giup e vs please :<<
Bài tập 3: Rút gọn các biểu thức sau:
\[ A = \sin(x + 14^\circ)\sin(x + 74^\circ) + \sin^2(x - 76^\circ)\sin^4(x - 16^\circ) \]
Bài tập 4: Cho \(\cos 2x = -\frac{4}{5}\), với \(\frac{\pi}{4} < x < \frac{\pi}{2}\). Tính \(\sin x, \cos x, \sin \left( x + \frac{\pi}{3} \right), \cos \left( 2x - \frac{\pi}{4} \right)\).
Bài tập 5: Đơn giản biểu thức sau:
\[ A = \frac{\cos a + 2 \cos 2a + \cos 3a}{\sin a + \sin 2a + \sin 3a} \]
Câu 5:
\(cos\alpha+cos3\alpha=2\cdot cos\left(\frac{\alpha+3\alpha}{2}\right)\cdot cos\left(\frac{\alpha-3\alpha}{2}\right)\)
\(=2\cdot cos\left(\frac{4\alpha}{2}\right)\cdot cos\left(-\frac{2\alpha}{2}\right)=2\cdot cos2\alpha\cdot cos\alpha\)
\(\sin\alpha+\sin3\alpha=2\cdot\sin\left(\frac{\alpha+3\alpha}{2}\right)\cdot cos\left(\frac{3\alpha-\alpha}{2}\right)\)
\(=2\cdot\sin\left(\frac{4\alpha}{2}\right)\cdot cos\left(\frac{2\alpha}{2}\right)=2\cdot\sin2\alpha\cdot cos\alpha\)
\(A=\frac{cos\alpha+cos2\alpha+cos3\alpha}{\sin\alpha+\sin2\alpha+\sin3\alpha}\)
\(=\frac{2\cdot cos2\alpha\cdot cos\alpha+cos2\alpha}{2\cdot\sin2\alpha\cdot cos\alpha+\sin2\alpha}=\frac{cos2\alpha\left(2\cdot cos\alpha+1\right)}{\sin2\alpha\left(2\cdot cos\alpha+1\right)}\)
\(=\frac{cos2\alpha}{\sin2\alpha}=\cot2\alpha\)
Câu 4:
\(cos2x=-\frac45\)
=>\(2\cdot cos^2x-1=-\frac45\)
=>\(2\cdot cos^2x=-\frac45+1=\frac15\)
=>\(cos^2x=\frac{1}{10}\)
\(\frac{\pi}{4}
=>\(cosx>0;\sin x>0;\tan x>0;\cot x>0\)
=>\(cosx=\sqrt{\frac{1}{10}}=\frac{\sqrt{10}}{10}\)
Ta có: \(\sin^2x+cos^2x=1\)
=>\(\sin^2x=1-\frac{1}{10}=\frac{9}{10}\)
=>\(\sin x=\sqrt{\frac{9}{10}}=\frac{3}{\sqrt{10}}\)
Câu 3:
\(\sin\left(x+14^0\right)\cdot\sin\left(x+74^0\right)\)
\(=2\cdot\sin\left(\frac{x+74^0+x+14^0}{2}\right)\cdot cos\left(\frac{x+74^0-x-14^0}{2}\right)\)
\(=2\cdot\sin\left(x+44^0\right)\cdot cos30^0=2\cdot\sin60^0\cdot\sin\left(x+44^0\right)\)
\(\sin\left(x-76^0\right)\cdot\sin\left(x-16^0\right)\)
\(=2\cdot\sin\left(\frac{x-76^0+x-16^0}{2}\right)\cdot cos\left(\frac{x-76^0-x+16^0}{2}\right)\)
\(=2\cdot\sin\left(x-46^0\right)\cdot cos\left(-30^0\right)=2\cdot\sin\left(x-46^0\right)\cdot\sin60^0\)
Do đó: \(A=2\cdot\sin60^0\cdot\left\lbrack\sin\left(x+44^0\right)+\sin\left(x-46^0\right)\right\rbrack\)
\(=2\cdot\frac{\sqrt3}{2}\cdot\left\lbrack\sin\left(x+44^0\right)+\sin\left(x+44^0-90^0\right)\right\rbrack\)
\(=\sqrt3\cdot\left\lbrack\sin\left(x+44^0\right)-cos\left(x+44^0\right)\right\rbrack\)
\(=\sqrt3\cdot\sqrt2\cdot\sin\left(x+44^0-45^0\right)=\sqrt6\cdot\sin\left(x-1^0\right)\)
