[문제 3-2]
+3점 · 교점 B( x 0 , y 0 ) \displaystyle (x_0,y_0) ( x 0 , y 0 ) x 0 = 2 4 + tan 2 2 θ \displaystyle x_0=\frac2{\sqrt{4+\tan^22\theta}} x 0 = 4 + tan 2 2 θ 2 y 0 = 2 tan 2 θ 4 + tan 2 2 θ \displaystyle y_0=\frac{2\tan2\theta}{\sqrt{4+\tan^22\theta}} y 0 = 4 + tan 2 2 θ 2 tan 2 θ 교점 B( x 0 , y 0 ) \displaystyle (x_0,y_0) ( x 0 , y 0 ) x 2 + y 2 4 = 1 \displaystyle x^2+\frac{y^2}4=1 x 2 + 4 y 2 = 1 y = ( tan 2 θ ) x \displaystyle y=(\tan2\theta)x y = ( tan 2 θ ) x ( θ ≠ π 4 ) \displaystyle \left(\theta\ne\frac\pi4\right) ( θ = 4 π ) x 0 2 = 4 4 + tan 2 ( 2 θ ) \displaystyle x_0^2=\frac4{4+\tan^2(2\theta)} x 0 2 = 4 + tan 2 ( 2 θ ) 4 ( x 0 , y 0 ) \displaystyle (x_0,y_0) ( x 0 , y 0 ) y = ( tan θ ) x \displaystyle y=(\tan\theta)x y = ( tan θ ) x ( θ ≠ π 2 ) \displaystyle \left(\theta\ne\frac\pi2\right) ( θ = 2 π )
∣ x 0 tan θ − y 0 ∣ 1 + tan 2 θ = ∣ x 0 tan θ − x 0 tan ( 2 θ ) ∣ 1 + tan 2 θ \displaystyle \frac{|x_0\tan\theta-y_0|}{\sqrt{1+\tan^2\theta}}=\frac{|x_0\tan\theta-x_0\tan(2\theta)|}{\sqrt{1+\tan^2\theta}} 1 + tan 2 θ ∣ x 0 tan θ − y 0 ∣ = 1 + tan 2 θ ∣ x 0 tan θ − x 0 tan ( 2 θ ) ∣
+5점 · 삼각형의 넓이 1 2 tan θ 1 + tan 2 θ ( 1 − tan 2 θ ) 2 + tan 2 θ \displaystyle \frac12\frac{\tan\theta\sqrt{1+\tan^2\theta}}{\sqrt{(1-\tan^2\theta)^2+\tan^2\theta}} 2 1 ( 1 − tan 2 θ ) 2 + tan 2 θ tan θ 1 + tan 2 θ 이다. 삼각형 AOB의 넓이는
1 2 ∣ tan θ − tan ( 2 θ ) ∣ 1 + tan 2 θ ∣ x ∣ = ∣ tan θ − tan ( 2 θ ) ∣ 1 + tan 2 θ 1 4 + tan 2 ( 2 θ ) \displaystyle \frac12\frac{|\tan\theta-\tan(2\theta)|}{\sqrt{1+\tan^2\theta}}|x|=\frac{|\tan\theta-\tan(2\theta)|}{\sqrt{1+\tan^2\theta}}\frac1{\sqrt{4+\tan^2(2\theta)}} 2 1 1 + tan 2 θ ∣ tan θ − tan ( 2 θ ) ∣ ∣ x ∣ = 1 + tan 2 θ ∣ tan θ − tan ( 2 θ ) ∣ 4 + tan 2 ( 2 θ ) 1
이고 tan 2 θ = 2 tan θ 1 − tan 2 θ \displaystyle \tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta} tan 2 θ = 1 − tan 2 θ 2 tan θ 1 2 tan θ 1 + tan 2 θ ( 1 − tan 2 θ ) 2 + tan 2 θ \displaystyle \frac12\frac{\tan\theta\sqrt{1+\tan^2\theta}}{\sqrt{(1-\tan^2\theta)^2+\tan^2\theta}} 2 1 ( 1 − tan 2 θ ) 2 + tan 2 θ tan θ 1 + tan 2 θ θ = π 4 \displaystyle \theta=\frac\pi4 θ = 4 π 2 2 \displaystyle \frac{\sqrt2}2 2 2 f ( θ ) \displaystyle f(\theta) f ( θ )
f ( θ ) = { 1 2 tan θ 1 + tan 2 θ ( 1 − tan 2 θ ) 2 + tan 2 θ ( θ ≠ π 2 ) 1 2 ( θ = π 2 ) \displaystyle f(\theta)=\begin{cases}\frac12\frac{\tan\theta\sqrt{1+\tan^2\theta}}{\sqrt{(1-\tan^2\theta)^2+\tan^2\theta}}&\left(\theta\ne\frac\pi2\right)\\\frac12&\left(\theta=\frac\pi2\right)\end{cases} f ( θ ) = { 2 1 ( 1 − t a n 2 θ ) 2 + t a n 2 θ t a n θ 1 + t a n 2 θ 2 1 ( θ = 2 π ) ( θ = 2 π )
+7점 · tan 2 θ = 1 + 3 2 \displaystyle \tan^2\theta=\frac{1+\sqrt3}2 tan 2 θ = 2 1 + 3 M 2 = 3 + 2 3 12 \displaystyle M^2=\frac{3+2\sqrt3}{12} M 2 = 12 3 + 2 3 f 2 \displaystyle f^2 f 2 s = tan 2 θ \displaystyle s=\tan^2\theta s = tan 2 θ ( s ≥ 0 ) \displaystyle (s\ge0) ( s ≥ 0 )
f 2 = s ( 1 + s ) 4 { ( 1 − s ) 2 + s } = s 2 + s 4 ( s 2 − s + 1 ) \displaystyle f^2=\frac{s(1+s)}{4\{(1-s)^2+s\}}=\frac{s^2+s}{4(s^2-s+1)} f 2 = 4 {( 1 − s ) 2 + s } s ( 1 + s ) = 4 ( s 2 − s + 1 ) s 2 + s
이고 미분하면 − 2 s 2 + 2 s + 1 4 ( s 2 − s + 1 ) 2 \displaystyle \frac{-2s^2+2s+1}{4(s^2-s+1)^2} 4 ( s 2 − s + 1 ) 2 − 2 s 2 + 2 s + 1 s = 1 + 3 2 \displaystyle s=\frac{1+\sqrt3}2 s = 2 1 + 3 M 2 = 1 4 + 3 6 = 3 + 2 3 12 \displaystyle M^2=\frac14+\frac{\sqrt3}6=\frac{3+2\sqrt3}{12} M 2 = 4 1 + 6 3 = 12 3 + 2 3
수리논술, 배움에서 논증의 완성까지.