nếu sin $\alpha$ =2/7 và $\alpha$ thuộc (0; $\pi$/2) thì sin(15 $\pi$ /2 + $\alpha$) bằng 12/10/2021 Bởi Kinsley nếu sin $\alpha$ =2/7 và $\alpha$ thuộc (0; $\pi$/2) thì sin(15 $\pi$ /2 + $\alpha$) bằng
Đáp án: $-\dfrac{3\sqrt{5}}{7}$ Giải thích các bước giải: $\sin^2\alpha+\cos^2\alpha=1\\\Rightarrow \cos^2\alpha=1-\sin^2\alpha\\=1-\left ( \dfrac{2}{7} \right )^2=\dfrac{45}{49}\\\Rightarrow \cos \alpha=\pm \dfrac{3\sqrt{5}}{7}$Do $0<\alpha<\dfrac{\pi}{2}\Rightarrow \cos \alpha>0$$\Rightarrow \cos \alpha=\dfrac{3\sqrt{5}}{7}\\+) \sin\left ( \dfrac{15\pi}{2}+\alpha \right )\\=\sin\dfrac{15\pi}{2}\cos \alpha+\cos \dfrac{15\pi}{2}\sin\alpha\\=(-1).\cos \alpha+0.\sin\alpha\\=-\dfrac{3\sqrt{5}}{7}$ Bình luận
$0<\alpha<\dfrac{\pi}{2}$ $\Rightarrow \cos\alpha>0$ $\Rightarrow \cos\alpha=\sqrt{1-\sin^2\alpha}=\dfrac{3\sqrt5}{7}$ $\sin(\dfrac{15\pi}{2}+\alpha)$ $=\sin(-\dfrac{\pi}{2}+\alpha)$ $=-\sin(\dfrac{\pi}{2}-\alpha)$ $=-\cos\alpha$ $=-\dfrac{3\sqrt5}{7}$ Bình luận
Đáp án:
$-\dfrac{3\sqrt{5}}{7}$
Giải thích các bước giải:
$\sin^2\alpha+\cos^2\alpha=1\\
\Rightarrow \cos^2\alpha=1-\sin^2\alpha\\
=1-\left ( \dfrac{2}{7} \right )^2=\dfrac{45}{49}\\
\Rightarrow \cos \alpha=\pm \dfrac{3\sqrt{5}}{7}$
Do $0<\alpha<\dfrac{\pi}{2}\Rightarrow \cos \alpha>0$
$\Rightarrow \cos \alpha=\dfrac{3\sqrt{5}}{7}\\
+) \sin\left ( \dfrac{15\pi}{2}+\alpha \right )\\
=\sin\dfrac{15\pi}{2}\cos \alpha+\cos \dfrac{15\pi}{2}\sin\alpha\\
=(-1).\cos \alpha+0.\sin\alpha\\
=-\dfrac{3\sqrt{5}}{7}$
$0<\alpha<\dfrac{\pi}{2}$
$\Rightarrow \cos\alpha>0$
$\Rightarrow \cos\alpha=\sqrt{1-\sin^2\alpha}=\dfrac{3\sqrt5}{7}$
$\sin(\dfrac{15\pi}{2}+\alpha)$
$=\sin(-\dfrac{\pi}{2}+\alpha)$
$=-\sin(\dfrac{\pi}{2}-\alpha)$
$=-\cos\alpha$
$=-\dfrac{3\sqrt5}{7}$