Question

4sin(α)+5cos(α)
Найдите наибольшее значение выражения​

Answer 1

$y=4sina+5cosa=\sqrt{41}\cdot \Big(\dfrac{4}{\sqrt{41}}\cdot sina+\dfrac{5}{\sqrt{41}}\cdot cosa\Big)=\\\\\\=\sqrt{41}\cdot (cos\beta \cdot sina+sin\beta \cdot cosa)=\sqrt{41}\cdot sin(a+\beta )\; ,\\\\-1\leq sin(a+\beta )\leq 1\;\; \; \Rightarrow \; \; \; \underline {-\sqrt{41}\leq \sqrt{41}\cdot sin(a+\beta )\leq \sqrt{41}\; }\\\\ \sqrt{41}\cdot sin(a+\beta )\in [-\sqrt{41}\, ;\, \sqrt{41}\; ]\\\\\underline {y(naibol.)=\sqrt{41}}$

$\star \; \; \Big(\dfrac{4}{\sqrt{41}}\Big)^2 +\Big(\dfrac{5}{\sqrt{41}}\Big)^2=\dfrac{16+25}{41}=1\; \; \Rightarrow \; \; \dfrac{4}{\sqrt{41}}=cos\beta \; \; ,\; \; \dfrac{5}{\sqrt{41}}=sin\beta ,\\\\\\sin^2\beta +cos^2\beta =1\; \; \star$

Answer 2

Ответ:

Объяснение:*******************