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

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

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

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

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

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

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

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

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

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

Biểu thức này ko thể rút gọn tiếp được

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

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

\(A=\frac{\sqrt{2}}{2}cos^252+\frac{\sqrt{2}}{2}sin^252=\frac{\sqrt{2}}{2}\left(sin^252+cos^252\right)=\frac{\sqrt{2}}{2}\)

\(B=\sqrt{3}.cos^247+\sqrt{3}.sin^247=\sqrt{3}\left(sin^247+cos^247\right)=\sqrt{3}\)

Botay.com.vn

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

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

.......

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

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

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

\(P=4\left[\left(cos^21^0+cos^289^0\right)+\left(cos^22^0+cos^288^0\right)+...+\left(cos^244^0+cos^246^0\right)+cos^245^0\right]\)

\(=4\left[\left(cos^21^0+sin^21^0\right)+\left(cos^22^0+sin^22^0\right)+...+\left(cos^244^0+sin^244^0\right)+cos^245^0\right]\)

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

Bạn kiểm tra lại đề, có vẻ như trong 2 cái \(sin^2\) kia phải có 1 cái là \(cos^2\) mới hợp lý

à, cái đầu là cos

a) Ta có: \(sin\alpha=cos\left(90-\alpha\right)\Rightarrow sin42=cos48\)

\(\Rightarrow sin42-cos48=0\)

b) Ta có: \(sin\alpha=cos\left(90-\alpha\right)\Rightarrow sin61=cos29\Rightarrow sin^261=cos^229\)

\(\Rightarrow sin^261+sin^229=sin^229+cos^229=1\)

c) Ta có: \(tan\alpha=\dfrac{1}{tan\left(90-\alpha\right)}\Rightarrow tan40=\dfrac{1}{tan50}\)

\(\Rightarrow tan40.tan50=1\) mà \(tan45=1\Rightarrow tan40.tan45.tan50=1\)

\(sin42^0-cos48^0=sin42^0-sin\left(90^0-48^0\right)=sin42^0-sin42^0=0\)

\(sin^261^0+sin^229^0=sin^261^0+cos^2\left(90^0-29^0\right)=sin^261^0+cos^261^0=1\)

\(tan40^0.tan50^0.tan45^0=tan40^0.cot\left(90^0-50^0\right).1=tan40^0.cot40^0=1\)

Sử dụng các công thức:

\(cosa=sin\left(90^0-a\right)\) ; \(sina=cos\left(90^0-a\right)\) ; \(tana=cot\left(90^0-a\right)\) ; \(tana.cota=1\)