October 13, 2023 – Physicspupil

If $\cos A=\frac{4}{5}$ and $\cos B=\frac{12}{13}$ where $A$ and $B$ are both acute angles, find, without using tables or a calculator,

(a) $\sin (A+B)$ ,
(b) $\cos (A-B)$ ,
(c) $\tan (A+B)$ .

_{Extracted from J. F. Talbert & H. H. Heng. (1995). Additional Mathematics Pure and Applied (6e).}

Roughwork.

(a)

Not yet to approach a solution, I wish to get a feel of the problem from three perspectives— i. algebraic, ii. geometric, and iii. differential:

(i. algebraic)

$\begin{aligned} \cos A & =\frac{4}{5} \\ \cos B & = \frac{12}{13} \\ & \\ \cos^2A & = \frac{16}{25} \\ \cos^2B & = \frac{144}{169} \\ & \\ \sin^2A & = 1-\cos^2A\\ & = \frac{9}{25} \\ \sin^2B & = 1-\cos^2B\\ & = \frac{25}{169} \\ & \\ \sin A & = \sqrt{\frac{9}{25}} \\ & = \frac{3}{5} \\ \sin B & = \sqrt{\frac{25}{169}} \\ & = \frac{5}{13} \\ \end{aligned}$

(ii. geometric)

(iii. differential)

Let $f(x)=\sin x$ . Then, recall that

$\displaystyle{f(x)\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots }$

because of Taylor series/expansion of a function $f(x)$ at some point $a$ :

$\displaystyle{f(x)\big|_{x=a} = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\bigg|_{x=a}}$

such that in our situation, you know, we have $a=0$ (the Maclaurin series):

$\displaystyle{f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n}$

Now that we have enough, here begins the

Solution.

(i. algebraic)

$\begin{aligned} \sin (A+B) & = \sin A\cos B + \cos A\sin B \\ & = \bigg(\frac{3}{5}{\bigg)\bigg(\frac{12}{13}\bigg) + \bigg(\frac{4}{5}\bigg)\bigg(\frac{5}{13}\bigg) \\ & = \frac{56}{65} \\ \end{aligned}$

(ii. geometric)

$\begin{aligned} \frac{h}{14} & = \sin (90^\circ -B) \\ h & = 14\cos B \\ & \\ \sin (A+B) & = \frac{h}{15} \\ & = \frac{14\cos B}{15} \\ & = \frac{14}{15}\bigg( \frac{12}{13} \bigg) \\ & = \frac{56}{65} \\ \end{aligned}$

(iii. differential)

That angles $A$ and $B$ are acute angles, i.e.,

$0<B<A<90^\circ$

does not imply $A+B$ is also an acute angle, but that

$0<A+B<180^\circ$ .

such that

$\begin{aligned} 0 & <\sin (A+B)<1 \\ -1 & <\cos (A+B)<1 \\ 0 & <\sin B<\sin A<\cos A<\cos B < 1\\ \end{aligned}$

Letting $f(x)=\sin x$ ,

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [0,A+B] \\ & = \int_{0}^{A+B}f(x)\,\mathrm{d}x \\ & = \int_{0}^{A+B}\sin (x)\,\mathrm{d}x \\ & = \big[-\cos x\big]_{0}^{A+B} \\ & = 1-\cos (A+B) \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [0,A] \\ & = \int_{0}^{A}f(x)\,\mathrm{d}x \\ & = \int_{0}^{A}\sin (x)\,\mathrm{d}x \\ & = \big[-\cos x\big]_{0}^{A} \\ & = 1-\cos A \\ & = 1-\frac{4}{5} \\ & = \frac{1}{5} \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [0,B] \\ & = \int_{0}^{B}f(x)\,\mathrm{d}x \\ & = \int_{0}^{B}\sin (x)\,\mathrm{d}x \\ & = \big[-\cos x\big]_{0}^{B} \\ & = 1-\cos B \\ & = 1-\frac{12}{13} \\ & = \frac{1}{13} \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [B,A] \\ & = \frac{1}{5}-\frac{1}{13} \\ & = \frac{8}{65} \\ \end{aligned}$

$\begin{aligned} &\quad \textrm{Area under curve }f(x)\textrm{ in }x\in [A,A+B] \\ & = 1-\cos (A+B) - \frac{1}{5} \\ & = \frac{4}{5}-\cos (A+B) \\ \end{aligned}$

then letting $g(x)=\cos x$ , and noting in a right-angled triangle the cosine of an angle admits of no negative values, so can it further be said

$\begin{aligned} & \quad\enspace 0<g(x)<1 \\ &\Rightarrow 0<g(A+B)<1 \\ &\Rightarrow 0<A+B<90^\circ \\ \end{aligned}$

I lost my way…

Let $w(u(t),v(t))=u(t)-v(t)$ be the difference in degrees between time-varying angles $u$ and $v$ where assumed is $u(t)>v(t)$ without loss of generality.

$\begin{aligned} f(u,v,w) & \stackrel{\textrm{def}}{=} \sin (u+v) \\ & = \sin (2v+w ) \\ & = \sin (2u-w ) \\ & \\ g(u,v,w) &\stackrel{\textrm{def}}{=} \cos (u+v) \\ & = \cos (2v+w ) \\ & = \cos (2u-w ) \\ \end{aligned}$

Taking partial derivatives wrt $u$ , $v$ , $w$ , and $t$ :

$\begin{aligned} \partial_uf(u,v,w) & = 2\cos (2u-w) \\ \partial_vf(u,v,w) & = 2\cos (2v+w) \\ \partial_wf(u,v,w) & = 0 \\ \partial_tf(u,v,w) & = (u'+v')\cos (u+v)\\ \end{aligned}$

$\begin{aligned} \mathrm{d}f(u,v,w) & = \partial_uf\,\mathrm{d}u + \partial_vf\,\mathrm{d}v + \partial_wf\,\mathrm{d}w \\ & = 2\cos (2u-w )\,\mathrm{d}u + 2\cos (2v+w )\,\mathrm{d}v + 0 \\ \dots\,\mathrm{d}u & = \mathrm{d}v+\mathrm{d}w\,\dots \\ & = 4\cos (u+v)\,\mathrm{d}v + 2\cos (u+v)\,\mathrm{d}w \\ & = 2\big(\cos (u+v)\big)(2\,\mathrm{d}v+\mathrm{d}w) \\ \end{aligned}$

I lost my faith…

$f(x,y)\stackrel{\textrm{(1)}}{=}\sin (x+y)$

then

$f(x'=x+y,0)\stackrel{\textrm{(2)}}{=} f(x,y)\stackrel{\textrm{(3)}}{=} f(0,x+y=y')$ .

By equality $\textrm{(1)}$ it states that the output value $f(x,y)$ of $f(\texttt{var1},\texttt{var2})$ is determined by input values of two arguments $(\texttt{var1},\texttt{var2})=(x,y)$ ;

by equality $\textrm{(2)}$ that the output value $f(x,y)$ of $f(\texttt{var1},\texttt{var2})$ is determined by input values $(\texttt{var1},\texttt{var2})=(x+y,0)$ , i.e., by one significant argument $\texttt{var1}=x+y$ , and another trivial argument $\texttt{var2}=0$ ; and

by equality $\textrm{(3)}$ that the output value $f(x,y)$ of $f(\texttt{var1},\texttt{var2})$ is determined by input values $(\texttt{var1},\texttt{var2})=(0,x+y)$ , i.e., by one trivial argument $\texttt{var1}=0$ , and another significant argument $\texttt{var2}=x+y$ .

Hence, assuming a partial variation

$\begin{aligned} &\quad\enspace \sin (\texttt{var1}+\texttt{var2}) \\ & = k_1(\texttt{var1},\texttt{var2})\sin (\texttt{var1}) + k_2(\texttt{var1},\texttt{var2})\sin (\texttt{var2}) \\ \end{aligned}$

we WTS the following

$\begin{aligned} k_1(\texttt{var1},\texttt{var2}) & = \cos (\texttt{var2}) \\ k_2(\texttt{var1},\texttt{var2}) & = \cos (\texttt{var1}) \\ \end{aligned}$

LHS:

$\begin{aligned} &\quad\enspace \frac{\mathrm{d}}{\mathrm{d}t}\sin (\texttt{var1}+\texttt{var2}) \\ & = \big[\cos (\texttt{var1}+\texttt{var2})\big] (\texttt{var1}'+\texttt{var2}') \\ \end{aligned}$

RHS:

$\begin{aligned} &\quad\enspace \frac{\mathrm{d}}{\mathrm{d}t}(k_1\sin (\texttt{var1})+k_2\sin (\texttt{var2})) \\ & = \big(k_1'\cos (\texttt{var1})\big) \texttt{var1}' + \big(k_2'\cos (\texttt{var2})\big) \texttt{var2}' \\ \end{aligned}$

and LHS=RHS implies:

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}k_1 & = \frac{\cos (\texttt{var1}+\texttt{var2})}{\cos (\texttt{var1})} \\ \frac{\mathrm{d}}{\mathrm{d}t}k_2 & = \frac{\cos (\texttt{var1}+\texttt{var2})}{\cos (\texttt{var2})} \\ \dots & \dots \dots \dots \\ k_1' & = \bigg( \frac{\cos (\texttt{var2})}{\cos (\texttt{var1})}\bigg) k_2' \\ \end{aligned}$

(to be continued)

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Day: October 13, 2023

202310130920 Exercise 14.1.3