Question Number 210126 by mnjuly1970 last updated on 31/Jul/24
Answered by Frix last updated on 31/Jul/24
![These substitutions make it easy to see what′s going on: Let x=sin α ⇒^([differentiate]) −1≤4x^2 −4x≤8 Let 4x^2 −4x=((7/2)+(9/2)sin β) ∈[−1; 8] −(7/2)<k ⇒ max(∣k+(7/2)+(9/2)sin β∣)=8+k>(9/2) k=−(7/2) ⇒ max(∣k+(7/2)+(9/2)sin β∣)=(9/2) ★ k<−(7/2) ⇒ max(∣k+(7/2)+(9/2)sin β∣)=1−k>(9/2)](https://www.tinkutara.com/question/Q210134.png)
$$\mathrm{These}\:\mathrm{substitutions}\:\mathrm{make}\:\mathrm{it}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see} \\ $$$$\mathrm{what}'\mathrm{s}\:\mathrm{going}\:\mathrm{on}: \\ $$$$ \\ $$$$\mathrm{Let}\:{x}=\mathrm{sin}\:\alpha\:\overset{\left[\mathrm{differentiate}\right]} {\Rightarrow} \\ $$$$−\mathrm{1}\leqslant\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}\leqslant\mathrm{8} \\ $$$$\mathrm{Let}\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}=\left(\frac{\mathrm{7}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}\mathrm{sin}\:\beta\right)\:\in\left[−\mathrm{1};\:\mathrm{8}\right] \\ $$$$−\frac{\mathrm{7}}{\mathrm{2}}<{k}\:\Rightarrow\:\mathrm{max}\left(\mid{k}+\frac{\mathrm{7}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}\mathrm{sin}\:\beta\mid\right)=\mathrm{8}+{k}>\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${k}=−\frac{\mathrm{7}}{\mathrm{2}}\:\Rightarrow\:\mathrm{max}\left(\mid{k}+\frac{\mathrm{7}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}\mathrm{sin}\:\beta\mid\right)=\frac{\mathrm{9}}{\mathrm{2}}\:\bigstar \\ $$$${k}<−\frac{\mathrm{7}}{\mathrm{2}}\:\Rightarrow\:\mathrm{max}\left(\mid{k}+\frac{\mathrm{7}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{2}}\mathrm{sin}\:\beta\mid\right)=\mathrm{1}−{k}>\frac{\mathrm{9}}{\mathrm{2}} \\ $$