\(\cos^25^o+\cos^210^o+....+\cos^285^o\\ =\left(\cos^25^o+\cos^285^o\right)+\left(\cos^210^o+\cos^280^o\right)+...+\left(\cos^240^o+\cos^250^o\right)+\cos^245^o\\ \\ =\left(\cos^25^o+\sin^25^o\right)+\left(\cos^210^o+\sin^210^o\right)+...+\left(\cos^240^o+\sin^240^o\right)+\frac{1}{2}\\ =1+1+...+1+\frac{1}{2}=16+\frac{1}{2}=\frac{33}{2}\)

a: \(=\left(\sin^210^0+\sin^280^0\right)+\left(\sin^220^0+\sin^270^0\right)+\left(\sin^230^0+\sin^260^0\right)+\left(\sin^240^0+\sin^250^0\right)\)

=1+1+1+1

=4

b: \(=\left(\cos^25^0+\cos^285^0\right)+\left(\cos^215^0+\cos^275^0\right)+\left(\cos^225^0+\cos^265^0\right)+\left(\cos^235^0+\cos^255^0\right)+\cos^245^0\)

\(=1+1+1+1+\dfrac{1}{2}=4+\dfrac{1}{2}=\dfrac{9}{2}\)

Có

A=\(\left(sin^215^o+sin^275^o\right)+\left(sin^240^o+sin^250^o\right)+\left(sin^260^o+sin^230^o\right)\)

\(=\left(sin^215^o+cos^215^o\right)+...\)

\(=1\cdot3=3\)

Câu c tương tự mà mk nghĩ đề sai dấu - trước cos^245độ

Nói chung nếu: a+b=90 độ

thì: \(sin^2a+sin^2b=1\)

b) thì áp dụng nếu a+b=90 độ:

\(tana=cotb\) và ngược lại

Mà \(tana\cdot cota=1\)

Nói chung là công thức......

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

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

.......

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

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

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

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

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

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

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

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

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

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

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

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

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

Biểu thức này ko thể rút gọn tiếp được

Ta có sin10⁰=cos80⁰(vì 10⁰+80⁰=90⁰)⇒sin²10⁰=cos²80⁰

sin20⁰=cos70⁰(vì 20⁰+70⁰=90⁰)⇒sin²20⁰=cos²70⁰

Ta có công thức sin²a+cos²a=1

\(P=cos^210^0+cos^220^0+cos^270^0+cos^280^0=cos^210^0+cos^220^0+sin^220^0+sin^210^0=\left(cos^210^0+sin^210^0\right)+\left(cos^220^0+sin^220^0\right)=1+1=2\)

a) \(cos^275+cos^253+cos^217+cos^237\)

ta áp dụng: \(sin^2a+cos^2a=1\)

ta được: \(\left(cos^275+cos^2\left(90-75\right)\right)+\left(cos^253+cos^2\left(90-53\right)\right)\)

=\(1+1=2\)

b) \(\frac{tan^215-1}{cot75-1}-cos75\)

=\(\frac{\left(tan15-1\right)\left(tan15+1\right)}{tan15-1}-cos75\)

=\(tan15+1-sin15\)=sin15\(\left(\frac{1}{cos15}-1+\frac{1}{sin15}\right)\)

a) \(cos^273^o+cos^253^o+cos^217^o+cos^237^o=\left(cos^273^o+cos^217^o\right)+\left(cos^253^o+cos^237^o\right)\)

\(=\left(cos^273^o+sin^273^o\right)+\left(cos^253^o+sin^253^o\right)=1+1=2\)

b) \(\frac{tan^215^o-1}{cotg75^o-1}-cos75^o=\frac{\left(tan15^o-1\right)\left(tan15^o+1\right)}{tan15^o-1}-cos75^o=tan15^o+1-cos75^o\)

Theo mình thì làm như zầy :
A=cos² 73⁰+cos² 53⁰+cos² 17⁰+cos² 37⁰

   = cos² 73⁰ +sin² 73⁰+cos² 53⁰+ sin² 53⁰

   = 1+1 = 2
 

B=^{tan2150−1cot750−1−cos750}

   =(cot² 75⁰ -1)/(cot 75⁰ -1) - cos 75⁰

   ⁼(cot 75⁰ +1)(cot 75⁰ -1)/(cot 75⁰ -1) -cos 75⁰ ( HĐT :a² +b²=(a+b)(a-b) )

   = cot 75⁰ +1 - cos 75⁰ (Cái này mình bấm máy tính nó ra =)) )

   = 3- √3 - (√6 -√2 )/4  

   =√3 (√3 -1) - √2 (√3 -1)/4     

   =   [ 4√3 (√3 -1) - √2 (√3 -1) ]     

   =   [ ( √3 -1 ) ( 4√3 - √2 ) ] /4 (chỗ này mình để phân số,nếu không thì bạn để là  (12 - 4√3 - √6 +√2)/4 cũng được)

√ c os

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

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