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If theta is real then 3 - cos theta + co...

If `theta` is real then `3 - cos theta + cos (theta + (pi)/(3))` lies in the interval

A

`[-2, 3]`

B

`[-2, 1]`

C

`[2, 4]`

D

`[1, 5]`

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the expression \(3 - \cos \theta + \cos\left(\theta + \frac{\pi}{3}\right)\) and determine the interval in which it lies. ### Step 1: Expand the expression We start with the expression: \[ 3 - \cos \theta + \cos\left(\theta + \frac{\pi}{3}\right) \] Using the cosine addition formula, we can expand \(\cos\left(\theta + \frac{\pi}{3}\right)\): \[ \cos\left(\theta + \frac{\pi}{3}\right) = \cos \theta \cos\frac{\pi}{3} - \sin \theta \sin\frac{\pi}{3} \] Substituting \(\cos\frac{\pi}{3} = \frac{1}{2}\) and \(\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\), we get: \[ \cos\left(\theta + \frac{\pi}{3}\right) = \cos \theta \cdot \frac{1}{2} - \sin \theta \cdot \frac{\sqrt{3}}{2} \] ### Step 2: Substitute back into the expression Now substituting this back into our original expression: \[ 3 - \cos \theta + \left(\frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta\right) \] Combining like terms: \[ 3 - \cos \theta + \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta = 3 - \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \] ### Step 3: Rewrite the expression Now we can rewrite this as: \[ 3 - \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \] ### Step 4: Identify coefficients for the interval calculation Let’s denote: - \(a = -\frac{1}{2}\) - \(b = -\frac{\sqrt{3}}{2}\) - \(c = 3\) ### Step 5: Calculate the range We can find the maximum and minimum values of the expression \(c - \sqrt{a^2 + b^2}\) and \(c + \sqrt{a^2 + b^2}\). Calculating \(a^2 + b^2\): \[ a^2 + b^2 = \left(-\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1 \] Thus, \(\sqrt{a^2 + b^2} = \sqrt{1} = 1\). ### Step 6: Find the minimum and maximum values Now we can find the minimum and maximum values: - Minimum value: \[ c - \sqrt{a^2 + b^2} = 3 - 1 = 2 \] - Maximum value: \[ c + \sqrt{a^2 + b^2} = 3 + 1 = 4 \] ### Conclusion The expression \(3 - \cos \theta + \cos\left(\theta + \frac{\pi}{3}\right)\) lies in the interval \([2, 4]\).
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