Giải bài 1, 2, 3, 4 trang 154 sách giáo khoa đại số 10 - Bài trang sgk đại số

Bài 1 trang 153 sgk đại số 10

Tính

a) \(\cos {225^0},\sin {240^0},cot( - {15^0}),tan{75^0}\);

b) \(\sin \frac{7\pi}{12}\), \(\cos \left ( -\frac{\pi}{12} \right )\), \(\tan\left ( \frac{13\pi}{12} \right )\)

Giải

a)

+) \(\cos{225^0} = \cos({180^0} +{45^0})= - \cos{45^{0}}= -\frac{\sqrt{2}}{2}\)

+) \(\sin{240^0} = \sin({180^0} +{60^0}) = - \sin{60^0}= -\frac{\sqrt{3}}{2}\)

+) \(\cot( - {15^0})= - \cot{15^0} = - \tan{75^0} =- \tan({30^0} +{45^0})\)

\( =\frac{-\tan30^{0}-\tan45^{0}}{1-\tan30^{0}\tan45^{0}}=\frac{-\frac{1}{\sqrt{3}}-1}{1-\frac{1}{\sqrt{3}}}=-\frac{\sqrt{3}+1}{\sqrt{3}-1}=-\frac{(\sqrt{3}+1)^{2}}{2} = -2 - \sqrt 3\)

+) \(\tan 75^0= \cot15^0=2 + \sqrt3\)

b)

+) \(\sin \frac{7\pi}{12} = \sin \left ( \frac{\pi}{3}+\frac{\pi}{4} \right ) =\sin\frac{\pi }{3}\cos\frac{\pi}{4}+ \cos \frac{\pi }{3}\sin\frac{\pi}{4}\)

\( =\frac{\sqrt{2}}{2}\left ( \frac{\sqrt{3}}{2} +\frac{1}{2}\right )=\frac{\sqrt{6}+\sqrt{2}}{4}\)

+) \(\cos \left ( -\frac{\pi }{12} \right ) = \cos \left ( \frac{\pi }{4} -\frac{\pi }{3}\right ) = \cos \frac{\pi }{4}\cos\frac{\pi }{3} + sin \frac{\pi }{3}sin \frac{\pi }{4}\)

\( =\frac{\sqrt{2}}{2}\left ( \frac{\sqrt{3}}{2} +\frac{1}{2}\right )=0,9659\)

+) \(\tan \left ( \frac{13\pi }{12} \right ) = \tan(π + \frac{\pi }{12}) = \tan \frac{\pi }{12} = \tan \left ( \frac{\pi }{3}-\frac{\pi}{4} \right )\)

\(= \frac{\tan\frac{\pi }{3}-\tan\frac{\pi }{4}}{1+\tan\frac{\pi }{3}\tan\frac{\pi }{4}}=\frac{\sqrt{3}-1}{1+\sqrt{3}}= 2 - \sqrt3\)

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Bài 2 trang 154 sgk đại số 10

Tính

a) \(\cos(α + \frac{\pi}{3}\)), biết \(\sinα = \frac{1}{\sqrt{3}}\)và \(0 < α < \frac{\pi }{2}\).

b) \(\tan(α - \frac{\pi }{4}\)), biết \(\cosα = -\frac{1}{3}\)và\( \frac{\pi }{2} <α < π\)

c) \(\cos(a + b), \sin(a - b)\) biết \(\sin a = \frac{4}{5}\), \(0^0< a < 90^0\) và \(\sin b = \frac{2}{3}\), \(90^0< b < 180^0\)

Giải

a) Do \(0 <α < \frac{\pi}{2}\)nên \(\sinα > 0, \cosα > 0\)

\(\cosα = \sqrt{1-\sin^{2}\alpha }=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}\)

\(cos(α + \frac{\pi}{3}) = \cosα\cos \frac{\pi }{3} - \sinα\sin \frac{\pi}{3}\)

\( = \frac{\sqrt{6}}{3}.\frac{1}{2}-\frac{1}{\sqrt{3}}.\frac{\sqrt{3}}{2}=\frac{\sqrt{6}-3}{6}\)

b) Do\( \frac{\pi}{2}<α <π\) nên \(\sinα > 0, \cosα < 0, \tanα < 0, \cotα < 0\)

\(\tanα = -\sqrt{\frac{1}{cos^{2}\alpha }-1}=-\sqrt{3^{3}-1} = -2\sqrt2\)

\(tan(α - \frac{\pi}{4}) = \frac{\tan\alpha -\tan\frac{\pi}{4}}{1+\tan\alpha tan\frac{\pi}{4}}=\frac{-1-2\sqrt{2}}{1-2\sqrt{2}}=\frac{2\sqrt{2}+1}{2\sqrt4{2}-1}\)

c) \(0^0< a < 90^0\Rightarrow \sin a > 0, \cos a > 0\)

\(90^0< b < 180^0\Rightarrow\sin b > 0, \cos b < 0\)

\(\cos a = \sqrt{1-sin^{2}a}=\sqrt{1-\left ( \frac{4}{5} \right )^{2}}=\frac{3}{5}\)

\(\cos b = -\sqrt{1-sin^{2}a}=-\sqrt{1-\left ( \frac{2}{3} \right )^{2}}=-\frac{\sqrt{5}}{3}\)

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

\( =\frac{3}{5}\left ( -\frac{\sqrt{5}}{3} \right )-\frac{4}{5}.\frac{2}{3}=-\frac{3\sqrt{5}+8}{15}\)

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Bài 3 trang 154 sgk đại số 10

Rút gọn các biểu thức

a) \(\sin(a + b) + \sin(\frac{\pi}{2}- a)\sin(-b)\).

b) \(cos(\frac{\pi }{4} + a)\cos( \frac{\pi}{4} - a) + \frac{1 }{2} \sin^2a\)

c) \(\cos( \frac{\pi}{2} - a)\sin( \frac{\pi}{2} - b) - \sin(a - b)\)

Giải

a) \(\sin(a + b) + \sin( \frac{\pi }{2} - a)\sin(-b) = \sin a\cos b + \cos a\sin b - \cos a\sin b = \sin a\cos b\)

b) \(\cos( \frac{\pi }{4} + a)\cos(\frac{\pi }{4}- a) + \frac{1 }{2}\sin^2a\)

\( =\frac{1 }{2}\cos\left [ \frac{\pi }{4}+a+\frac{\pi}{4} -a\right ]+\frac{1}{2}\cos\left [ \left ( \frac{\pi }{4} +a\right ) -\left ( \frac{\pi}{4}-a \right )\right ]+\frac{1}{2}\left ( \frac{1-\cos 2a}{2} \right )\)

\( =\frac{1}{2}\cos 2a + \frac{1}{4}(1 - \cos 2a) = \frac{1+\cos 2a}{4 }= \frac{1 }{2}\cos^2 a\)

c) \(\cos( \frac{\pi}{2} - a)\sin( \frac{\pi}{2} - b) - \sin(a - b) = \sin a\cos b - \sin a\cos b + \sin b\cos a\)

\(= \sin b\cos a\)

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Bài 4 trang 154 sgk đại số 10

Chứng minh các đẳng thức

a)\( \frac{cos(a-b)}{cos(a+b)}=\frac{cotacotb+1}{cotacotb-1}\)

b) \(\sin(a + b)\sin(a - b) = \sin^2a \sin^2b = \cos^2b \cos^2a\)

c) \(\cos(a + b)\cos(a - b) = \cos^2a - \sin^2b = \cos^2b \sin^2a\)

Giải

a) \(VT = {{\cos a\cos b+\sin a\sin b}\over{\cos a\cos b-\sin a\sin b}}=\frac{\frac{\cos a\cos b}{\sin a\sin b}+1}{\frac{\cos a\cos b}{\sin a\sin b}-1}=\frac{\cot a\cot b+1}{\cot a\cot b-1}\)

b) \(VT = [\sin a\cos b + \cos a\sin b][\sin a\cos b - \cos a\sin a]\)

\(=(\sin a\cos b)^2 (\cos a\sin b)^2= \sin^2a(1 \sin^2b) (1 \sin^2 a)\sin^2b\)

\(= \sin^2a \sin^2b = \cos^2b( 1 \cos^2a) \cos^2 a(1 \cos^2 b) = \cos^2 b \cos^2 a\)

c) \(VT = (\cos a\cos b - \sin a\sin b)(\cos a\cos b + \sin a\sin b)\)

\(= (\cos a\cos b)^2 (\sin a\sin b)^2\)

\(= \cos^2a(1 \sin^2b) (1 \cos^2a)\sin^2b = \cos^2a \sin^2b\)

\(= \cos^2b(1 \sin^2a) (1 \cos^2b)\sin^2a = \cos^2b \sin^2a\)