Question 179

The minimum value of $$2sin^{2}$$ θ + $$3cos^{2}$$ θ is

Solution

Expression : $$2sin^{2}$$ θ + $$3cos^{2}$$ θ

We know that, $$sin^2 \theta = 1 - cos^2 \theta$$

=> $$2 (1- cos^2 \theta) + 3cos^2 \theta$$

= $$cos^2 \theta + 2$$

Using the formula, $$cos^2 \theta = \frac{cos 2\theta + 1}{2}$$

=> $$\frac{cos 2\theta + 1}{2} + 2$$

= $$\frac{cos 2\theta}{2} + \frac{5}{2}$$

$$\because$$ minimum value of $$cos 2\theta = -1$$

=> min value = $$\frac{5}{2} - \frac{1}{2} = 2$$

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