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![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\cos^215+\cos^225+\cos^235+\cos^245+\sin^235+\sin^225+\sin^215\)
=\(\left(\cos^215+\sin^215\right)+\left(\cos^225+\sin^225\right)+\left(\cos^235+\sin^235\right)+\cos^245\)
=\(1+1+1+\frac{1}{2}=\frac{7}{2}\)
b.\(\sin^210-\sin^220-\sin^230-\sin^240-\cos^240-\cos^220+\cos^210\)
=\(\left(\sin^210+\cos^210\right)-\left(\sin^220+\cos^220\right)-\left(\sin^240+\cos^240\right)-\sin^230\)
=\(1-1-1-\frac{1}{4}=-\frac{5}{4}\)
c,\(\sin15+\sin75-\sin75-\cos15+\sin30=\sin30=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
áp dụng \(sin^2a+\cos^2a=1\)
ta có \(\sin^275^o+sin^215^o-\cos^250^o-\cos^240^o+\)\(cot45^o.cot45^o\)\(=sin^275^o+\cos^275^o-\left(\cos^250^o+sin^250^o\right)\)\(+cot^245^o\)\(=1-1+1=1\)
vì đây là tam giác vuông, hai góc nhọn phụ nhau nên sin góc này bằng cosin góc kia
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : sin\(^2\)12o=cos278o=> sin212o+sin278o=1.
tương tự => A=3
b) tương tự câu (a) ta có: cos215o=sin275o ( do 15+75=90 nha bạn ) => cos215o+cos275o=1. Tương tự => B=0
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ADCT:\sin^2\alpha+\cos^2\alpha=1\)
\(A=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin45^0\)
\(A=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\frac{\sqrt{2}}{2}\)
\(A=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}\)
Câu b lm tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\dfrac{\sin25^0}{\cos25^0}+2\left(\sin^215^0+\cos^215^0\right)-\tan25^0+4\cdot\dfrac{1}{2}\\ =\tan25^0+2\cdot1-\tan25^0+2=4\)