Айниятро исбот намоед: \(\cos{\alpha}+\cos{2\alpha}+\cos{6\alpha}+\cos{7\alpha}=4\cos{\frac{\alpha}{2}}\cos{\frac{5\alpha}{2}}\cos{4\alpha}\)
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- Автор: Раҳимҷон Ҳакимов
- Категория: Тригонометрия
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Айниятро исбот намоед:
\(\cos{\alpha}+\cos{2\alpha}+\cos{6\alpha}+\cos{7\alpha}=4\cos{\frac{\alpha}{2}}\cos{\frac{5\alpha}{2}}\cos{4\alpha}\)
\(\cos{\alpha}+\cos{7\alpha}=2\cos{\frac{\alpha+7\alpha}{2}}\cos{\frac{\alpha-7\alpha}{2}}=2\cos{4\alpha}\cos{3\alpha}\)
\(\cos{2\alpha}+\cos{6\alpha}=2\cos{\frac{2\alpha+6\alpha}{2}}\cos{\frac{2\alpha-6\alpha}{2}}=2\cos{4\alpha}\cos{2\alpha}\)
\(\cos{\alpha}+\cos{2\alpha}+\cos{6\alpha}+\cos{7\alpha}=2\cos{4\alpha}\cos{2\alpha}+2\cos{4\alpha}\cos{3\alpha}=\)
\(=2\cos{4\alpha}(\cos{2\alpha}+\cos{3\alpha})\)
\(\cos{2\alpha}+\cos{3\alpha}=2\cos{\frac{2\alpha+3\alpha}{2}}\cos{\frac{2\alpha-3\alpha}{2}}=2\cos{\frac{5\alpha}{2}}\cos{\frac{\alpha}{2}}\)
\(\cos{\alpha}+\cos{2\alpha}+\cos{6\alpha}+\cos{7\alpha}=2\cos{4\alpha}\cdot2\cos{\frac{5\alpha}{2}}\cos{\frac{\alpha}{2}}=\)
\(=4\cos{\frac{\alpha}{2}}\cos{\frac{5\alpha}{2}}\cos{4\alpha}\)
\(\cos{\alpha}+\cos{2\alpha}+\cos{6\alpha}+\cos{7\alpha}=4\cos{\frac{\alpha}{2}}\cos{\frac{5\alpha}{2}}\cos{4\alpha}\)
Айният исбот шуд.
- Таҳқиқи функсияи \(y = \frac{x^3-1}{4x^2}\)
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- Таҳқиқи функсияи \(y = \frac{e^x}{x}\)
- Таҳқиқи функсияи \(y = -\frac{1}{4}(x^3-3x^2+4)\)
- Соҳаи муайянии функсияи \(y = \frac{x^2}{1+x}\)
- Соҳаи муайянии функсияи \(y = \sqrt{\cos x^2}\)
- Ҳисоб карда шавад: \(\lim\limits_{n \rightarrow \infty}\left(\frac{1}{n^2} + \frac{2}{n^2} + ... + \frac{n-1}{n^2} \right)\)
- Соҳаи муайянии функсияи \(y = \sqrt{\sin\left(\sqrt{x}\right)}\)
- Ҳисоб карда шавад: \(\lim\limits_{n \rightarrow \infty}\frac{1 + a + a^2 + ... + a^n}{1 + b + b^2 + ... + b^n}\)
- Соҳаи муайянии функсияи \(y = \log(x+2) + \log(x-2)\)