Max and min

Geometry Level 4

cos ⁡ 2 θ − 6 sin ⁡ θ cos ⁡ θ + 3 sin ⁡ 2 θ + 2 \cos^2\theta - 6\sin\theta\cos\theta+3\sin^2\theta +2

Let θ \theta be a real number. Find the sum of the maximum and minimum values of the expression above.


The answer is 8.

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Rakshit Joshi by Rakshit Joshi , 26, India

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1 solution

Chew-Seong Cheong
May 16, 2016

X = cos ⁡ 2 θ − 6 sin ⁡ θ cos ⁡ θ + 3 sin ⁡ 2 θ + 2 sin ⁡ 2 θ + cos ⁡ 2 θ = 1 = 1 − 6 sin ⁡ θ cos ⁡ θ + 2 sin ⁡ 2 θ + 2 = 3 − 3 ( 2 sin ⁡ θ cos ⁡ θ ) + 2 sin ⁡ 2 θ − 1 + 1 = 4 − 3 sin ⁡ ( 2 θ ) − cos ⁡ ( 2 θ ) = 4 − 10 ( 3 10 sin ⁡ ( 2 θ ) − 1 10 cos ⁡ ( 2 θ ) ) = 4 − 10 ( sin ⁡ ( 2 θ ) cos ⁡ ϕ + cos ⁡ ( 2 θ ) sin ⁡ ϕ ) Let sin ⁡ ϕ = 1 10 , cos ⁡ ϕ = 3 10 ⟹ ϕ = tan ⁡ − 1 1 3 ⟹ X = 4 − 10 sin ⁡ ( 2 θ + tan ⁡ − 1 1 3 ) X m a x = 4 − 10 ( − 1 ) = 4 + 10 X m i n = 4 − 10 ( 1 ) = 4 − 10 \begin{aligned} X & = \color{#3D99F6}{\cos^2 \theta} - 6 \sin \theta \cos \theta + \color{#3D99F6}{3 \sin^2 \theta} + 2 \quad \quad \small \color{#3D99F6}{\sin^2 \theta + \cos^2 \theta = 1} \\ & = \color{#3D99F6}{1} - 6 \sin \theta \cos \theta + \color{#3D99F6}{2 \sin^2 \theta} + 2 \\ & = 3 - 3 (2 \sin \theta \cos \theta) + 2 \sin^2 \theta - 1 + 1 \\ & = 4 - 3 \sin (2 \theta) - \cos (2 \theta) \\ & = 4 - \sqrt{10} \left( \frac{3}{\sqrt{10}} \sin (2 \theta) - \frac{1}{\sqrt{10}} \cos (2 \theta) \right) \\ & = 4 - \sqrt{10} \left(\sin (2 \theta) \cos \phi + \cos (2 \theta) \sin \phi \right) \quad \quad \small \color{#3D99F6}{\text{Let }\sin \phi = \frac{1}{\sqrt{10}}, \ \cos \phi = \frac{3}{\sqrt{10}} \implies \phi = \tan^{-1} \frac{1}{3}} \\ \implies X & = 4 - \sqrt{10} \sin \left( 2 \theta + \tan^{-1} \frac{1}{3} \right) \\ X_{max} & = 4 - \sqrt{10} \left(-1 \right) = 4 + \sqrt{10} \\ X_{min} & = 4 - \sqrt{10} \left(1 \right) = 4 - \sqrt{10} \end{aligned}

⟹ X m a x + X m i n = 8 \implies X_{max} + X_{min} = \boxed{8}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

This time I got it perfectly;)

Rakshit Joshi - 5 years, 1 month ago

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It is the same trick.

Chew-Seong Cheong - 5 years, 1 month ago

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well you could also use + − a 2 + b 2 +-\sqrt{a^2+b^2} i.e + − − 3 2 + − 2 2 +-\sqrt{-3^2+-2^2} = + − 10 +-\sqrt{10} .

then max is 4 + 10 4+\sqrt{10} and min. is 4 − 10 4-\sqrt{10} instead of taking t a n − 1 tan^-1 ;) isn't it

Rakshit Joshi - 5 years ago

Since we know that the minimum value of sin and cos is -1 and maximum value of sin and cos is 1

If we put sinx=cosx=-1 in the equation we get 1-6+3+2=0

And sinx=cosx=1 we get 1-6+3+2=0

Sum of minimum and maximum value is 0

Sid Rana - 5 years ago

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When sin ⁡ x = − 1 \sin x = -1 , cos ⁡ x = 0 \cos x = 0 and when cos ⁡ x = − 1 \cos x = -1 , sin ⁡ x = 0 \sin x = 0 . It is impossible to have sin ⁡ x = cos ⁡ x = − 1 \sin x = \cos x = -1 . sin ⁡ x \sin x and cos ⁡ x \cos x are not independent. In fact, cos ⁡ x = sin ⁡ ( π / 2 − x ) \cos x = \sin (\pi/2 -x) or sin ⁡ x = cos ⁡ ( π / 2 − x ) \sin x = \cos (\pi/2 -x) . We can never have sin ⁡ x = sin ⁡ ( π / 2 − x ) = − 1 \sin x = \sin (\pi/2 - x) = -1 . sin ⁡ x = cos ⁡ x = ± 1 / 2 \sin x = \cos x = \pm 1/\sqrt{2} , when x = ( 2 k + 1 / 4 ) π x = (2k + 1/4) \pi , where k k is an integer.

Chew-Seong Cheong - 5 years ago

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Thanks for correcting me

Can you solve it the way I wantedto do

Sid Rana - 5 years ago

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@Sid Rana – No, your method is not useful. Because sin ⁡ x \sin x and cos ⁡ x \cos x do not have maximum and minimum together. That is why we have to change the form with sin ⁡ x \sin x and cos ⁡ x \cos x to only sin ⁡ x \sin x or only cos ⁡ x \cos x as I have done.

Chew-Seong Cheong - 5 years ago

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