Cho sin$\alpha$ + cos$\alpha$ = 5/4 Tính A= sin$\alpha$ – cos$\alpha$ 19/07/2021 Bởi Quinn Cho sin$\alpha$ + cos$\alpha$ = 5/4 Tính A= sin$\alpha$ – cos$\alpha$
Đáp án: $A=±\dfrac{\sqrt{7}}{4}$ Giải thích các bước giải: $\sin\alpha+\cos\alpha=\dfrac{5}{4}$ $⇒(\sin\alpha+\cos\alpha)^2=\dfrac{25}{16}$ $⇒\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha=\dfrac{25}{16}$ $⇒1+2\sin\alpha\cos\alpha=\dfrac{25}{16}$ $⇒ 2\sin\alpha\cos\alpha=\dfrac{9}{16}$ $A=\sin\alpha-\cos\alpha$ $A^2=(\sin\alpha-\cos\alpha)^2$ $=\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha$ $=1-2\sin\alpha\cos\alpha$ $=1-\dfrac{9}{16}=\dfrac{7}{16}$ $⇒A=±\dfrac{\sqrt{7}}{4}$. Bình luận
Đây nhé !
Đáp án:
$A=±\dfrac{\sqrt{7}}{4}$
Giải thích các bước giải:
$\sin\alpha+\cos\alpha=\dfrac{5}{4}$
$⇒(\sin\alpha+\cos\alpha)^2=\dfrac{25}{16}$
$⇒\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha=\dfrac{25}{16}$
$⇒1+2\sin\alpha\cos\alpha=\dfrac{25}{16}$
$⇒ 2\sin\alpha\cos\alpha=\dfrac{9}{16}$
$A=\sin\alpha-\cos\alpha$
$A^2=(\sin\alpha-\cos\alpha)^2$
$=\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha$
$=1-2\sin\alpha\cos\alpha$
$=1-\dfrac{9}{16}=\dfrac{7}{16}$
$⇒A=±\dfrac{\sqrt{7}}{4}$.