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I have questions regarding trigonometry used in the solution of this problem:

Discuss the differentiability of $f(x)$ at the point $x=\frac{\pi}{2}$. $$f(x) = \begin{cases} 1, & x<0 \\[4pt] 1+\sin x, &0\leq x<\frac\pi2 \\[2pt] 2+\left(x-\frac{\pi}{2}\right)^2, &x\geq\frac\pi2 \end{cases}$$
$$\begin{align} Lf'\left(\frac{\pi}{2}\right)&=\lim_{h\to0^-} \frac{f(\frac{\pi}{2}+h)-f(\frac{\pi}{2})}{h} \\[4pt] &=\lim_{h\to0^-} \frac{1+\sin(\frac{\pi}{2}+h)-2}{h} \\[4pt] &=\lim_{h\to0^-} \frac{\cos h-1}{h} \\[4pt] &=\lim_{h\to0^-} \frac{-2 \sin^2 (\frac{h}{2})}{h} \\[4pt] &=-\lim_{h\to0^-} \left(\frac{\sin\frac{h}{2}}{\frac{h}{2}}\right)^2\cdot\frac{h}{2} \\[4pt] &=0 \end{align}$$

The last two line was little bit weird (It was looking just like magic) to me.

  1. How they converted $$\sin(\frac{\pi}{2}+h)=\cos h$$?
  2. How they wrote $$\cos h-1 = -2 \sin^2 (\frac{h}{2})$$

I think I am missing something on trigonometry.

Blue
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1 Answers1

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Both of your questions are answered by formulas which can be derived from the angle sum formulas for sine and cosine given by \begin{align} \sin(\alpha \pm \beta) &= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\ \cos(\alpha \pm \beta) &= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \end{align}

For the first part, using the angle sum formula for sine we get $$ \sin\left(\frac{\pi}{2}+h\right) = \underbrace{\sin\left(\frac{\pi}{2}\right)}_{\color{blue}{1}}\cos(h) + \sin(h) \underbrace{\cos\left(\frac{\pi}{2}\right)}_{\color{blue}{0}} = \cos(h) $$ For the second part, you can use the double angle formula for cosine (which is just the angle sum formula when $\alpha = \beta$) to obtain said relationship: \begin{align*} &\cos(2u) = \underbrace{\cos^2(u)}_{\color{blue}{1 - \sin^2(u)}} - \sin^2(u) = 1-2\sin^2(u) \\ \implies& -2\sin^2(u) =\cos(2u) - 1\overset{\color{purple}{u = h/2}}{\implies}-2\sin^2\left(\frac{h}{2}\right) =\cos(h) - 1 \end{align*}

Robert Lee
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  • Is ± same as ∓? ∓ symbol is new to me. But, it is looking like minus-plus.. ? –  Jul 05 '21 at 07:13
  • What happens if there 3 values inside $\sin$ or, $\cos$? $$\sin(\alpha + \beta + \gamma)$$ Can I use angle formula at that time? –  Jul 05 '21 at 07:18
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    @Istiak $a\pm b\mp c$ means that you either take $a+b-c$ or $a-b+c$. For $\sin(a+b+c)$ you write $\sin((a+b)+c)$ and expand as if $a+b$ was a single angle. Then you will have terms involving $\sin(a+b)$ and $\cos(a+b)$. You expand these again. – Gary Jul 05 '21 at 08:30
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    @Istiak, the symbol is indeed a minus-plus. It is written in this way in the cosine rule because when you have a plus on the LHS you will have a minus on the RHS, and viceversa. So you match either both the "top symbols" or both the "bottom symbols" depending on which case you're in. – Robert Lee Jul 05 '21 at 09:29
  • And as @Gary pointed out, you can indeed obtain a formula for adding three values (or any number of values you want) by repeated use of the normal sum identities. The only problem is that these formulas become really long really fast. – Robert Lee Jul 05 '21 at 09:34
  • For instante, the case you're interested in gives \begin{align} \sin([\alpha + \beta] + \gamma) &= \sin(\alpha + \beta)\cos(\gamma) + \cos(\alpha + \beta)\sin(\gamma) \ &=[ \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)]\cos(\gamma) + [ \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)]\sin(\gamma) \ &= \sin(\alpha)\cos(\beta)\cos(\gamma) + \cos(\alpha)\sin(\beta) \cos(\gamma)+ \cos(\alpha)\cos(\beta)\sin(\gamma) - \sin(\alpha)\sin(\beta)\sin(\gamma) \end{align} – Robert Lee Jul 05 '21 at 09:35