a:\(a\cdot sin0+b\cdot cos0+c\cdot sin90\)

\(=a\cdot0+b\cdot1+c\cdot1\)

=b+c

b: \(a\cdot cos90+b\cdot sin90+c\cdot sin180\)

\(=a\cdot0+b\cdot1+c\cdot0\)

=b

c: \(a^2\cdot sin90+b^2\cdot cos90+c^2\cdot cos180\)

\(=a^2\cdot1+b^2\cdot0+c^2\left(-1\right)\)

\(=a^2-c^2\)

a: \(=\dfrac{\sqrt{2}}{2}\left(cos^252^0+sin^252^0\right)=\dfrac{\sqrt{2}}{2}\)

b: \(=\dfrac{\sqrt{2}}{2}\left(cos^247^0+sin^247^0\right)=\dfrac{\sqrt{2}}{2}\)

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\(\cos^21^o+\cos^289^o=\cos^21^o+\cos^2\left(90^o-1^o\right)=\cos^21^o+\sin^21^o=1\)

\(\cos^22^o+\cos^288^o=\cos^22^o+\cos^2\left(90^o-2^o\right)=\cos^22^o+\sin^22^o=1\)

.......

\(\cos^244^o+\cos^246^o=\cos^244^o+\cos^2\left(90^o-44^o\right)=\cos^244^o+\sin^244^o=1\)

\(\cos^245^o=\left(\frac{\sqrt{2}}{2}\right)^2=\frac{1}{2}\)

=> \(A=1.44+\frac{1}{2}-\frac{1}{2}=44\)

\(A = \cos {75^0}\cos {15^0} = \frac{1}{2}\left[ {\cos \left( {{{75}^0} - {{15}^0}} \right) + \cos \left( {{{75}^0} + {{15}^0}} \right)} \right] \\= \frac{1}{2}.\cos {60^0}.\cos {90^0} = 0\)

\(B = \sin \frac{{5\pi }}{{12}}\cos \frac{{7\pi }}{{12}} = \frac{1}{2}\left[ {\sin \left( {\frac{{5\pi }}{{12}} - \frac{{7\pi }}{{12}}} \right) + \sin \left( {\frac{{5\pi }}{{12}} + \frac{{7\pi }}{{12}}} \right)} \right] \\= \frac{1}{2}\sin \left( { - \frac{{2\pi }}{{12}}} \right).\sin \left( {\frac{{12\pi }}{{12}}} \right) =  - \frac{1}{2}\sin \frac{\pi }{6}\sin \pi  = 0\)

\(A=cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\left(-cos\left(\pi-\dfrac{5\pi}{7}\right)\right)=-cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)

\(\Rightarrow A.sin\left(\dfrac{\pi}{7}\right)=-sin\left(\dfrac{\pi}{7}\right).cos\left(\dfrac{\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)

\(=-\dfrac{1}{2}sin\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{2\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)=-\dfrac{1}{4}sin\left(\dfrac{4\pi}{7}\right)cos\left(\dfrac{4\pi}{7}\right)\)

\(=-\dfrac{1}{8}sin\left(\dfrac{8\pi}{7}\right)=\dfrac{1}{8}sin\left(\dfrac{\pi}{7}\right)\)

\(\Rightarrow A=\dfrac{1}{8}\)

\(B=\dfrac{\sqrt{3}}{2}.cos48^0.cos24^0.cos12^0\)

\(\Rightarrow B.sin12^0=\dfrac{\sqrt{3}}{2}sin12^0.cos12^0cos24^0.cos48^0\)

\(=\dfrac{\sqrt{3}}{4}sin24^0cos24^0cos48^0=\dfrac{\sqrt{3}}{8}sin48^0.cos48^0\)

\(=\dfrac{\sqrt{3}}{16}sin96^0=\dfrac{\sqrt{3}}{16}cos6^0\)

\(\Rightarrow2B.sin6^0.cos6^0=\dfrac{\sqrt{3}}{16}cos6^0\Rightarrow B=\dfrac{\sqrt{3}}{32.sin6^0}\)

Biểu thức này ko thể rút gọn tiếp được

Ta có: \(\sin {70^o} = \cos {20^o};\;\cos {110^o} =  - \cos {70^o} =  - \sin {20^o}\)

\(\begin{array}{l} \Rightarrow A = {(\sin {20^o} + \cos {20^o})^2} + {(\cos {20^o} - \sin {20^o})^2}\\ = ({\sin ^2}{20^o} + {\cos ^2}{20^o} + 2\sin {20^o}\cos {20^o}) + ({\cos ^2}{20^o} + {\sin ^2}{20^o} - 2\sin {20^o}\cos {20^o})\\ = 2({\sin ^2}{20^o} + {\cos ^2}{20^o})\\ = 2\end{array}\)

Ta có: \(\tan {110^o} =  - \tan {70^o} =  - \cot {20^o};\;\cot {110^o} =  - \cot {70^o} =  - \tan {20^o}.\)

\( \Rightarrow B = \tan {20^o} + \cot {20^o} + ( - \cot {20^o}) + ( - \tan {20^o}) = 0\)

\(a,A=\sin^234^0+\cos^234^0+\dfrac{\cot42^0}{\cot42^0}=1+1=2\\ b,B=\left(\cos^213^0+\sin^277^0\right)+\dfrac{3\cot64^0}{\cot64^0}+2\cot32^0\cdot\tan32^0\\ B=1+3+2\cdot1=6\\ c,B=\dfrac{5\cot35^0}{\cot35^0}-2\left(\sin^261^0-\cos^261^0\right)=5-2\cdot1=3\)

Bài 1: 

Ta có: \(A=\sin^6\alpha+3\cdot\sin^2\alpha\cdot\cos^2\alpha+\cos^6\alpha\)

\(=\left(\sin^2\alpha+\cos^2\alpha\right)^3-3\cdot\sin^2\alpha\cdot\cos\alpha\cdot\left(\sin^2\alpha+\cos^2\alpha\right)+3\cdot\sin^2\alpha\cdot\cos^2\alpha\)

\(=1^3\)

=1