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Geometry Level 3

[ ( 4 cos ⁡ 4 0 ∘ ) 2 − 3 ] × [ 3 − ( 4 sin ⁡ 4 0 ∘ ) 2 ] = A + B cos ⁡ 2 0 ∘ [ (4 \cos 40 ^\circ )^2-3] \times [3-( 4 \sin 40 ^\circ )^2 ]=A+ B \cos20 ^ \circ

Find A-B (given that they are both integers).


The answer is 39.

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Avn Bha by avn bha , 22, India

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1 solution

Chew-Seong Cheong
May 27, 2015

[ ( 4 cos ⁡ 4 0 ∘ ) 2 − 3 ] [ 3 − ( 4 2 sin ⁡ 2 4 0 ∘ ) 2 ] = ( 16 cos ⁡ 2 4 0 ∘ − 8 + 5 ) ( − 5 + 8 − 16 sin ⁡ 2 4 0 ∘ ) = ( 8 cos ⁡ 8 0 ∘ + 5 ) ( 8 cos ⁡ 8 0 ∘ − 5 ) = ( 64 cos ⁡ 2 8 0 ∘ − 25 ) = 32 cos ⁡ 16 0 ∘ + 7 = 7 − 32 cos ⁡ 2 0 ∘ \begin{aligned} \left[\left(4\cos{40^\circ}\right)^2-3\right] \left[3-\left(4^2\sin^2{40^\circ}\right)^2\right] & = \left(16\cos^2{40^\circ}-8+5\right)\left(-5+ 8 -16\sin^2{40^\circ}\right) \\ & = \left(8\cos{80^\circ}+5\right)\left(8\cos{80^\circ}-5\right) \\ & = \left(64\cos^2{80^\circ}-25\right) \\ & = 32\cos{160^\circ}+7 \\ & = 7 - 32\cos{20^\circ} \end{aligned}

⇒ A − B = 7 − ( − 32 ) = 39 \Rightarrow A - B = 7-(-32) = \boxed{39}

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Be careful with parenthesis!

a great solution !!

avn bha - 6 years ago

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I hope you can learn up how to use LaTex. Just hover over to see the keystrokes.

Chew-Seong Cheong - 6 years ago

According to brackets 4 is also in whole square with cos40 . Hence answer is 39

anchal shukla - 6 years ago

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Thanks. I have updated the answer to 39.

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