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


Similar Questions

Evaluate sin78 – sin66 – sin42 + sin6.

If A + B + C = \(3\pi\over 2\), then cos2A + cos2B + cos2C is equal to

\(sin5x + sin2x – sinx\over {cos5x + 2cos3x + 2cos^x + cosx}\) is equal to

Prove that \(2cos2A+1\over {2cos2A-1}\) = tan(\(60^{\circ}\) + A)tan(\(60^{\circ}\) – A)

Leave a Comment

Your email address will not be published.