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\(ADCT:\sin^2\alpha+\cos^2\alpha=1\)
\(A=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin45^0\)
\(A=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\frac{\sqrt{2}}{2}\)
\(A=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}\)
Câu b lm tương tự
Lời giải:
Ta biết rằng $\sin a=\cos (90-a)$ và $\sin ^2a+\cos ^2a=1$
Do đó:
\(A=\sin ^242+\sin ^243+....+\sin ^248=(\sin ^242+\sin ^248)+(\sin ^243+\sin ^247)+(\sin ^244+\sin ^246)+\sin ^245\)
\(=(\sin ^242+\cos ^242)+(\sin ^243+\cos ^243)+(\sin ^244+\cos ^244)+\sin ^245\)
\(=1+1+1+(\frac{\sqrt{2}}{2})^2=\frac{7}{2}\)
Ta có: \(A=\sin^25^0+\sin^225^0+\sin^245^0+\sin^265^0+\sin^285^0\)
\(=\left(\sin^25^0+\sin^285^0\right)+\left(\sin^225^0+\sin^265^0\right)+\dfrac{1}{2}\)
\(=2+\dfrac{1}{2}=\dfrac{5}{2}\)
\(\Rightarrow A=\left(sin^25^0+sin^285^0\right)+\left(sin^225^0+sin^265^0\right)+sin^245^0=\left(sin^25^0+cos^25^0\right)+\left(sin^225^0+cos^225^0\right)+\dfrac{1}{2}=1+1+\dfrac{1}{2}=\dfrac{5}{2}\)
Ta có B = sin245o + sin262o + sin227o - (sin247o = sin248o)
sin227o = cos263o
mà cos263o < cos262o
=> sin262o + cos263o < sin262o + cos262o
hay sin262o + sin227o <1 (1)
sin248o = cos242o
mà cos242o > cos247o
=> sin247o + cos242o > sin247o + cos247o
hay sin247o + sin248o > 1
=> - (sin247o + sin248o) <1 (2)
Từ (1) và (2) ta thấy:
sin262o + sin227o - (sin247o = sin248o) < 1
sin245o = 1/2 <1
=> B = sin245o + sin262o + sin227o - (sin247o = sin248o) <1
=> B < A
cái chỗ (sin247o = sin248o) thay thành (sin247o + sin248o) nha ^_^
Ta có: \(\cos33^o=\sin57^o\)
Và \(\sin^244^o=\cos^246^o\)
Thay vào A, ta có;
\(A=\sin57^o-\sin57^o+\cos^246^o+\sin^246^o\)
A=1
`sin^2 25^o + sin^2 65^o`
`=cos^2 65^o + sin^2 65^o`
=1`
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`***` Áp dụng công thức lượng giác: `sin^2 \alpha +cos^2 \alpha =1`
\(A=\left(\sin^25^0+\sin^285^0\right)+\left(\sin^225^0+\sin65^0\right)+\sin^245^0\)
\(=\left(\sin^25^0+\cos^25^0\right)+\left(\sin^225^0+\cos^225^0\right)+\frac{1}{2}\)
\(=1+1+\frac{1}{2}\)
\(=\frac{5}{2}\)
\(B=\left(\tan1^0.\tan89^0\right).\left(\tan2^0.\tan88^0\right).\left(\tan3^0.\tan87^0\right)...\tan45^0=\left(\tan1^0.\cot1^0\right).\left(\tan2^0.\cot2^0\right).\left(\tan3^0.\cot3^0\right)...1=1\)
Ta có \(\sin x=\cos\left(90^0-x\right)\)
\(\Rightarrow M=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin^245^0\)
\(=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\sin^245^0\)
\(=1+1+1+\left(\frac{\sqrt{2}}{2}\right)^2=3+\frac{1}{2}=\frac{7}{2}\)