Монгол Бодлогын Сан
Эх хэлээрээ суралцаж, эх хэлээрээ мэдлэгээ түгээе.
Бодлого №10578
$\sin\theta=s$, $\cos\theta=c$, $\tg\theta=t$ гэвэл
- $\cos4\theta$, $\cos5\theta$-ийг $c$-ээр илэрхийл.
- $\sin 5\theta$-ийг $s$-ээр илэрхийл.
- $\tg 3\theta$, $\tg 4\theta$-ийг $t$-ээр илэрхийл.
Бодлогын төрөл: Уламжлалт
Бодлогыг оруулсан: Балхүүгийн Батбаясгалан
Бодолт
Заавар: Муаврын томьёо ашигла.
Бодолт:
- Муаврын томьёогоор \begin{align*} \cos4\theta+i\sin4\theta&=(\cos\theta+i\sin\theta)^4\\ &=\cos^4\theta+C_4^1\cos^3\theta(i\sin\theta)+C_4^2\cos^2\theta(i\sin\theta)^2+C_4^3\cos\theta(i\sin\theta)^3+(i\sin\theta)^4\\ &=\cos^4\theta+iC_4^1\cos^3\theta\sin\theta-C_4^2\cos^2\theta\sin^2\theta-iC_4^3\cos\theta\sin^3\theta+\sin^4\theta\\ &=(\cos^4\theta-C_4^2\cos^2\theta\sin^2\theta+\sin^4\theta)+(C_4^1\cos^3\theta\sin\theta-C_4^3\cos\theta\sin^3\theta)i \end{align*} тул комплекс тоонуудын тэнцэх нөхцөлөөр \begin{align*} \cos4\theta&=\cos^4\theta-C_4^2\cos^2\theta\sin^2\theta+\sin^4\theta\\ &=\cos^4\theta-C_4^2\cos^2\theta(1-\cos^2)+(1-\cos^2\theta)^2\\ &=8\cos^4\theta-8\cos^2\theta+1=8c^2-8c+1 \end{align*} болно. Яг адилаар \begin{align*} \cos5\theta+i\sin5\theta&=(\cos\theta+i\sin\theta)^5\\ &=\cos^5\theta+C_5^1\cos^4\theta(i\sin\theta)+C_5^2\cos^3\theta(i\sin\theta)^2+\dots+(i\sin\theta)^5\\ &=\cos^5\theta+iC_5^1\cos^4\theta\sin\theta-C_5^2\cos^3\theta\sin^2\theta-iC_5^3\cos^2\theta\sin^3\theta+C_5^4\cos\theta\sin^4\theta+i\sin^5\theta\\ &=(\cos^5\theta-C_5^2\cos^3\theta\sin^2\theta+C_5^4\cos\theta\sin^4\theta)+(C_5^1\cos^4\theta\sin\theta-C_5^3\cos^2\theta\sin^3\theta+\sin^5\theta)i \end{align*} ба \begin{align*} \cos5\theta&=\cos^5\theta-C_5^2\cos^3\theta\sin^2\theta+C_5^4\cos\theta\sin^4\theta\\ &=\cos^5\theta-C_5^2\cos^3\theta(1-\cos^2)+C_5^4\cos\theta(1-\cos^2\theta)^2\\ &=16\cos^5\theta-20\cos^3\theta+5\cos\theta=16c^5-20c^3+5c \end{align*}
- Өмнөх хэсгээс \begin{align*} \sin5\theta&=C_5^1\cos^4\theta\sin\theta-C_5^3\cos^2\theta\sin^3\theta+\sin^5\theta\\ &=5(1–s^2)^2s-10(1-s^2)s^3+s^5\\ &=16s^5-20s^3+5s \end{align*}
- Муаврын томьёогоор \begin{align*} \cos3\theta+i\sin3\theta&=(\cos\theta+i\sin\theta)^3\\ &=\cos^3\theta+C_3^1\cos^2\theta(i\sin\theta)+C_3^2\cos\theta(i\sin\theta)^2+(i\sin\theta)^3\\ &=\cos^3\theta+iC_3^1\cos^2\theta\sin\theta-C_3^2\cos\theta\sin^2\theta-i\sin^3\theta\\ &=(\cos^3\theta-C_3^2\cos\theta\sin^2\theta)+(C_3^1\cos^2\theta\sin\theta-\sin^3\theta)i \end{align*} тул \[\tg3\theta=\dfrac{\sin3\theta}{\cos3\theta}=\dfrac{3\cos^2\theta\sin\theta-\sin^3\theta}{\cos^3\theta-3\cos\theta\sin^2\theta}=\dfrac{3t-t^3}{1-3t^2}\] байна. Төстэйгөөр \[\tg4\theta=\dfrac{\sin4\theta}{\cos4\theta}=\dfrac{4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta}{\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta}=\dfrac{4t-4t^3}{1-6t^2+t^4}\]