If ((sin 1°+sin 2°+sin 3°+...+sin 44°)/(cos 1°+cos 2°+cos 3°+...+cos 44°))=χ then χ^4 +4χ^3 +4χ^2 +4=


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Question Number 115328 by bobhans last updated on 25/Sep/20

$I f \frac{\sin 1 ° + \sin 2 ° + \sin 3 ° + . . . + \sin 44 °}{\cos 1 ° + \cos 2 ° + \cos 3 ° + . . . + \cos 44 °} = χ$ $t h e n χ^{4} + 4 χ^{3} + 4 χ^{2} + 4 =$
	Answered by bemath last updated on 25/Sep/20
	$sin 44°+sin 1°=2sin (((45°)/2)).cos (((43°)/2)) sin 43°+sin 2°=2sin (((45°)/2)).cos (((41°)/2)) cos 44°+cos 1° = 2cos (((45°)/2))cos (((43°)/2)) cos 43°+cos 2° = 2cos (((45°)/2))cos (((41°)/2)) ((2sin (((45°)/2)) {cos (((43°)/2))+cos (((41°)/2))+...})/(2cos (((45°)/2)) {cos (((43°)/2))+cos (((41°)/2))+...})) = χ so χ = tan (((45°)/2)) consider tan 2x = ((2tan x)/(1−tan^2 x)) put x = ((45°)/2)⇒1−tan^2 (((45°)/2))=2tan (((45°)/2)) let tan (((45°)/2)) = q → q^2 +2q−1=0 (q+1)^2 = 2 ⇒q = (√2) −1 hence χ = (√2) −1 or χ+1 = (√2) now consider χ^4 +4χ^3 +4χ+4 = χ^3 (χ+4)+4(χ+1) = χ^3 (3+(√2))+4(√2) = ((√2)−1)(3−2(√2))(3+(√2))+4(√2) = ((√2)−1)(9−3(√2)−4)+4(√2) = ((√2)−1)(5−3(√2))+4(√2) = 5(√2)−6−5+3(√2) +4(√2) =12(√2)−11$
	$\sin 44 ° + \sin 1 ° = 2 \sin (\frac{45 °}{2}) . \cos (\frac{43 °}{2})$ $\sin 43 ° + \sin 2 ° = 2 \sin (\frac{45 °}{2}) . \cos (\frac{41 °}{2})$ $\cos 44 ° + \cos 1 ° = 2 \cos (\frac{45 °}{2}) \cos (\frac{43 °}{2})$ $\cos 43 ° + \cos 2 ° = 2 \cos (\frac{45 °}{2}) \cos (\frac{41 °}{2})$ $\frac{2 \sin (\frac{45 °}{2}) {\cos (\frac{43 °}{2}) + \cos (\frac{41 °}{2}) + . . .}}{2 \cos (\frac{45 °}{2}) {\cos (\frac{43 °}{2}) + \cos (\frac{41 °}{2}) + . . .}} = χ$ $s o χ = \tan (\frac{45 °}{2})$ $c o n s i d e r \tan 2 x = \frac{2 \tan x}{1 - \tan^{2} x}$ $p u t x = \frac{45 °}{2} \Rightarrow 1 - \tan^{2} (\frac{45 °}{2}) = 2 \tan (\frac{45 °}{2})$ $l e t \tan (\frac{45 °}{2}) = q \to q^{2} + 2 q - 1 = 0$ ${(q + 1)}^{2} = 2 \Rightarrow q = \sqrt{2} - 1$ $h e n c e χ = \sqrt{2} - 1 o r χ + 1 = \sqrt{2}$ $n o w c o n s i d e r χ^{4} + 4 χ^{3} + 4 χ + 4 =$ $χ^{3} (χ + 4) + 4 (χ + 1) =$ $χ^{3} (3 + \sqrt{2}) + 4 \sqrt{2} =$ $(\sqrt{2} - 1) (3 - 2 \sqrt{2}) (3 + \sqrt{2}) + 4 \sqrt{2} =$ $(\sqrt{2} - 1) (9 - 3 \sqrt{2} - 4) + 4 \sqrt{2}$ $= (\sqrt{2} - 1) (5 - 3 \sqrt{2}) + 4 \sqrt{2}$ $= 5 \sqrt{2} - 6 - 5 + 3 \sqrt{2} + 4 \sqrt{2}$ $= 12 \sqrt{2} - 11$
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