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Question: The maximum value of \(3\cos\theta + 4\sin\theta\) equal to...

The maximum value of 3cos⁡θ+4sin⁡θ3\cos\theta + 4\sin\theta equal to

A

3

B

4

C

5

D

None of these

Answer

3

Explanation

Solution

cos⁡2A+cos⁡2B−cos⁡2C=cos⁡2A+(1−sin⁡2B)−cos⁡2C=1+[cos⁡2A−sin⁡2B]−cos⁡2C=1+cos⁡(A+B)cos⁡(A−B)−cos⁡2C=1+cos⁡(π−C)cos⁡(A−B)−cos⁡2C=1−cos⁡C[cos⁡(A−B)+cos⁡C]\cos^{2}A + \cos^{2}B - \cos^{2}C = \cos^{2}A + (1 - \sin^{2}B) - \cos^{2}C = 1 + \lbrack\cos^{2}A - \sin^{2}B\rbrack - \cos^{2}C = 1 + \cos(A + B)\cos(A - B) - \cos^{2}C = 1 + \cos(\pi - C)\cos(A - B) - \cos^{2}C = 1 - \cos C\lbrack\cos(A - B) + \cos C\rbrack

=1−cos⁡C[cos⁡(A−B)+cos⁡{π−(A+B)}]=1−cos⁡C[cos⁡(A−B)−cos⁡(A+B)]=1−cos⁡C[2sin⁡Asin⁡B]=1−2sin⁡Asin⁡Bcos⁡C= 1 - \cos C\lbrack\cos(A - B) + \cos\{\pi - (A + B)\}\rbrack = 1 - \cos C\lbrack\cos(A - B) - \cos(A + B)\rbrack = 1 - \cos C\lbrack 2\sin A\sin B\rbrack = 1 - 2\sin A\sin B\cos C.