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(請用六種不同高中數學方法解此題)

sin(a)=sin(a-Pi/6+Pi/6)=sin(a-Pi/6)*cos(Pi/6)+cos(a-Pi/6)*sin(Pi/6)
=(4/5)*(3^0.5/2)+(3/5)*(1/2) =(4*3^0.5+3)/10

101.6.4版主補充

101.6.18版主補充

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https://math.pro/db/attachment.php?aid=1197&k=cef9a7326187212d23c498ab88fa7fb9&t=1709180443

$$\displaystyle 0 < a < \frac{\pi}{2} \Rightarrow -\frac{\pi}{3} < a-\frac{\pi}{3} < \frac{\pi}{6}$$ , 因此  $$\displaystyle -\frac{\sqrt{3}}{2} < \sin (a-\frac{\pi}{3}) < \frac12$$

##### 引用:

$$0 < a < \frac{\pi}{2} \Rightarrow -\frac{\pi}{3} < a-\frac{\pi}{3} < \frac{\pi}{6}$$ , 因此  $$-\frac{\sqrt{3}}{2} < \sin (a-\frac{\pi}{3}) < \frac12$$ ...

https://math.pro/db/attachment.php?aid=1192&k=b326e0461cbd0916fcfe5a38f5757f0a&t=1709180443

2.

13.

thepaino 老師解了： http://www.shiner.idv.tw/teachers/viewtopic.php?f=53&p=7716

14.

10.

$$sin^2 x-(2a+1)cosx-a^2=0$$
$$1-cos^2 x-(2a+1)cosx-a^2=0$$
$$cos^2 x+(2a+1)cos x+(a^2-1)=0$$

1.
$$D\ge 0$$，$$\displaystyle (2a+1)^2-4(a^2-1)\ge 0 \Rightarrow a\ge -\frac{5}{4}$$
2.

(1)$$f(1)\ge 0 \Rightarrow a^2+2a+1 \ge 0 \Rightarrow a \in R$$
(2)$$f(-1)\ge 0 \Rightarrow a^2-2a-1 \ge 0 \Rightarrow a\ge 1+\sqrt{2}$$或$$a \le 1-\sqrt{2}$$
(3)$$\displaystyle -1<\frac{-2a-1}{2}<1 \Rightarrow -\frac{3}{2}\le a \le \frac{1}{2}$$

https://math.pro/db/attachment.php?aid=2629&k=0810276b46570c5b0889bcdbfa508392&t=1709180443

https://math.pro/db/attachment.php?aid=2630&k=d3deeae988b5a4c9945e6e3b41db7ba8&t=1709180443

8.

##### 引用:

$${x^2} = - a(y - a)$$

$$\displaystyle y' = - \frac{2}{a}x$$，故過P點之切線斜率為$$- 2t$$

$$\displaystyle \Delta = \frac{1}{2}\frac{{a + a{t^2}}}{{2t}}(a + a{t^2}) = \frac{1}{4}\frac{{{{(a + a{t^2})}^2}}}{t}$$
$$\displaystyle \Delta ' = \frac{1}{4}\frac{{2(a + a{t^2}) \cdot 2at \cdot t - (a + a{t^2})2}}{{{t^2}}} = 0$$

$$\displaystyle \Delta = \frac{1}{4}\sqrt 3 \cdot {a^2}{(1 + \frac{1}{3})^2} = \frac{{4\sqrt 3 }}{9}{a^2}$$

##### 引用:

#2
⊿ACF≈⊿AEC [AB=AC=小圓半徑 => AB弧=AC弧 =>∠AFC=∠BCA]

=> AC/AF = AE/AC

=> AB/ 4 = 1/AB

=> AB=2

a = -cosx ± √(1 - cosx)

a = 2t^2 - 1 ± (√2)t = 2(t ± √2/4)^2 - 5/4
-5/4 ≦ a ≦ 1 + √2

12.補充一個拙見
$$\displaystyle f(x)=-x \pm \sqrt{1-x}$$

case 1. if $$\displaystyle f(x)=-x + \sqrt{1-x} , -1\leq x\leq 1$$

case 2. if $$\displaystyle f(x)=-x + \sqrt{1-x} , -1\leq x\leq 1$$

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