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]
+] \[\sin \frac{7\pi}{12} = \sin \left [ \frac{\pi}{3}+\frac{\pi}{4} \right ] =\sin\frac{\pi }{3}\cos\frac{\pi}{4}+ \cos \frac{\pi }{3}\sin\frac{\pi}{4}\]
\[ =\frac{\sqrt{2}}{2}\left [ \frac{\sqrt{3}}{2} +\frac{1}{2}\right ]=\frac{\sqrt{6}+\sqrt{2}}{4}\]
+] \[\cos \left [ -\frac{\pi }{12} \right ] = \cos \left [ \frac{\pi }{4} -\frac{\pi }{3}\right ] = \cos \frac{\pi }{4}\cos\frac{\pi }{3} + sin \frac{\pi }{3}sin \frac{\pi }{4}\]
\[ =\frac{\sqrt{2}}{2}\left [ \frac{\sqrt{3}}{2} +\frac{1}{2}\right ]=0,9659\]
+] \[\tan \left [ \frac{13\pi }{12} \right ] = \tan[π + \frac{\pi }{12}] = \tan \frac{\pi }{12} = \tan \left [ \frac{\pi }{3}-\frac{\pi}{4} \right ]\]
\[= \frac{\tan\frac{\pi }{3}-\tan\frac{\pi }{4}}{1+\tan\frac{\pi }{3}\tan\frac{\pi }{4}}=\frac{\sqrt{3}-1}{1+\sqrt{3}}= 2 - \sqrt3\]
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\[ \frac{\pi }{2} 0\]
\[\cosα = \sqrt{1-\sin^{2}\alpha }=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}\]
\[cos[α + \frac{\pi}{3}] = \cosα\cos \frac{\pi }{3} - \sinα\sin \frac{\pi}{3}\]
\[ = \frac{\sqrt{6}}{3}.\frac{1}{2}-\frac{1}{\sqrt{3}}.\frac{\sqrt{3}}{2}=\frac{\sqrt{6}-3}{6}\]
b] Do\[ \frac{\pi}{2} 0, \cos a > 0\]
\[90^0< b < 180^0\Rightarrow\sin b > 0, \cos b < 0\]
\[\cos a = \sqrt{1-sin^{2}a}=\sqrt{1-\left [ \frac{4}{5} \right ]^{2}}=\frac{3}{5}\]
\[\cos b = -\sqrt{1-sin^{2}a}=-\sqrt{1-\left [ \frac{2}{3} \right ]^{2}}=-\frac{\sqrt{5}}{3}\]
\[\cos[a + b] = \cos a\cos b - \sin a\sin b\]
\[ =\frac{3}{5}\left [ -\frac{\sqrt{5}}{3} \right ]-\frac{4}{5}.\frac{2}{3}=-\frac{3\sqrt{5}+8}{15}\]
\[\eqalign{
& \sin[a - b] = \sin a\cos b - \cos a\sin b \cr
& = {4 \over 5}.\left[ { - {{\sqrt 5 } \over 3}} \right] - {3 \over 5}.{2 \over 3} = - {{4\sqrt 5 + 6} \over {15}} \cr} \]
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] \[\cos[ \frac{\pi}{2} - a]\sin[ \frac{\pi}{2} - b] - \sin[a - b] = \sin a\cos b - \sin a\cos b + \sin b\cos a\]
\[= \sin b\cos a\]
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] \[VT = [\cos a\cos b - \sin a\sin b][\cos a\cos b + \sin a\sin b]\]
\[= [\cos a\cos b]^2 [\sin a\sin b]^2\]
\[= \cos^2a[1 \sin^2b] [1 \cos^2a]\sin^2b = \cos^2a \sin^2b\]
\[= \cos^2b[1 \sin^2a] [1 \cos^2b]\sin^2a = \cos^2b \sin^2a\]