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,\(\sin15+\sin75-\sin75-\cos15+\sin30=\sin30=\frac{1}{2}\)

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

a) Ta có : sin\(^2\)12^o=cos²78^o=> sin²12^o+sin²78^o=1.

tương tự => A=3

b) tương tự câu (a) ta có: cos²15^o=sin²75^o ( do 15+75=90 nha bạn ) => cos²15^o+cos²75^o=1. Tương tự => B=0

c) Ta có: C= 4cos²a -6sin²a=4(cos²a+sin²a)-10sin²a=4-10.\(\dfrac{1}{5^2}\)=4-0,4=3,6

\(ADCT:\sin^2\alpha+\cos^2\alpha=1\)

\(A=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin45^0\)

\(A=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\frac{\sqrt{2}}{2}\)

\(A=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}\)

Câu b lm tương tự

\(=0+5\cdot1+2\cdot1=7\)