Đề bài

Chứng minh rằng

a) \(\cos a - \sin a = \sqrt 2 \cos \left( {a + \frac{\pi }{4}} \right)\);           

b) \(\sin a + \sqrt 3 \cos a = 2\sin \left( {a + \frac{\pi }{3}} \right)\).

Phương pháp giải

Áp dụng công thức cộng:

\(\sin (a + b) = \sin a\cos b + \cos a\sin b\).

\(\cos (a + b) = \cos a\cos b - \sin a\sin b\).

Lời giải của GV Loigiaihay.com

a) Ta có:

\(\begin{array}{l}\sqrt 2 \cos \left( {a + \frac{\pi }{4}} \right) = \sqrt 2 \left( {\cos a\cos \frac{\pi }{4} - \sin a\sin \frac{\pi }{4}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 2 \left( {\cos a.\frac{{\sqrt 2 }}{2} - \sin a.\frac{{\sqrt 2 }}{2}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 2 .\cos a.\frac{{\sqrt 2 }}{2} - \sqrt 2 .\sin a.\frac{{\sqrt 2 }}{2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \cos a - \sin a.\end{array}\)

b) Ta có:

\(\begin{array}{l}{\rm{VT}} = 2\sin \left( {a + \frac{\pi }{3}} \right)\,\\\,\,\,\,\,\,\,\,\, = 2\left( {\sin a\cos \frac{\pi }{3} + \cos a\sin \frac{\pi }{3}} \right)\\\,\,\,\,\,\,\,\,\, = 2\left( {\sin a.\frac{1}{2} + \cos a.\frac{{\sqrt 3 }}{2}} \right)\\\,\,\,\,\,\,\,\,\, = 2\sin a.\frac{1}{2} + 2\cos a.\frac{{\sqrt 3 }}{2}\\\,\,\,\,\,\,\,\,\, = \sin a + \sqrt 3 \cos a.\end{array}\)