Question 62

The maximum value of sin θ + cos θ is

Solution

for asin θ + bcos θ + c,

maximum value = $$c+\sqrt{a^{2}+b^{2}}$$

minimum value = $$c-\sqrt{a^{2}+b^{2}}$$

for sin θ + cos θ , a = 1, b = 1, c = 0

maximum value = $$c+\sqrt{a^{2}+b^{2}}=0+\sqrt{1^{2}+1^{2}}=\sqrt{2}$$

so the answer is option B.

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