Hi there!
Today let me own up to disgracing myself in public in my kindergarten days. To my consternation, I swear it was the first but the last public humiliation.
It was March 3 when the Dolls’ Festival took place in my nursery school. Almost all my classmates’ parents came to see our stage performance, making all kinds of racket, saying “That is my son,” such and such.
I was about to play the role of the obina called “odairi-sama” of the dairi-bina, literally “Imperial Palace Dolls.” The parents erupted into applause as the curtain opened gradually.
At the center of the stage was me, who suddenly felt awkward without something in my hands.
―― “Where is my scepter?” I completely forgot to bring it amid escalating tensions.
―― “What the hell are you doing!?” screamed my teacher running to me with a scandalous look on her
face.
I had no choice but to improvise somehow, but how?
The teacher in the so-called “Sazae-san hairstyle” surprisingly proposed I should grab my hands in front of me as if they were a scepter. I don’t remember exactly, but I was apparently forced to behave erratically at that time.
In retrospect, it was not a stigma but the greatest disgrace for me in my entire life, feeling disgraced almost “double” in public.
Thus, solving a double-angle formula word problem today, though too much of a stretch, will be convincing to you, given the story’s drift.
Evaluate \(sin\,2\theta\) at \(sin\,\theta = \frac{2}{3}（90°<\theta<180°）\)
Here is my solution to the problem.
Let me use the formula \(sin^{2}\theta + cos^{2}\theta = 1\).
\(cos^{2}\theta = 1 – sin^{2}\theta = 1 – (\frac{2}{3})^{2} = \frac{5}{9}\)
Therefore, \(cos\,\theta = -\frac{\sqrt{5}}{3}\) given the condition of \((cos\,\theta<0)\) based on \((90°<\theta<180°)\). \(sin\,2\theta = 2sin\,\theta cos\,\theta = 2・\frac{2}{3}・(-\frac{\sqrt{5}}{3}) = -\frac{4\sqrt{5}}{9}\)
Stay tuned, and expect to see my next post.
Keep well.
Frank Yoshida
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【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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