# Find the maximum value of 1 + $$sin({\pi\over 4} + \theta)$$ + $$2cos({\pi\over 4} – \theta)$$

## Solution :

We have 1 + $$sin({\pi\over 4} + \theta)$$ + $$2cos({\pi\over 4} – \theta)$$

= 1 + $$1\over sqrt{2}$$$$(cos\theta + cos\theta)$$ + $$\sqrt{2}$$$$(cos\theta + cos\theta)$$

= 1 + $$({1\over \sqrt{2}} + \sqrt{2})$$ + $$(cos\theta + cos\theta)$$

= 1 + $$({1\over \sqrt{2}} + \sqrt{2})$$.$$\sqrt{2}cos(\theta – {pi\over 4})$$

$$\therefore$$   Maximum Value = 1 + $$({1\over \sqrt{2}} + \sqrt{2})$$.$$\sqrt{2}$$ = 4

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