Câu trả lời này có tính chất là ko giúp được cái gì cả !

\(A=sin^21^o+c\text{os}^22^o+sin^23^o+c\text{os}^24^o+...+sin^2179^o+c\text{os}^2180^o\)

\(=sin^21^o+c\text{os}^22^o+sin^23^o+c\text{os}^24^o+...+c\text{os}^290^o-sin^289^o-c\text{os}^288^o-...-sin^21^o-c\text{os}^20^o\)

\(=c\text{os}^290^o-c\text{os}^20^o\)

\(=-1\)

Chúc bn học tốt

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

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

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

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

1.

Kẻ \(MH\perp NP\) tại H

Ta có: \(S_{MNP}=\dfrac{1}{2}MH.NP\) (1)

\(S_{MNK}=\dfrac{1}{2}MH.KN\) (2)

Ta lại có: KN=MN mà NM<NP

\(\Rightarrow KN< NP\) (3)

Từ (1),(2) và (3) suy ra: \(S_{MNP}>S_{MNK}\)

2.

\(Sin^21^o+Sin^22^o+Sin^23^o+...+Sin^287^o+Sin^288^o+Sin^298^o\)

\(=\left(Sin^21^o+Sin^289^o\right)\left(Sin^22^o+Sin^288^o\right)+...+Sin^245^o\\ =\left(Sin^21^o+Cos^21^o\right)\left(Sin^22^o+Cos^22^o\right)+....+Sin^245^o\\ =44+Sin^245^o\\ =44+\dfrac{1}{2}=44,5\)

Lời giải:

a) Ta có tính chất quen thuộc là nếu \(\alpha+\beta=90^0\Rightarrow \cos \alpha=\sin \beta\)(có thể thấy rất rõ khi xét một tam giác vuông)

Tức là \(\sin \beta=\cos (90-\beta)\)

Do đó:

\(A=(\sin ^22^0+\sin ^288^0)+(\sin ^24^0+\sin ^286^0)+...+(\sin ^244^0+\sin ^246^0)\)

\(=\underbrace{(\sin ^22^0+\cos ^22^0)+(\sin ^24^0+\cos ^24^0)+...+(\sin ^244^0+\cos ^244^0)}_{22\text{cặp}}\)

\(=\underbrace{1+1+...+1}_{22}=22\) (tổng 2 bình phương sin và cos của một góc thì bằng 1)

b)

\(P=1994(\sin ^6x+\cos ^6x)-2991(\sin ^4x+\cos ^4x)\)

\(=1994[(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos^2 x+\cos ^4x)]-2991(\sin ^4x+\cos ^4x)\)

\(=1994(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-2991(\sin ^4x+\cos ^4x)\)

\(=-1994\sin ^2x\cos ^2x-997\sin ^4x-997\cos ^4x\)

\(=-997(\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x) \)

\(=-997(\sin ^2x+\cos ^2x)^2=-997\)

Do đó biểu thức không phụ thuộc vào $x$

Ta có \(\sin x=\cos\left(90^0-x\right)\)

\(\Rightarrow M=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin^245^0\)

\(=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\sin^245^0\)

\(=1+1+1+\left(\frac{\sqrt{2}}{2}\right)^2=3+\frac{1}{2}=\frac{7}{2}\)

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

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

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

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

\(A=\left(sin^212^o+sin^278^o\right)+\left(sin^21^o+sin^289^o\right)+\left(sin^273^o+sin^217^o\right)\)

\(A=\left(sin^290^o\right)+\left(sin^290^o\right)+\left(sin^290^o\right)\)

\(A=1+1+1=3\)