20.
高2第2回
15 (1) Cos 20 = C05²0-Sinzo
= (1 - sin²0) -Sinza
= 1-2 sin ²00
(2) -25in²0 + 2 (50-1) sing - 1₂0² + (0-1=0
a=0ac -25in²0 — 2 sin0 /1 = 0
-25ing (sino + 1) = 0
再
Sing=0₁
0=0.π 310
(3) -2sin² + 2(50-1) sind -120²³ + 60-2=0
Sinz0 - (50-1) sin@ +60²²-30+1=0
Sind=tegicy
f(t)= +² - (5a-1)t + (a²-30+ [ ) = 0
0 ≤ 0 < 2π F4-lets /
~[t_50+)² (50-1)²
2
4
2
+6a²-30+1
2
= (t_52-1) ² 250 ²
4
240²² 20² +4
250²-100+1
+ 4
= (t_50²-¹)²-a²-20² + 3
4
1) pr 1 ≤t ≤ / apr
la
範囲中で
異なる実数解でつつもてばよい
(i) 頂点のy座標=204350
(²²) ₁ x = 30-15" | ate
2
2
(₁₁²) F(-1) >0 f1² f(1) > 0
-0²-20+3.
(^)
KO (12²
4
33
a²+ 20-370 1/11
(a +3/a-1)>0
as-3. Ka
50+
(ii) - | ≤ 2
≤
-255a-152
60²-82²+3=0
A=4=√16-18
5a = 3
3
(ii) f(-1) = 1 + (5α - ) +-la²³ -70+ [ 20
60² +20 + 1 = 0
60²+2a+1=0 f(1) = 1-5″ + | +f6²70+|
a=-1± √1-6
70
6
= 60²-80²+3>0