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a)Theo định lí tỉ số lượng giác của hai góc phụ nhau, ta có:
\(\sin1=\cos89....\sin89=\cos1\)
Vậy \(A=0\)
b) Theo định lí tỉ số lượng giác của 2 góc phụ nhau, ta có:
\(\tan1=\cot89...\tan2=\cot88...\)
\(\Rightarrow B=\tan45\cdot\tan46\cdot\cot46\cdot...\cdot\tan89\cdot\cot89\)
Mà \(\tan\lambda\cdot\cot\lambda=1\)
\(\Rightarrow B=\tan45\cdot1=1\)
c) Bạn làm tương tự dựa vào CT \(\sin^2\lambda+\cos^2\lambda=1\)
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,\(\sin15+\sin75-\sin75-\cos15+\sin30=\sin30=\frac{1}{2}\)
Ta có : \(cos^215^o=sin^275^o;cos^225^o=sin^265^o;cos^235^o=sin^255^o;\frac{cos^245^o}{2}=\frac{sin^245^o}{2}\)
Khi đó \(N=sin^275^o+cos^275^o-\left(sin^265^o+cos^265^o\right)+sin^255^o+cos^255^o-\left(\frac{sin^245^0+cos^245^o}{2}\right)\)
Áp dụng công thức \(sin^2a+cos^2a=1\)ta được
\(N=1-1+1-\frac{1}{2}=\frac{1}{2}\)
Vậy N = 1/2
câu b chờ chút mình làm cho nhé <33
Ta có : \(cos^21^o=sin^289^o;cos^22^o=sin^288^o;...;cos^244^o=sin^246^o;\frac{cos^245^o}{2}=\frac{sin^245^o}{2}\)
Khi đó \(A=\frac{sin^245^o+cos^245^o}{2}+\left(sin^246^0+cos^246^o\right)+...+\left(sin^289^o+cos^289^o\right)\)
Áp dụng ct \(sin^2a+cos^2a=1\)ta được \(A=\frac{1}{2}+1+1+...+1=...\)
P/S : bạn tự đếm xem bao nhiêu cặp nhé ;) tìm ssh á
mk bỏ dấu độ nha . trong toán người ta cho phép
a) ta có : \(cos^215+cos^225+cos^235+cos^245+cos^255+cos^265+cos^275\)
\(=cos^215+cos^275+cos^225+cos^265+cos^235+cos^255+cos^245\) \(=cos^215+cos^2\left(90-15\right)+cos^225+cos^2\left(90-25\right)+cos^235+cos^2\left(90-35\right)+cos^245\) \(=cos^215+sin^215+cos^225+sin^225+cos^235+sin^235+cos^245\)\(=1+1+1+\dfrac{1}{2}=\dfrac{7}{2}\)
b) ta có : \(sin^210-sin^220+sin^230-sin^240-sin^250-sin^270+sin^280\)
\(=sin^210+sin^280-sin^220-sin^270-sin^240-sin^250+sin^230\) \(=sin^210+sin^2\left(90-10\right)-sin^220-sin^2\left(90-20\right)-sin^240-sin^2\left(90-40\right)+sin^230\) \(=sin^210+cos^210-sin^220-cos^220-sin^240-cos^240+sin^230\) \(=1-1-1+\dfrac{1}{4}=\dfrac{-3}{4}\)
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
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 : \(B=tan1.tan2.tan3...tan87.tan88.tan89\)
\(=\left(tan1.tan89\right).\left(tan2.tan88\right).\left(tan3.tan87\right)...\left(tan44.tan46\right).tan45\)
\(=\left(tan1.tan\left(90-1\right)\right).\left(tan2.tan\left(90-2\right)\right).\left(tan3.tan\left(90-3\right)\right)...\left(tan44.tan\left(90-44\right)\right).tan45\)
\(=\left(tan1.cot1\right).\left(tan2.cot2\right).\left(tan3.cot3\right)...\left(tan44.cot44\right).tan45\) \(=tan45=1\)3) bạn xem lại đề nha
4) ta có : \(D=sin^21+sin^22+sin^23+...+sin^289\)
\(=\left(sin^21+sin^289\right)+\left(sin^22+sin^288\right)+...+\left(sin^244+sin^246\right)+sin^245\)
\(=\left(sin^21+sin^2\left(90-1\right)\right)+\left(sin^22+sin^2\left(90-2\right)\right)+...+\left(sin^244+sin^2\left(90-44\right)\right)+sin^245\)
\(=\left(sin^21+cos^21\right)+\left(sin^22+cos^22\right)+...+\left(sin^244+cos^244\right)+sin^245\)\(=44+sin^245=44+\dfrac{1}{2}=\dfrac{89}{2}\)
5) ta có : \(E=cos^21+cos^22+cos^23+...+cos^289\)
\(=\left(cos^21+cos^289\right)+\left(cos^22+cos^288\right)+...+\left(cos^244+cos^246\right)+cos^245\)
\(=\left(cos^21+cos^2\left(90-1\right)\right)+\left(cos^22+cos^2\left(90-2\right)\right)+...+\left(cos^244+cos^2\left(90-44\right)\right)+cos^245\)
\(=\left(cos^21+sin^21\right)+\left(cos^22+sin^22\right)+...+\left(cos^244+sin^244\right)+cos^245\)\(=44+cos^245=44+\dfrac{1}{2}=\dfrac{89}{2}\)
\(1.A=\left(sin1^o+sin2^o+sin3^o+...+sin88^o+sin89^o\right)-\left(cos1^o+cos2^o+cos3^o+...+cos88^o+cos89^o\right)=sin1^o+sin2^o+sin3^o+...+sin88^o+sin89^o-sin89^o-sin88^o-sin87^o-...-sin2^o-sin1^o=0\)
2.\(B=tan1^o.tan2^o.tan3^o...tan87^o.tan88^o.tan89^o\Rightarrow B=tan1^o.tan2^o.tan3^o...cot3^o.cot2^o.cot1^o=1^{45}=1\)