Category: Arithmetic

Question Number 211873 by Spillover last updated on 22/Sep/24 Answered by IbtisamAdnan last updated on 23/Sep/24 $${\mathbf{lim}}_{\mathbf{x}\to 4}\frac{{\left(\mathbf{cos}\mathit{\alpha }\right)}^{\mathrm{x}}-{\left(\sin \alpha \right)}^{\mathrm{x}}-\cos 2\alpha }{\mathrm{x}-4}$$$$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{4}} \:\frac{\left(\boldsymbol{\mathrm{cos}\alpha}\right)^{\mathrm{x}} .\:\mathrm{ln}\:\mathrm{cos}\alpha\:\:−\:\left(\mathrm{sin}\:\alpha\right)^{\mathrm{x}} .\mathrm{ln}\:\mathrm{sin}\alpha}{\mathrm{1}}\left[\mathrm{L}\:\mathrm{hospital}\:\mathrm{rule}\right]…

Question Number 211636 by BaliramKumar last updated on 15/Sep/24 Answered by Rasheed.Sindhi last updated on 15/Sep/24 $$▸A=x^{8k+3}+x^{8k+6}+x^{8k+9}+x^{8k+12}$$$$\:\:\:\:\:\:\:\:={x}^{\mathrm{8}{k}+\mathrm{3}} \left(\mathrm{1}+{x}^{\mathrm{3}} +{x}^{\mathrm{6}} +{x}^{\mathrm{9}}…

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Question Number 211019 by peter frank last updated on 26/Aug/24 Answered by som(math1967) last updated on 26/Aug/24 $$letx=sin\alpha y=sin\beta$$$$cos\alpha +cos\beta =a(sin\alpha -sin\beta )$$$$\Rightarrow \frac{2cos\frac{\alpha +\beta }{2}cos\frac{\alpha -\beta }{2}}{2sin\frac{\alpha -\beta }{2}cos\frac{\alpha +\beta }{2}}=a$$$$\Rightarrow{cot}\frac{\alpha−\beta}{\mathrm{2}}={a} \…