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\(A=\frac{\cos57}{\cos57}+\frac{\cot58}{\cot58}-2\left(1+1\right)\)\()\)
=1+1-4
=-2
4. \(D=sin^21^o+sin^22^o+sin^23^o+...+sin^287^o+sin^288^o+sin^289^o=\left(sin^21^o+sin^289^o\right)+\left(sin^22^o+sin^288^o\right)+...+\left(sin^244^o+sin^246^o\right)+sin^245^o=1+1+1+...+1+1+0,5=44,5\)
\(5.E=cos^21^o+cos^22^o+cos^23^o+...+cos^287^o+cos^288^o+cos^289^o=\left(cos^21^o+cos^289^o\right)+\left(cos^22^o+cos^288^o\right)+...+\left(cos^244^o+cos^246^o\right)+cos^245^o=1+1+1+...+1+0,5=1.44+0,5=44,5\)
a) ta có : \(A=\dfrac{sin33}{cos57}+\dfrac{tan32}{cot58}-2\left(sin20.cos70+cos20.sin70\right)\)
\(\Leftrightarrow A=\dfrac{sin33}{cos\left(90-33\right)}+\dfrac{tan32}{cot\left(90-32\right)}-2\left(sin20.cos\left(90-20\right)+cos20.sin\left(90-20\right)\right)\)
\(\Leftrightarrow A=\dfrac{sin33}{sin33}+\dfrac{tan32}{tan32}-2\left(sin20.sin20+cos20.cos20\right)\)\(\Leftrightarrow A=1+1-2\left(sin^220+cos^220\right)=1+1-2=0\)
b) sữa đề chút nha
ta có : \(B=\dfrac{sin^215+sin^275-sin^212-sin^278}{cos^213+cos^277+cos^21+cos^289}+\dfrac{2tan55}{cot35}\)
\(\Leftrightarrow B=\dfrac{sin^215+sin^2\left(90-15\right)-sin^212-sin^2\left(90-12\right)}{cos^213+cos^2\left(90-13\right)+cos^21+cos^2\left(90-1\right)}+\dfrac{2tan\left(90-35\right)}{cot35}\)
\(\Leftrightarrow B=\dfrac{sin^215+cos^215-sin^212-cos^212}{cos^213+sin^213+cos^21+sin^21}+\dfrac{2cot35}{cot35}\) \(\Leftrightarrow B=\dfrac{sin^215+cos^215-\left(sin^212+cos^212\right)}{cos^213+sin^213+cos^21+sin^21}+\dfrac{2cot35}{cot35}\)\(\Leftrightarrow B=\dfrac{1-1}{cos^213+sin^213+cos^21+sin^21}+2=0+2=2\)
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
\(ADCT:\sin^2\alpha+\cos^2\alpha=1\)
\(A=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin45^0\)
\(A=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\frac{\sqrt{2}}{2}\)
\(A=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}\)
Câu b lm tương tự
Có
A=\(\left(sin^215^o+sin^275^o\right)+\left(sin^240^o+sin^250^o\right)+\left(sin^260^o+sin^230^o\right)\)
\(=\left(sin^215^o+cos^215^o\right)+...\)
\(=1\cdot3=3\)
Câu c tương tự mà mk nghĩ đề sai dấu - trước cos^245độ
Nói chung nếu: a+b=90 độ
thì: \(sin^2a+sin^2b=1\)
b) thì áp dụng nếu a+b=90 độ:
\(tana=cotb\) và ngược lại
Mà \(tana\cdot cota=1\)
Nói chung là công thức......
b) \(sin^23^o+sin^215^o+sin^275^o+sin^287^o\)
\(=\left(sin^23^o+cos^23^o\right)+\left(sin^215^o+cos^215^o\right)\)
\(=1+1=2\)
a) \(cos^212^o+cos^278^o+cos^21^o+cos^289^o\)
\(=\left(sin^278^o+cos^278^o\right)+\left(sin^289^o+cos^289^o\right)\)
\(=1+1=2\)