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

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

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

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

\(\begin{array}{l}\cos 75^\circ  = \frac{{\sqrt 6  - \sqrt 2 }}{4}\\\tan \left( { - \frac{{19\pi }}{6}} \right) =  - \frac{{\sqrt 3 }}{3}\end{array}\)

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

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

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

a) Ta có A=\dfrac{\tan \alpha+3 \dfrac{1}{\tan \alpha}}{\tan \alpha+\dfrac{1}{\tan \alpha}}=\dfrac{\tan ^{2} \alpha+3}{\tan ^{2} \alpha+1}=\dfrac{\dfrac{1}{\cos ^{2} \alpha}+2}{\dfrac{1}{\cos ^{2} \alpha}}=1+2 \cos ^{2} \alphaA=tanα+tanα1​tanα+3tanα1​​=tan2α+1tan2α+3​=cos2α1​cos2α1​+2​=1+2cos2α Suy ra A=1+2 \cdot \dfrac{9}{16}=\dfrac{17}{8}A=1+2⋅169​=817​.

b) B=\dfrac{\dfrac{\sin \alpha}{\cos ^{3} \alpha}-\dfrac{\cos \alpha}{\cos ^{3} \alpha}}{\dfrac{\sin ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{3 \cos ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{2 \sin \alpha}{\cos ^{3} \alpha}}=\dfrac{\tan \alpha\left(\tan ^{2} \alpha+1\right)-\left(\tan ^{2} \alpha+1\right)}{\tan ^{3} \alpha+3+2 \tan \alpha\left(\tan ^{2} \alpha+1\right)}B=cos3αsin3α​+cos3α3cos3α​+cos3α2sinα​cos3αsinα​−cos3αcosα​​=tan3α+3+2tanα(tan2α+1)tanα(tan2α+1)−(tan2α+1)​.

Suy ra B=\dfrac{\sqrt{2}(2+1)-(2+1)}{2 \sqrt{2}+3+2 \sqrt{2}(2+1)}=\dfrac{3(\sqrt{2}-1)}{3+8 \sqrt{2}}B=22​+3+22​(2+1)2​(2+1)−(2+1)​=3+82​3(2​−1)​.