Minimario's equations

Note that $\begin{aligned} a \sin x + b \cos x &= \sqrt{a^2 + b^2}\left(\frac{a}{\sqrt{a^2 + b^2}} \sin x + \frac{b}{\sqrt{a^2 + b^2}}\cos x\right) \\ &= \sqrt{a^2 + b^2}(\cos y \cdot \sin x + \sin y \cdot \cos x) = \sqrt{a^2 + b^2}\sin(x + y) \end{aligned}$ where $\cos y = \frac{a}{\sqrt{a^2 + b^2}}$ and $\sin y = \frac{b}{\sqrt{a^2 + b^2}}$ . Hence $$-\sqrt{a^2 + b^2} \leq a\sin x + b\cos x \leq \sqrt{a^2 + b^2}$$ So $$-\sqrt{13^2 + 11^2} \leq 13 \sin x + 11 \cos x \leq \sqrt{13^2 + 11^2}$$ i.e. $$-\sqrt{290} \leq 13\sin x + 11\cos x \leq \sqrt{290}$$ Now $17^2 = 289$ , hence $\left\lfloor\sqrt{290} \right\rfloor = 17$ . So $2k + 1 \in \{-17,-15,\dots,15,17\}$ , i.e. $k \in \{-9,-8,\dots,7,8\}$ . So there are $$9 + 8 + 1 = \boxed{18}$$ such values of $k$ .

• To find range of f(x) differentiate:
• Therefore dy/dx= 13cos(x)-11sin(x)
• Looking for max/min therefore where dy/dx= 0
• We can then arrange to show that tan(x)=13/11
• From this we can deduce 13sin(x)=+/-169/sqrt(290) and 11cos(x)=+/-121/169 .
• We can then put these expressions into f(x) to show max at +sqrt(290) and minimum at -sqrt(290). This is because the solutions to tan(x)=13/11 are such that cos(x) and sin(x) are positive and negative at the same times.
• We now need to find the value of sqrt(290). 16^2 is 256 (2^8), so try one higher to reach 17^2=289. We now know that sqrt(290) is just greater than 17.
• The range of this function is now from just below -17 to just above +17, covering every integer in between.
• Only odd numbers can be expressed in the form 2k+1 where k is an integer, so we count the odd numbers between (including) -17 and +17, reaching the correct answer of 18 solutions.

11cosx + 13sinx ranges from $- \sqrt{11^{2}+ 13^2}$ to $\sqrt{11^{2}+ 13^2}$

=> - $\sqrt{121+ 169}$ </= 2k+1</= $\sqrt{121 + 169}$

=> - $\sqrt{290}$ </= 2k+1</= $\sqrt{290}$

=> -17.02 </=2k+1 </= 17.02

Number of integral values of 'k' satisfying this inequality is 18.

11cosx+13sinx=V290 sin(0+40.2) V290 sin(0+40.2)=2k+1 sin(0+40.2)=2k+1/V290 Since sine lies between -1 and 1 inclusive, -17<=2k+1<=17 -9<=k<=8 Total number of integers that lie between -9 and 8 inclusive=18, hence total number of possible integer values of k=18