Chú ý 2 điều: \(\cos45^o=\sin45^o=\frac{\sqrt{2}}{2}\) và \(\cos^2a+\sin^2a=1\)

Do đó: 

a) \(A=\cos^252^o.\frac{\sqrt{2}}{2}+\sin^252^o.\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\left(\cos^252^o+\sin^252^o\right)=\frac{\sqrt{2}}{2}.1=\frac{\sqrt{2}}{2}\)

b) \(B=\frac{\sqrt{2}}{2}.\cos^247^o+\frac{\sqrt{2}}{2}.\sin^247^o=\frac{\sqrt{2}}{2}\left(\cos^247^o+\sin^247^o\right)=\frac{\sqrt{2}}{2}.1=\frac{\sqrt{2}}{2}\)

a)
\(A=cos^230^o-sin^230^o=\left(\dfrac{\sqrt{3}}{2}\right)^2-\left(\dfrac{1}{2}\right)^2=\dfrac{1}{2}\);
\(B=cos60^o+sin45^o=\dfrac{1}{2}+\dfrac{\sqrt{2}}{2}\).
Vì vậy \(A< B\).
b)
\(C=\dfrac{2tan30^o}{1-tan^230^o}=\dfrac{2\dfrac{\sqrt{3}}{2}}{1-\left(\dfrac{\sqrt{3}}{2}\right)^2}=\sqrt{3}\).
\(D=\left(-tan135^o\right)tan60^o=-\left(-1\right).\sqrt{3}=\sqrt{3}\).
Vậy \(C=D\).

Vì sin(\(\alpha\) ) = cos (\(90-\alpha\)) nên \(sin^2\alpha=cos^2\left(90-\alpha\right)\)

a/ \(sin^230-sin^240-sin^250+sin^260=\left(cos^260+sin^260\right)-\left(cos^250+sin^250\right)=1-1=0\)

b/ \(cos^225-cos^235+cos^245-cos^255+cos^265=\left(sin^265+cos^265\right)-\left(sin^255+cos^255\right)+cos^245=1-1+cos^245=cos^245=\dfrac{1}{2}\)

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

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

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

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......

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

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