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Phan Lê Kim Chi
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alibaba nguyễn
27 tháng 8 2021 lúc 16:34

a/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)

\(=\frac{\frac{cos^2a}{sin^2a}-cos^2a}{\frac{cos^2a}{sin^2a}}-\frac{sina.cosa}{\frac{cosa}{sina}}\)

\(=\left(1-sin^2a\right)-sin^2a=1\)

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alibaba nguyễn
27 tháng 8 2021 lúc 16:38

b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)

\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)

\(=2+\left(cos^2a-sin^2a\right)-2cos^2a\)

\(=2-sin^2a-cos^2a=2-1=1\)

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alibaba nguyễn
27 tháng 8 2021 lúc 16:41

c/ \(C=sin^6x+cos^6x+3sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3sin^2x.cos^2x\)

\(=sin^4x-sin^2x.cos^2x+cos^4x+3sin^2x.cos^2x\)

\(=sin^4x+cos^4x+2sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2=1\)

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Sách Giáo Khoa
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Bùi Thị Vân
18 tháng 5 2017 lúc 11:18

a)
\(A=\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=1+2sin\alpha cos\alpha+1-2sin\alpha cos\alpha=2\) (không phụ thuộc vào \(\alpha\)).
b)
\(B=sin^4\alpha-cos^4\alpha-2sin^2\alpha+1\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-cos^2\alpha\right)-2sin^2\alpha+1\)
\(=sin^2\alpha-cos^2\alpha-2sin^2\alpha+1\)
\(=-sin^2\alpha-cos^2\alpha+1\)
\(=-\left(sin^2\alpha+cos^2\alpha\right)+1=-1+1=0\).

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Min YoongMin
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Thảo Phạm
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Trần Văn Tú
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le thi khanh huyen
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Hoàng Đức
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An Thy
30 tháng 7 2021 lúc 10:24

\(\left(\sqrt{\dfrac{1+sin\alpha}{1-sin\alpha}}+\sqrt{\dfrac{1-sin\alpha}{1+sin\alpha}}\right).\dfrac{1}{\sqrt{1+tan^2\alpha}}\)

\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{\left(1+sin\alpha\right)\left(1-sin\alpha\right)}}\right).\dfrac{1}{\sqrt{1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2}}\)

\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{1-sin^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{1-sin^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}}}\)

\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{cos^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{cos^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{1}{cos^2\alpha}}}\)

\(=\left(\dfrac{1+sin\alpha}{cos\alpha}+\dfrac{1-sin\alpha}{cos\alpha}\right).\dfrac{1}{\dfrac{1}{cos\alpha}}=\dfrac{2}{cos\alpha}.cos\alpha=2\)

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Sách Giáo Khoa
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Bùi Thị Vân
11 tháng 5 2017 lúc 9:41

a) \(A=2\left(sin^6\alpha+cos^6\alpha\right)-3\left(sin^4\alpha+cos^4\alpha\right)\)
\(=2\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha-sin^2\alpha cos^2\alpha+cos^4\alpha\right)\)\(-3\left(sin^4\alpha+cos^4\alpha\right)\)
\(=2\left(sin^4\alpha+cos^4\alpha-sin^2\alpha cos^2\alpha\right)-3\left(sin^4\alpha+cos^4\alpha\right)\)
\(=-\left(sin^4\alpha+cos^4\alpha+2sin^2\alpha cos^2\alpha\right)\)
\(=-\left(sin^2\alpha+cos^2\alpha\right)^2=-1\) (Không phụ thuộc vào \(\alpha\)).

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Bùi Thị Vân
11 tháng 5 2017 lúc 9:47

b) \(B=4\left(sin^4\alpha+cos^4\alpha\right)-cos4\alpha\)
\(=4\left(sin^4\alpha+cos^4\alpha+2sin^2\alpha cos^2\alpha\right)-8sin^2\alpha cos^2\alpha\)\(-\left(1-2sin^22\alpha\right)\)
\(=4.\left(sin^2\alpha+cos^2\alpha\right)^2-2sin^22\alpha-1+2sin^22\alpha\)
\(=4-1=3\).

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Bùi Thị Vân
11 tháng 5 2017 lúc 10:07

c) \(8\left(cos^8\alpha-sin^8\alpha\right)-cos6\alpha-7cos2\alpha\)
\(=8\left(cos^4\alpha-sin^4\alpha\right)\left(sin^4\alpha+cos^4\alpha\right)-cos6\alpha-7cos2\alpha\)
\(=8\left(cos^2\alpha-sin^2\alpha\right)\left(sin^4\alpha+cos^4\alpha\right)-cos6\alpha-7cos2\alpha\)
\(=8cos2\alpha\left(sin^4\alpha+cos^4\alpha\right)-cos6\alpha-7cos2\alpha\)
\(=8cos2\alpha\left(sin^4\alpha+cos^4\alpha\right)-8cos2\alpha+cos2\alpha-cos6\alpha\)
\(=8cos2\alpha\left(sin^4\alpha+cos^4-1\right)+sin4\alpha sin2\alpha\)
\(=8cos2\alpha\left[\left(sin^4\alpha-sin^2\alpha\right)+\left(cos^4\alpha-cos^2\alpha\right)\right]+\)\(sin4\alpha sin2\alpha\)
\(=8cos2\alpha.\left[sin^2\alpha\left(sin^2\alpha-1\right)+cos^2\alpha\left(cos^2\alpha-1\right)\right]\)\(+sin4\alpha sin2\alpha\)
\(=8.cos2\alpha.\left(-2sin^2\alpha cos^2\alpha\right)+2sin2\alpha cos2\alpha sin2\alpha\)
\(=-2cos2\alpha.\left(sin2\alpha\right)^2+2cos2\alpha.\left(sin2\alpha\right)^2\)
\(=sin^22\alpha\left(-cos2\alpha+cos2\alpha\right)=sin^22\alpha.0=0\) (không phụ thuộc vào \(\alpha\)).

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Nguyễn Ngọc
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Nguyễn Hoàng Tiến
8 tháng 7 2016 lúc 13:56

Có: \(\sin^2+\cos^2=1\)

=> \(\sin^2=1-\cos^2\)

Ta có:

\(\cos^4a+\sin^2a\cos^2a+\sin^2a=\cos^4a+\left(1-\cos^2\right)a\cos^2a+\sin^2\)

\(=\cos^4a-\cos^4a+\cos^2a+\sin^2a=\cos^2a+\sin^2a=1\)

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