\(\begin{array}{l}A = \sin \left( {a - 17^\circ } \right)\cos \left( {a + 13^\circ } \right) - \sin \left( {a + 13^\circ } \right)\cos \left( {a - 17^\circ } \right)\\A = \sin \left( {a - 17^\circ  - a - 13^\circ } \right) = \sin \left( { - 30^\circ } \right) =  - \frac{1}{2}\end{array}\)

\(\begin{array}{l}B = \cos \left( {b + \frac{\pi }{3}} \right)\cos \left( {\frac{\pi }{6} - b} \right) - \sin \left( {b + \frac{\pi }{3}} \right)\sin \left( {\frac{\pi }{6} - b} \right)\\B = \cos \left( {b + \frac{\pi }{3} + \frac{\pi }{6} - b} \right) = \cos \frac{\pi }{2} = 0\end{array}\)

\(a,cos\left(\dfrac{21\pi}{6}\right)=cos\left(3\pi+\dfrac{\pi}{2}\right)=cos\left(\pi+\dfrac{\pi}{2}\right)=-cos\left(\dfrac{\pi}{2}\right)=0\\ b,sin\left(\dfrac{129\pi}{4}\right)=sin\left(32\pi+\dfrac{\pi}{4}\right)=sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\\ c,tan\left(1020^o\right)=tan\left(5\cdot180^o+120^o\right)=tan\left(120^o\right)=-\sqrt{3}\)

\(\begin{array}{l}\sin \left( { - \frac{{2\pi }}{3}} \right) =  - \frac{{\sqrt 3 }}{2}\\\tan 495^\circ  =  - 1\end{array}\)

a) \({\cos ^2}\frac{\pi }{8} + {\cos ^2}\frac{{3\pi }}{8} = {\cos ^2}\frac{\pi }{8} + {\cos ^2}\left( {\frac{\pi }{2} - \frac{\pi }{8}} \right) = {\cos ^2}\frac{\pi }{8} + {\sin ^2}\frac{\pi }{8} = 1\)

b)

\(\begin{array}{l}\tan {1^ \circ }.\tan {2^ \circ }.\tan {45^ \circ }.\tan {88^ \circ }.\tan {89^ \circ }\\ = (\tan {1^ \circ }.\tan {89^ \circ }).(\tan {2^ \circ }.\tan {88^ \circ }).\tan {45^ \circ }\\ = (\tan {1^ \circ }.\cot {1^ \circ }).(\tan {2^ \circ }.\cot {2^ \circ }).\tan {45^ \circ }\\ = 1\end{array}\)

a) Ta có: \(sin^2x+sin^2\left(90-x\right)=sin^2x+cos^2x=1.\)

áp dụng: A = 2

b)Ta có: \(cos\left(x\right)=-cos\left(180-x\right)\)

áp dụng: B = 0

c) Ta có: \(tan\left(x\right)\cdot tan\left(90-x\right)=\frac{sinx}{cosx}\cdot\frac{sin\left(90-x\right)}{cos\left(90-x\right)}=\frac{sinx}{cosx}\cdot\frac{cosx}{sinx}=1\)

áp dụng: C = 1

quá sai

trả dép em về

a) Vì \(\frac{\pi }{2} < a < \pi \) nên \(\cos a < 0\)

Ta có: \({\sin ^2}a + {\cos ^2}a  = 1\)

 \(\Leftrightarrow \frac{1}{9} + {\cos ^2}a  = 1\)

\(\Leftrightarrow {\cos ^2}a =  1 - \frac{1}{9}= \frac{8}{9}\)

\(\Leftrightarrow \cos a  =\pm\sqrt { \frac{8}{9}}  =  \pm \frac{{2\sqrt 2 }}{3}\)

Vì \(\cos a < 0\) nên \(cos a =-\frac{{2\sqrt 2 }}{3}\)

Suy ra \(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{{\frac{1}{3}}}{{ - \frac{{2\sqrt 2 }}{3}}} =  - \frac{{\sqrt 2 }}{4}\)

Ta có: \(\sin 2a = 2\sin a\cos a = 2.\frac{1}{3}.\left( { - \frac{{2\sqrt 2 }}{3}} \right) =  - \frac{{4\sqrt 2 }}{9}\)

\(\cos 2a = 1 - 2{\sin ^2}a = 1 - \frac{2}{9} = \frac{7}{9}\)

\(\tan 2a = \frac{{2\tan a}}{{1 - {{\tan }^2}a}} = \frac{{2.\left( { - \frac{{\sqrt 2 }}{4}} \right)}}{{1 - {{\left( { - \frac{{\sqrt 2 }}{4}} \right)}^2}}} =  - \frac{{4\sqrt 2 }}{7}\)

b) Vì \(\frac{\pi }{2} < a < \frac{{3\pi }}{4}\) nên \(\sin a > 0,\cos a < 0\)

\({\left( {\sin a + \cos a} \right)^2} = {\sin ^2}a + {\cos ^2}a + 2\sin a\cos a = 1 + 2\sin a\cos a = \frac{1}{4}\)

Suy ra \(\sin 2a = 2\sin a\cos a = \frac{1}{4} - 1 =  - \frac{3}{4}\)

Ta có: \({\sin ^2}a + {\cos ^2}a = 1\;\)

\( \Leftrightarrow \left( {\frac{1}{2} - {\cos }a} \right)^2 + {\cos ^2}a - 1 = 0\)

\( \Leftrightarrow \frac{1}{4} - \cos a + {\cos ^2}a + {\cos ^2}a - 1 = 0\)

\( \Leftrightarrow 2{\cos ^2}a - \cos a - \frac{3}{4} = 0\)

\( \Rightarrow \cos a = \frac{{1 - \sqrt 7 }}{4}\) (Vì \(\cos a < 0)\)

\(\cos 2a = 2{\cos ^2}a - 1 = 2.{\left( {\frac{{1 - \sqrt 7 }}{4}} \right)^2} - 1 =  - \frac{{\sqrt 7 }}{4}\)

\(\tan 2a = \frac{{\sin 2a}}{{\cos 2a}} = \frac{{ - \frac{3}{4}}}{{ - \frac{{\sqrt 7 }}{4}}} = \frac{{3\sqrt 7 }}{7}\)

A=a2sin90∘+b2cos90∘+c2cos180∘A=a2sin⁡90∘+b2cos⁡90∘+c2cos⁡180∘

=a2*1+b2* 0 +c2* (-1

=a2 - c2

B=3−sin290∘+2cos260∘−3tan245∘B=3−sin2⁡90∘+2cos2⁡60∘−3tan2⁡45∘.

= 3 - 1 + 1/2 - 3 = -1/2

What did you see at the zoo?

 I saw crocodiles.

toán mà

a) \(\cos \frac{{3\pi }}{7} = 0,22252\); \(\tan ( - {37^ \circ }25') = 0,765018\)      

b) \(179^o23'30"\approx3,130975234\left(rad\right)\)

c) \(\frac{{7\pi }}{9} = {140^ \circ }\)