Question Number 20505 by Tinkutara last updated on 27/Aug/17
$${Simplify}: \\ $$$$\mathrm{cos}^{−\mathrm{1}} \:\left(\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{cos}\:{x}\:+\:\frac{\mathrm{4}}{\mathrm{5}}\:\mathrm{sin}\:{x}\right),\:{where} \\ $$$$−\frac{\mathrm{3}\pi}{\mathrm{4}}\:\leqslant\:{x}\:\leqslant\:\frac{\pi}{\mathrm{4}} \\ $$
Answered by ajfour last updated on 27/Aug/17
![let tan α=(4/3) ⇒ sin α=(4/5), and cos α=(3/5) ; α > x θ=cos^(−1) [cos αcos x+sin αsin x] =cos^(−1) [cos (α−x)] =α−x if α−x ≤ π =2π−(α−x) if α−x > π To summarise: θ= { ((cos^(−1) ((3/5))−x ; x ≥ −π+cos^(−1) ((3/5)))),((2π+x−cos^(−1) ((3/5)) ; x < −π+cos^(−1) ((3/5)) )) :}](https://www.tinkutara.com/question/Q20512.png)
$${let}\:\mathrm{tan}\:\alpha=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\:\mathrm{sin}\:\alpha=\frac{\mathrm{4}}{\mathrm{5}}, \\ $$$${and}\:\mathrm{cos}\:\alpha=\frac{\mathrm{3}}{\mathrm{5}}\:\:\:;\:\:\alpha\:>\:{x} \\ $$$$\theta=\mathrm{cos}^{−\mathrm{1}} \left[\mathrm{cos}\:\alpha\mathrm{cos}\:{x}+\mathrm{sin}\:\alpha\mathrm{sin}\:{x}\right] \\ $$$$\:\:\:\:\:=\mathrm{cos}^{−\mathrm{1}} \left[\mathrm{cos}\:\left(\alpha−{x}\right)\right] \\ $$$$\:\:\:\:\:=\alpha−{x}\:\:\:\:{if}\:\:\alpha−{x}\:\leqslant\:\pi \\ $$$$\:\:\:\:\:=\mathrm{2}\pi−\left(\alpha−{x}\right)\:{if}\:\alpha−{x}\:>\:\pi \\ $$$${To}\:{summarise}: \\ $$$$\theta=\:\begin{cases}{\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)−{x}\:\:;\:\:{x}\:\geqslant\:−\pi+\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)}\\{\mathrm{2}\pi+{x}−\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)\:;\:{x}\:<\:−\pi+\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)\:}\end{cases} \\ $$$$ \\ $$
Commented by Tinkutara last updated on 27/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$