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Question Number 168663 by DomaPeti last updated on 15/Apr/22

how to solve?  ((cos(30°)∙sin(45°)−cos(30°+dx)∙sin(45°+dx))/dx)=?  dx→0

$${how}\:{to}\:{solve}? \\ $$$$\frac{{cos}\left(\mathrm{30}°\right)\centerdot{sin}\left(\mathrm{45}°\right)−{cos}\left(\mathrm{30}°+{dx}\right)\centerdot{sin}\left(\mathrm{45}°+{dx}\right)}{{dx}}=? \\ $$$${dx}\rightarrow\mathrm{0} \\ $$

Commented by cortano1 last updated on 15/Apr/22

lim_(dx→0) (((1/4)(√6)−cos (30°+dx).sin (45°+dx))/dx)  = lim_(dx→0) ((sin (30°+dx)sin (45°+dx)−cos (30°+dx)cos (45°+dx))/1)  = sin 30° sin 45°−cos 30° cos 45°  =(1/4)(√2)−(1/4)(√6)  =(1/4)(√2)(1−(√3))

$$\underset{{dx}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{6}}−\mathrm{cos}\:\left(\mathrm{30}°+{dx}\right).\mathrm{sin}\:\left(\mathrm{45}°+{dx}\right)}{{dx}} \\ $$$$=\:\underset{{dx}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\left(\mathrm{30}°+{dx}\right)\mathrm{sin}\:\left(\mathrm{45}°+{dx}\right)−\mathrm{cos}\:\left(\mathrm{30}°+{dx}\right)\mathrm{cos}\:\left(\mathrm{45}°+{dx}\right)}{\mathrm{1}} \\ $$$$=\:\mathrm{sin}\:\mathrm{30}°\:\mathrm{sin}\:\mathrm{45}°−\mathrm{cos}\:\mathrm{30}°\:\mathrm{cos}\:\mathrm{45}° \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{6}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{2}}\left(\mathrm{1}−\sqrt{\mathrm{3}}\right) \\ $$

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