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Author: Tinku Tara

Function-is-a-b-a-b-AB-a-b-4-b-4-4-b-b-a-1-a-1-2-b-2-a-3-Function-is-a-b-sin-a-b-a-b-sin-a-b-a-b-1-b-5-a-4-9-a-b-9-b-a-Funcion-is-sin-a-sin-b-sin-1-a-sin-1-b-a-b-sin-a-sin-b-

Question Number 4790 by Dnilka228 last updated on 10/Mar/16 $${Function}\:\Gamma\:{is}\:{a}+{b} \\ $$$${a}+{b}={AB} \\ $$$${a}\neq{b}−\mathrm{4} \\ $$$${b}−\mathrm{4}=\mathrm{4}+{b} \\ $$$${b}={a}−\mathrm{1} \\ $$$${a}−\mathrm{1}=\mathrm{2} \\ $$$${b}=\mathrm{2} \\ $$$${a}=\mathrm{3} \\…

cos-cos-cos-cos-1-cos-1-sin-a-b-sin-a-sin-b-sin-1-a-sin-1-b-a-b-tan-a-a-b-k-tan-a-b-k-tan-1-a-b-k-

Question Number 4783 by Dnilka228 last updated on 10/Mar/16 $$\mathrm{cos}\:\alpha+\beta\:\approx\left(\frac{\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta}{\mathrm{cos}^{−\mathrm{1}} \alpha+\mathrm{cos}^{−\mathrm{1}} \beta}\right)^{\alpha+\beta} \\ $$$$\mathrm{sin}\:{a}+{b}\approx\left(\frac{\mathrm{sin}\:{a}+\mathrm{sin}\:{b}}{\mathrm{sin}^{−\mathrm{1}} {a}+\mathrm{sin}^{−\mathrm{1}} {b}}\right)^{{a}+{b}} \\ $$$$\mathrm{tan}\:\left({a}+\frac{{a}}{{b}}\right)^{{k}} \approx\left(\frac{\mathrm{tan}\:\left({a}+{b}\right)×{k}}{\mathrm{tan}^{−\mathrm{1}} \left({a}+{b}\right)×{k}}\right) \\ $$ Commented by Dnilka228…

What-are-the-possible-value-of-cos-sin-if-sin-cos-1-2-

Question Number 135855 by liberty last updated on 16/Mar/21 $${What}\:{are}\:{the}\:{possible}\:{value}\:{of} \\ $$$$\mathrm{cos}\:\alpha×\mathrm{sin}\:\beta\:\:{if}\:\mathrm{sin}\:\alpha×\mathrm{cos}\:\beta=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by EDWIN88 last updated on 16/Mar/21 $$\mathrm{We}\:\mathrm{have}\:−\mathrm{1}\leqslant\mathrm{sin}\:\left(\alpha+\beta\right)\leqslant\mathrm{1}\:\mathrm{and}\:−\mathrm{1}\leqslant\mathrm{sin}\:\left(\alpha−\beta\right)\leqslant\mathrm{1} \\ $$$$\mathrm{now}\:\mathrm{from}\:−\mathrm{1}\leqslant\mathrm{sin}\:\left(\alpha+\beta\right)\leqslant\mathrm{1}\:\mathrm{we}\:\mathrm{get}\: \\…

If-1-a-2-1-b-2-1-c-2-1-ab-1-bc-1-ca-then-prove-that-a-b-c-

Question Number 70312 by Shamim last updated on 03/Oct/19 $$\mathrm{If},\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{ab}}+\frac{\mathrm{1}}{\mathrm{bc}}+\frac{\mathrm{1}}{\mathrm{ca}}\:\mathrm{then}\:\mathrm{prove}\: \\ $$$$\mathrm{that},\:\mathrm{a}=\mathrm{b}=\mathrm{c}. \\ $$ Commented by MJS last updated on 03/Oct/19 $$\mathrm{this}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:{a},\:{b},\:{c}\:\in\mathbb{R}…