Спешна помощ за задача

[Гост]

Зад 1. ( cosd[tex]x^{2 }[/tex] + 2 sind[tex]x^{2 }[/tex].cosd - sind[tex]x^{2 }[/tex])[tex]x^{2 }[/tex]

Зад2 cos [tex]\frac{n}{7}[/tex] . cos [tex]\frac{4n}{7}[/tex] . cos [tex]\frac{5n}{7}[/tex].

Ако може да ми помогнете за задачите. Стигам до някъде и не мога да продължа

[Гост]

Извинете х не трябва да го има . Самите cos d и sin d са на квадрат

[Гост]

Първа задача трябва да е ( cos [tex]x^{d }[/tex] + 2 sin[tex]d^{2 }[/tex].cosd - sin[tex]x^{d }[/tex])[tex]^{2 }[/tex]

[ammornil]

$\cos{d^{2}}+2\cdot{\sin{d^{2}}}\cdot{\cos{d}}-\sin{d^{2}}= \cos{d^{2}}+\sin{d^{2}}\cdot{(2\cdot{\cos{d}}-1)}\\[6pt]$ Повече не виждам какво може да се опрости. Ако обаче задачата е да се опрости израза $$ \cos^{2}{d}+2\cdot{\sin^{2}{d}}\cdot{\cos{d}}-\sin^{2}{d} $$ тогава $\\[6pt] \cos^{2}{d}-\sin^{2}{d}+2\cdot{}\sin{d}\cdot{}\cos{d}\cdot{}\sin{d}= \cos{2d}+\sin{2d}\cdot{}\sin{d} \\[24pt] \begin{array}{l}\hline \hspace{120pt} \end{array} \\[12pt] $ Второто прилича на добре познат израз ако сте искали да напишете $\pi$ вместо $n$, тогава $$ A=\cos{\dfrac{\pi}{7}}\cdot{}\cos{\dfrac{4\pi}{7}}\cdot{}\cos{\dfrac{5\pi}{7}} $$ $\\[12pt] \\[12pt] \begin{cases} 0 < \dfrac{\pi}{7} < \dfrac{\pi}{2} \Rightarrow \text{I кв.} \Rightarrow \cos{\dfrac{\pi}{7}} > 0 \\[6pt] \dfrac{\pi}{2} < \dfrac{4\pi}{7} < \pi \Rightarrow \text{II кв.} \Rightarrow \cos{\dfrac{4\pi}{7}} < 0 \\[6pt] \dfrac{\pi}{2} < \dfrac{5\pi}{7} < \pi \Rightarrow \text{II кв.} \Rightarrow \cos{\dfrac{5\pi}{7}} < 0 \end{cases} \quad \Rightarrow A > 0 \\[12pt] \cos{\dfrac{5\pi}{7}}\cdot{}\cos{\dfrac{4\pi}{7}}= \dfrac{1}{2}\begin{bmatrix} \cos{ \begin{pmatrix} \dfrac{5\pi}{7} - \dfrac{4\pi}{7} \end{pmatrix} } +\cos{ \begin{pmatrix} \dfrac{5\pi}{7} + \dfrac{4\pi}{7} \end{pmatrix} } \end{bmatrix}= \dfrac{1}{2}\left( \cos{\dfrac{\pi}{7}} +\cos{\dfrac{9\pi}{7}} \right) \\[12pt] A=\cos{\dfrac{\pi}{7}}\cdot{}\dfrac{1}{2}\left( \cos{\dfrac{\pi}{7}} +\cos{\dfrac{9\pi}{7}} \right)$ $$ A=\dfrac{1}{2}\left(\cos^{2}{\dfrac{\pi}{7}}+\cos{\dfrac{\pi}{7}}\cdot{}\cos{\dfrac{9\pi}{7}} \right) $$ $ \\[12pt] \cos^{2}{\dfrac{\pi}{7}}= \cos{\dfrac{\pi}{7}} \cdot{} \cos{\dfrac{\pi}{7}}= \dfrac{1}{2}\begin{bmatrix} \cos{ \begin{pmatrix} \dfrac{\pi}{7} - \dfrac{\pi}{7} \end{pmatrix} } +\cos{ \begin{pmatrix} \dfrac{\pi}{7} + \dfrac{\pi}{7} \end{pmatrix} } \end{bmatrix}= \dfrac{1}{2}\left( \cos{0} +\cos{\dfrac{2\pi}{7}} \right) \\[12pt] \cos{\dfrac{9\pi}{7}}\cdot{}\cos{\dfrac{\pi}{7}}= \dfrac{1}{2}\begin{bmatrix} \cos{ \begin{pmatrix} \dfrac{9\pi}{7} - \dfrac{\pi}{7} \end{pmatrix} } +\cos{ \begin{pmatrix} \dfrac{9\pi}{7} + \dfrac{\pi}{7} \end{pmatrix} } \end{bmatrix}= \dfrac{1}{2}\left( \cos{\dfrac{8\pi}{7}} +\cos{\dfrac{10\pi}{7}} \right) \\[12pt] A= \dfrac{1}{2} \begin{bmatrix} \dfrac{1}{2}\left( 1 +\cos{\dfrac{2\pi}{7}} \right) +\dfrac{1}{2}\left( \cos{\dfrac{8\pi}{7}} +\cos{\dfrac{10\pi}{7}} \right) \end{bmatrix}$ $$ A= \dfrac{1}{4}\begin{pmatrix} 1 +\cos{\dfrac{2\pi}{7}} +\cos{\dfrac{8\pi}{7}} +\cos{\dfrac{10\pi}{7}} \end{pmatrix} $$ $ \cos{\dfrac{8\pi}{7}}= \cos{\begin{pmatrix} \pi+\dfrac{\pi}{7} \end{pmatrix}}=-\cos{\dfrac{\pi}{7}} \\[6pt] \cos{\dfrac{10\pi}{7}}= \cos{\begin{pmatrix} \pi+\dfrac{3\pi}{7} \end{pmatrix}}=-\cos{\dfrac{3\pi}{7}} \\[12pt] $ $$A=\dfrac{1}{4}\begin{pmatrix} 1 +\cos{\dfrac{2\pi}{7}} -\cos{\dfrac{\pi}{7}} -\cos{\dfrac{3\pi}{7}} \end{pmatrix}$$ $\\[24pt] \begin{array}{l}\hline \hspace{120pt} \end{array} \\[12pt] $ Вашето условие може да се преобразува по подобен начин с малко полагане: $\\[12pt] B=\cos{\dfrac{n}{7}}\cdot{}\cos{\dfrac{4n}{7}}\cdot{}\cos{\dfrac{5n}{7}} \\[6pt] \dfrac{n}{7}=k \Rightarrow B= \cos{k}\cdot{}\cos{4k}\cdot{\cos{5k}} \\[12pt] \cos{4k}\cdot{}\cos{k}=\dfrac{1}{2}\begin{bmatrix} \cos{\begin{pmatrix} 4k -k \end{pmatrix}} +\cos{\begin{pmatrix} 4k +k \end{pmatrix}} \end{bmatrix}=\dfrac{1}{2}\left(\cos{3k}+\cos{5k} \right) \\[6pt] \Rightarrow B=\dfrac{1}{2}\left(\cos{5k}\cdot{}\cos{3k}+ \cos{5k}\cdot{}\cos{5k} \right) \\[12pt] \cos{5k}\cdot{}\cos{3k}= \dfrac{1}{2}\begin{bmatrix} \cos{\begin{pmatrix} k5 -3k \end{pmatrix}} +\cos{\begin{pmatrix} 5k +3k \end{pmatrix}} \end{bmatrix}=\dfrac{1}{2}\left(\cos{2k} +\cos{8k} \right) \\[6pt] \cos{5k}\cdot{}\cos{5k}= \dfrac{1}{2}\begin{bmatrix} \cos{\begin{pmatrix} k5 -5k \end{pmatrix}} +\cos{\begin{pmatrix} 5k +5k \end{pmatrix}} \end{bmatrix}=\dfrac{1}{2}\left(\cos{0} +\cos{10k} \right) \\[12pt] \Rightarrow B=\dfrac{1}{4}\left(1+ \cos{2k} +\cos{8k} +\cos{10k} \right), \quad \because{} k=\dfrac{n}{7} \Rightarrow $ $$ B=\dfrac{1}{4}\left(1+ \cos{2\dfrac{n}{7}} +\cos{8\dfrac{n}{7}} +\cos{10\dfrac{n}{7}} \right) $$