Trigonometric Function

Algebra Level 2

If f ( x ) = 8 8 cos x + 8 + 8 cos x f(x)=\sqrt{8-8\cos{x}}+\sqrt{8+8\cos{x}} and the Domain is restricted to [ 0 , π ] [0,π] , then the range of f ( x ) f(x) is [ a , b ] [a,b] where a a and b b are real numbers. Determine the approximate value of ( a + b ) (a+b) . Note: You can use a Scientific Calculator.


The answer is 9.65685.

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1 solution

Yashas Ravi
Oct 9, 2019

The expression f ( x ) = 8 8 cos x + 8 + 8 cos x f(x)=\sqrt{8-8\cos{x}}+\sqrt{8+8\cos{x}} can be rewritten as f ( x ) = 4 sin 0.5 x + 4 cos 0.5 x = 4 2 ( sin ( 0.5 x + 0.25 π ) ) f(x)=4\sin{0.5x}+4\cos{0.5x}=4\sqrt{2}(\sin{(0.5x+0.25π)}) using the Half Angle identities and the R-Method. Since the Domain is [ 0 , π ] [0,π] , sin 0.5 x \sin{0.5x} and cos 0.5 x \cos{0.5x} are both positive so the minimum value of sin ( 0.5 x + 0.25 π ) \sin{(0.5x+0.25π)} is 0.5 2 0.5\sqrt{2} when x = 0 x=0 and maximum at 1 1 when x = 0.5 π x=0.5π . Thus, 4 2 ( 1 + 0.5 2 ) 4\sqrt{2}*(1+0.5\sqrt{2}) is approximately 9.65685 9.65685 , which is the final answer.

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