x) $$ Now, let's plug in $x = \frac{\pi}{60}$: $$ 120x = 120 \cdot \frac{\pi}{60} = 2\pi $$ Since $2\pi$ is a full rotation around the unit circle, the cosine and sine values will be the same as for $0$ radians: $$ \cos(2\pi) = 1 $$ $$ \sin(2\pi) = 0 $$ Therefore, the expression simplifies to: $$ \cos(2\pi) + i\sin(2\pi) = 1 + i \cdot 0 = 1 $$ So, the final result of the given expression is: $$ \left(\frac{1 + \cos(2x) + i\sin(2x)}{1 + \cos(2x) - i\sin(2x)}\right)^{30} = 1 $$ for $x = \frac{\pi}{60}$. |
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| kimi | To solve this problem, we can use Euler's formula, which states that \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \). We can rewrite the given expression using this formula. First, let's rewrite the numerator and denominator using Euler's formula: \[ \frac{1 + \cos(2x) + i\sin(2x)}{1 + \cos(2x) - i\sin(2x)} = \frac{1 + (\cos(2x) + i\sin(2x))}{1 + (\cos(2x) - i\sin(2x))} \] Using Euler's formula, we can see that \( \cos(2x) + i\sin(2x) = e^{i2x} \) and \( \cos(2x) - i\sin(2x) = e^{-i2x} \). So the expression becomes: \[ \frac{1 + e^{i2x}}{1 + e^{-i2x}} \] Now, let's simplify this expression by multiplying the numerator and denominator by \( e^{i2x} \): \[ \frac{e^{i2x}(1 + e^{i2x})}{e^{i2x}(1 + e^{-i2x})} = \frac{e^{i2x} + e^{2i2x}}{e^{i2x} + e^{0}} = \frac{e^{i2x}(1 + e^{i2x})}{e^{i2x}(1 + e^{-i2x})} \] This simplifies to: \[ \frac{e^{i2x} + e^{3i2x}}{1 + e^{i2x}} = e^{i2x} \cdot \frac{1 + e^{i2x}}{1 + e^{i2x}} = e^{i2x} \] So the original expression simplifies to \( e^{i2x} \). Now we need to raise this to the 30th power: \[ (e^{i2x})^{30} = e^{i60x} \] Now we substitute \( x = \pi / 60 \) into the expression: \[ e^{i60(\pi / 60)} = e^{i\pi} = \cos(\pi) + i\sin(\pi) \] Since \( \cos(\pi) = -1 \) and \( \sin(\pi) = 0 \), the expression simplifies to: \[ -1 + i \cdot 0 = -1 \] So the solution to the given problem is \( -1 \). |
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| wenxinyiyan | 首先，我们注意到给定的表达式是一个复数的乘方。为了简化这个表达式，我们可以使用复数的三角形式。 复数的三角形式为 $r(\cos \theta + i\sin \theta)$，其中 $r$ 是复数的模，$\theta$ 是幅角。 给定的表达式可以重写为： $\left(\frac{1 + \cos(2x) + i\sin(2x)}{1 + \cos(2x) - i\sin(2x)}\right)^{30}$这可以进一步简化为： $\left(\frac{2\cos^2(x) + 2i\sin(x)\cos(x)}{2\cos^2(x) - 2i\sin(x)\cos(x)}\right)^{30}$现在我们可以将分子和分母都除以 $2\cos^2(x)$，得到： $\left(\frac{1 + i\tan(x)}{1 - i\tan(x)}\right)^{30}$这可以看作是复数 $\frac{1 + i\tan(x)}{1 - i\tan(x)}$ 的 30 次方。 我们可以将这个复数转换为三角形式，其中模 $r = 1$，幅角 $\theta = 2x$。 因此，给定的表达式变为： $[\cos(2x) + i\sin(2x)]^{30}$使用 De Moivre 定理，我们可以将其展开为： $\cos(60x) + i\sin(60x)$现在我们将 $x = \frac{\pi}{60}$ 代入，得到： $\cos(\pi) + i\sin(\pi) = -1$因此，最终答案是 $-1$。 |
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| zhipu | To solve the given expression \((\frac{1 + \cos(2x) + i\sin(2x)}{1 + \cos(2x) - i\sin(2x)})^{30}\) with \(x = \pi / 60\), we can use some trigonometric identities and properties of complex numbers. First, let's simplify the expression inside the parentheses. ### Step 1: Substitute \(x = \pi / 60\) \[ 2x = 2 \cdot \frac{\pi}{60} = \frac{\pi}{30} \] So, the expression becomes: \[ \frac{1 + \cos(\frac{\pi}{30}) + i\sin(\frac{\pi}{30})}{1 + \cos(\frac{\pi}{30}) - i\sin(\frac{\pi}{30})} \] ### Step 2: Use Euler's Formula Euler's formula states that: \[ e^{i\theta} = \cos(\theta) + i\sin(\theta) \] Thus, we can rewrite the numerator and denominator using Euler's formula: \[ 1 + \cos(\frac{\pi}{30}) + i\sin(\frac{\pi}{30}) = 1 + e^{i\frac{\pi}{30}} \] \[ 1 + \cos(\frac{\pi}{30}) - i\sin(\frac{\pi}{30}) = 1 + e^{-i\frac{\pi}{30}} \] ### Step 3: Simplify the Fraction The fraction now looks like: \[ \frac{1 + e^{i\frac{\pi}{30}}}{1 + e^{-i\frac{\pi}{30}}} \] To simplify this, multiply the numerator and the denominator by \(e^{i\frac{\pi}{30}}\): \[ \frac{(1 + e^{i\frac{\pi}{30}})e^{i\frac{\pi}{30}}}{(1 + e^{-i\frac{\pi}{30}})e^{i\frac{\pi}{30}}} = \frac{e^{i\frac{\pi}{30}} + e^{i\frac{2\pi}{30}}}{e^{i\frac{\pi}{30}} + 1} \] Notice that \(e^{i\frac{2\pi}{30}} = e^{i\frac{\pi}{15}}\), so the expression becomes: \[ \frac{e^{i\frac{\pi}{30}} + e^{i\frac{\pi}{15}}}{e^{i\frac{\pi}{30}} + 1} \] ### Step 4: Recognize the Pattern The expression can be simplified further by recognizing that it involves complex numbers on the unit circle. Specifically, we can use the fact that: \[ \frac{e^{i\theta} + e^{i\phi}}{e^{i\theta} + 1} \] for certain angles \(\theta\) and \(\phi\) can often simplify to a form involving a single exponential term. ### Step 5: Simplify Using Symmetry Given the symmetry and periodicity of the trigonometric functions, we can simplify the expression to a form involving a power of a complex number on the unit circle. \[ \frac{e^{i\frac{\pi}{30}} + e^{i\frac{\pi}{15}}}{e^{i\frac{\pi}{30}} + 1} \] This simplifies to a complex number of the form \(e^{i\alpha}\) for some \(\alpha\). After some algebraic manipulation and recognizing the periodic nature, we find that: \[ \frac{1 + e^{i\frac{\pi}{30}}}{1 + e^{-i\frac{\pi}{30}}} = e^{i\frac{\pi}{30}} \] ### Step 6: Raise to the 30th Power Now, we need to raise this simplified expression to the 30th power: \[ (e^{i\frac{\pi}{30}})^{30} \] Using the property of exponents: \[ (e^{i\theta})^n = e^{i\theta n} \] So, \[ (e^{i\frac{\pi}{30}})^{30} = e^{i\frac{\pi}{30} \cdot 30} = e^{i\pi} \] ### Step 7: Simplify \(e^{i\pi}\) Using Euler's formula again: \[ e^{i\pi} = \cos(\pi) + i\sin(\pi) = -1 + 0i = -1 \] ### Final Answer Thus, the value of the given expression is: \[ \boxed{-1} \] |
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| gpto1 | To simplify the given expression and find its value, let's start by expressing the numerator and denominator in exponential form using Euler's formula. **Given:** \[ \left( \frac{1 + \cos(2x) + i\sin(2x)}{1 + \cos(2x) - i\sin(2x)} \right)^{30}, \text{ where } x = \frac{\pi}{60} \] **Step 1: Express numerator and denominator using trigonometric identities.** Recall the double-angle identities: \[ \cos(2x) = 2\cos^2 x - 1 \] \[ \sin(2x) = 2\sin x \cos x \] Therefore, \[ 1 + \cos(2x) = 2\cos^2 x \] \[ \sin(2x) = 2\sin x \cos x \] **Step 2: Rewrite the numerator and denominator.** Let’s write the numerator \(A\) and denominator \(B\) as: \[ A = 1 + \cos(2x) + i\sin(2x) = 2\cos^2 x + i(2\sin x \cos x) = 2\cos x (\cos x + i\sin x) \] \[ B = 1 + \cos(2x) - i\sin(2x) = 2\cos^2 x - i(2\sin x \cos x) = 2\cos x (\cos x - i\sin x) \] **Step 3: Use Euler's formula.** Euler's formula states: \[ e^{i\theta} = \cos \theta + i\sin \theta \] \[ e^{-i\theta} = \cos \theta - i\sin \theta \] So we can rewrite \(A\) and \(B\) using exponentials: \[ A = 2\cos x \cdot e^{i x} \] \[ B = 2\cos x \cdot e^{-i x} \] **Step 4: Simplify the fraction.** Compute \(R\): \[ R = \frac{A}{B} = \frac{2\cos x \cdot e^{i x}}{2\cos x \cdot e^{-i x}} = e^{i x} \cdot e^{i x} = e^{2i x} \] **Step 5: Compute \(R^{30}\).** \[ R^{30} = (e^{2i x})^{30} = e^{60i x} \] **Step 6: Substitute \(x = \frac{\pi}{60}\).** \[ R^{30} = e^{60i \cdot \frac{\pi}{60}} = e^{i \pi} = \cos \pi + i\sin \pi = -1 + i(0) = -1 \] **Final Answer:** \[ \left( \frac{1 + \cos(2x) + i\sin(2x)}{1 + \cos(2x) - i\sin(2x)} \right)^{30} = -1 \] **Answer: -1** |
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