Ta có: \(A=\sin^25^0+\sin^225^0+\sin^245^0+\sin^265^0+\sin^285^0\)

\(=\left(\sin^25^0+\sin^285^0\right)+\left(\sin^225^0+\sin^265^0\right)+\dfrac{1}{2}\)

\(=2+\dfrac{1}{2}=\dfrac{5}{2}\)

\(\Rightarrow A=\left(sin^25^0+sin^285^0\right)+\left(sin^225^0+sin^265^0\right)+sin^245^0=\left(sin^25^0+cos^25^0\right)+\left(sin^225^0+cos^225^0\right)+\dfrac{1}{2}=1+1+\dfrac{1}{2}=\dfrac{5}{2}\)

4. \(D=sin^21^o+sin^22^o+sin^23^o+...+sin^287^o+sin^288^o+sin^289^o=\left(sin^21^o+sin^289^o\right)+\left(sin^22^o+sin^288^o\right)+...+\left(sin^244^o+sin^246^o\right)+sin^245^o=1+1+1+...+1+1+0,5=44,5\)

\(5.E=cos^21^o+cos^22^o+cos^23^o+...+cos^287^o+cos^288^o+cos^289^o=\left(cos^21^o+cos^289^o\right)+\left(cos^22^o+cos^288^o\right)+...+\left(cos^244^o+cos^246^o\right)+cos^245^o=1+1+1+...+1+0,5=1.44+0,5=44,5\)

mk bỏ dấu độ hết nha bn : (trong toán người ta cho phép)

1) ta có : \(A=\left(sin1+sin2+...+sin89\right)-\left(cos1+cos2+...+cos89\right)\)

\(=\left(sin1+sin2+...+sin89\right)-\left(cos\left(90-89\right)+cos\left(90-88\right)+...+cos\left(90-1\right)\right)\)

\(=\left(sin1+sin2+...+sin89\right)-\left(sin89+sin88+...+sin1\right)=0\)

2) ta có : \(B=tan1.tan2.tan3...tan87.tan88.tan89\)

\(=\left(tan1.tan89\right).\left(tan2.tan88\right).\left(tan3.tan87\right)...\left(tan44.tan46\right).tan45\)

\(=\left(tan1.tan\left(90-1\right)\right).\left(tan2.tan\left(90-2\right)\right).\left(tan3.tan\left(90-3\right)\right)...\left(tan44.tan\left(90-44\right)\right).tan45\)

3) bạn xem lại đề nha

4) ta có : \(D=sin^21+sin^22+sin^23+...+sin^289\)

\(=\left(sin^21+sin^289\right)+\left(sin^22+sin^288\right)+...+\left(sin^244+sin^246\right)+sin^245\)

\(=\left(sin^21+sin^2\left(90-1\right)\right)+\left(sin^22+sin^2\left(90-2\right)\right)+...+\left(sin^244+sin^2\left(90-44\right)\right)+sin^245\)

\(=44+sin^245=44+\dfrac{1}{2}=\dfrac{89}{2}\)

5) ta có : \(E=cos^21+cos^22+cos^23+...+cos^289\)

\(=\left(cos^21+cos^289\right)+\left(cos^22+cos^288\right)+...+\left(cos^244+cos^246\right)+cos^245\)

\(=\left(cos^21+cos^2\left(90-1\right)\right)+\left(cos^22+cos^2\left(90-2\right)\right)+...+\left(cos^244+cos^2\left(90-44\right)\right)+cos^245\)

\(=44+cos^245=44+\dfrac{1}{2}=\dfrac{89}{2}\)

b: \(\cos30^0=\dfrac{\sqrt{3}}{2}\)

k mk đi

ai k mk

mk k lại

thanks

deo tra loi thoi m vo chat linh tinh ak

ap dung \(\sin a=\cos b\)

va ap dung\(sin^2a+cos^2a=1\)

bài 2 là tính tan C nhá

mik vt nhầm

sina=cos(90-a) thay vào ta được

sin²15+sin²25+sin²35+cos²35+cos²25+cos²15=3

tương tự câu dưới ta được =3/2

`sin^2 25^o + sin^2 65^o`

`=cos^2 65^o + sin^2 65^o`

=1`

__________________________________________

`***` Áp dụng công thức lượng giác: `sin^2 \alpha +cos^2 \alpha =1`

Đề cs sai 0 nhỉ phải là `sin^2 25^o + sin^2 65^o`

 Hoặc `sin^2 35^o + sin^2 55^o` chứ

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

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

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

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

Câu b lm tương tự