Bài 1.11 trang 21

Đề bài

Chứng minh đẳng thức sau:

$\sin \left(a+b\right)\sin \left(a-b\right)$

$={\sin }^{2}a-{\sin }^{2}b={\cos }^{2}b-{\cos }^{2}a$.

Lời giải chi tiết

+) $\sin \left(a+b\right)\sin \left(a-b\right)$

$=\left(\sin a\cos b+\cos a\sin b\right).\left(\sin a\cos b-\cos a\sin b\right)$

$={\left(\sin a\cos b\right)}^2-{\left(\cos a\sin b\right)}^2$

$={\sin }^{2}a\left(1-{\sin }^{2}b\right)-\left(1-{\sin }^{2}a\right){\sin }^{2}b$

$={\sin }^{2}a-{\sin }^{2}a{\sin }^{2}b-{\sin }^{2}b+{\sin }^{2}a{\sin }^{2}b$

$={\sin }^{2}a-{\sin }^{2}b$ (đpcm).

+)  $\sin \left(a+b\right)\sin \left(a-b\right)$

$=\left(\sin a\cos b+\cos a\sin b\right).\left(\sin a\cos b-\cos a\sin b\right)$

$={\left(\sin a\cos b\right)}^2-{\left(\cos a\sin b\right)}^2$

$={\cos }^{2}b\left(1-{\cos }^{2}a\right)-{\cos }^{2}a\left(1-{\cos }^{2}b\right)$

$={\cos }^{2}b-{\cos }^{2}a{\cos }^{2}b-{\cos }^{2}a+{\cos }^{2}a{\cos }^{2}b$

$={\cos }^{2}b-{\cos }^{2}a$ (đpcm).