Bài 110608

Question

Biến đổi các biểu thức sau về dạng tính được bằng logarit ( biến đổi tổng thành tích ).
a) $\sin a + \sin 2a + \sin 3a + \sin 4a $
b) $ \sin a + \sin 2a + \sin 3a + \cos a + \cos 2a + \cos 3a$
c) $\cos a + \cos 2a + \cos 3a$
d) $ \sin a + \sin b + \sin (a + b)$
e)$ 1 + \frac{ 1}{\cos a } + \tan a $
f) $ 1 - 4 \sin ^2a$
g) $\cot^2a - \cot^2b$

cobedangyeu_pro97 · Answer 1 · 7/4/2012 2:37:18 PM

a) $\sin a+\sin 2a+\sin3a+\sin4a=2\sin\frac{3a}2\cos\frac a2+2\sin\frac{7a}2\cos\frac a2
\\=2\cos\frac a2.2\sin\frac{5a}2\cos a= 4 \cos a\sin \frac{ 5a}{2 } \cos \frac{a }{ 2} $

b) $\sin a+\sin2a+\sin3a+\cos a+\cos2a+\cos3a
\\=\frac1{\sqrt{2}}(\sin(a+\frac\pi4)+\sin(2a+\frac\pi4)+\sin(3a+\frac\pi4))
\\=\frac1{\sqrt{2}}(2\sin(2a+\frac\pi4)\cos a+\sin(2a+\frac\pi4))
\\=\frac1{\sqrt{2}}.\sin(2a+\frac\pi4)(2\cos a+1)=\sqrt2.\sin(2a+\frac\pi4)(\cos a+\cos \frac\pi3)
\\=2 \sqrt{2} \cos \left( {\frac{a}{2} + \frac{\pi }{6}} \right) \cos\left( {\frac{a}{2} - \frac{\pi }{6}} \right) \sin \left( {2a + \frac{\pi }{4}} \right)$

c) $\cos a+\cos2a+\cos3a=2\cos2a.\cos a+\cos2a=\cos2a(2\cos a+1)
\\=2\cos2a(\cos a+\cos\frac\pi3)=4 \cos 2a . \cos \left( {\frac{a}{2} + \frac{\pi }{6}} \right) \cos \left( {\frac{a}{2} - \frac{\pi }{6}} \right)$

d) $\sin a+\sin b+\sin(a+b)=2\sin\frac{a+b}2.\cos\frac{a-b}2+2\sin\frac{a+b}2.\cos\frac{a+b}2
\\=2\sin\frac{a+b}2(\cos\frac{a-b}2+\cos\frac{a+b}2)=4 \sin \frac{ a + b}{2 } \cos \frac{ a}{2 } \cos \frac{ b}{2 }$

e) $1+\frac1{\cos a}+\tan a=\frac{\cos a+1+\sin a}{\cos a}=\frac{\sqrt2\sin(a+\frac\pi4)+1}{\sin (\frac\pi 2-a)}
\\=\sqrt2\frac{\sin(a+\frac\pi4)+\sin\frac\pi4}{2\sin (\frac\pi 4-\frac a2)\cos(\frac\pi 4-\frac a2)}=\sqrt2\frac{2\sin(\frac a2+\frac\pi 4)\cos\frac a2}{2\sin (\frac\pi 4-\frac a2)\sin(\frac\pi 4+\frac a2)}= \frac{ \sqrt{2} \cos \frac{ a}{2 } }{\sin \left( {\frac{\pi }{4} - \frac{a}{2}} \right)} $
f) $1-4\sin^2a=1-4\frac{1-\cos 2a}{2}=2\cos 2a-1=2(\cos 2a-\cos\frac\pi3)
\\= 4 \sin \left( {a + \frac{\pi }{6}} \right) . \sin \left( {\frac{\pi }{6} - a} \right) $

g) $\cot^2a-\cot^2b=\frac1{\sin^2a}-1-(\frac1{\sin^2b}-1)=\frac{\sin^2b-\sin^2a}{\sin^2a.\sin^2b}
\\=\frac12\frac{(1-\cos2b)-(1-\cos 2a)}{\sin^2a.\sin^2b}=\frac12\frac{\cos 2a-\cos 2b}{\sin^2a.\sin^2b}
\\=\frac{ \sin ( a + b ) \sin ( b - a )}{\sin ^2a . \sin ^2b } $

Bài 110608

1 Đáp án

Lý thuyết liên quan

Bài 110608

1 Đáp án

Liên quan

Lý thuyết liên quan