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DQ
31 tháng 7 2018
bài 1
a) \(M=\sin^242^o+\sin^243^o+\sin^244^o+\sin^245^o+\sin^246^o+\sin^247^o+\sin^248^o\)
\(M=\cos^248^o+\cos^247^o+\cos^246^o+\sin^245^o+\sin^246^o+\sin^247^o+\sin^248^o\)
\(M=\left(\sin^248^o+\cos^248^o\right)+\left(\sin^247^o+\cos^247^o\right)+\left(\sin^246^o+\cos^246^o\right)+\sin^245^o\)
\(M=1+1+1+0,5\)
\(M=3,5\)
DQ
31 tháng 7 2018
bài 1
b) \(N=\cos^215^o-\cos^225^o+\cos^235^o-\cos^245^o+\cos^255^o-\cos^265^o+\cos^275^o\)
\(N=\sin^275^o-\sin^265^o+\sin^255^o-\cos^245^o+\cos^255^o-\cos^265^o+\cos^275^o\)
\(N=\left(\sin^275^o+\cos^275^o\right)-\left(\sin^265^o+\cos^265^o\right)+\left(\sin^255^o+\cos^255^o\right)-\cos^245^o\)
\(N=1-1+1-0,5\)
\(N=0,5\)
DT
1
7 tháng 8 2017
\(tan^2x-sin^2x=\frac{sin^2x}{cos^2x}-sin^2x\)
\(=sin^2x.\left(\frac{1}{cos^2x}-1\right)=sin^2x.\frac{sin^2x}{cos^2x}=tan^2x.sin^2x\)
DD
Đoàn Đức Hà
Giáo viên
22 tháng 6 2021
a) \(cos^4x-sin^4x=\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=cos^2x-sin^2x\)
b) \(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{tanxcotx}{tanxcotx+cotx}=\frac{1}{1+tanx}+\frac{tanx}{tanx+1}\)
\(=\frac{1+tanx}{1+tanx}=1\)
c) Ta có: \(1+tan^2x=1+\frac{sin^2x}{cos^2x}=\frac{cos^2x+sin^2x}{cos^2x}=\frac{1}{cos^2x}\)
\(\Rightarrow\frac{1}{1+tan^2x}=cos^2x\)
Tương tự \(\frac{1}{1+tan^2y}=cos^2y\)
\(\Rightarrow cos^2x-cos^2y=\frac{1}{1+tan^2x}-\frac{1}{1+tan^2y}\)
\(cos^2x-cos^2y=\left(1-sin^2x\right)-\left(1-sin^2y\right)=sin^2y-sin^2x\)
d) \(\frac{1+sin^2x}{1-sin^2x}=\frac{cos^2x+sin^2x+sin^2x}{cos^2x+sin^2x-sin^2x}=\frac{cos^2x+2sin^2x}{cos^2x}=1+2\left(\frac{sinx}{cosx}\right)^2=1+2tan^2x\)
29 tháng 10 2018
a) 1- \(sin^2\alpha\)= \(cos^2\alpha\)
b) (\(1-cos\alpha\))(\(1+cos\alpha\)) = 1 - cos2\(\alpha\) = sin2\(\alpha\)
c) 1 + cos2\(\alpha\) + sin2\(\alpha\) = \(1+1=2\)
d) sin\(\alpha\) - sin\(\alpha.cos^2\alpha\)
= \(sin\alpha\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)
e) \(sin^4\alpha+cos^4\alpha+2sin^2\alpha.cos^2\alpha\)
= \(\left(sin^2\alpha\right)^2+2sin^2\alpha.cos^2\alpha+\left(cos^2\alpha\right)^2\)
= \(\left(sin^2\alpha+cos^2\alpha\right)^2=1^2=1\)
f) \(tan^2\alpha-sin^2\alpha.tan^2\alpha\)
= \(tan^2\alpha\left(1-sin^2\alpha\right)=tan^2\alpha.cos^2\alpha=sin^2\alpha\)
g) \(cos^2\alpha+tan^2\alpha.cos^2\alpha\)
= \(cos^2\alpha\left(1+tan^2\alpha\right)=cos^2\alpha.\dfrac{1}{cos^2\alpha}=1\)
h) \(tan^2\alpha\left(2cos^2\alpha+sin^2\alpha-1\right)\)
= \(tan^2\alpha\left[cos^2\alpha+\left(cos^2\alpha+sin^2\alpha\right)-1\right]\)
= \(tan^2\alpha\left(cos^2\alpha+1-1\right)\)
= \(tan^2\alpha.cos^2\alpha=sin^2\alpha\)
21 tháng 10 2019
Ta có : \(\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}\)
\(\Leftrightarrow\tan^2\left(x\right)=\left(\frac{\sin\left(x\right)}{\cos\left(x\right)}\right)^2\)
\(\Leftrightarrow\tan^2\left(x\right)=\frac{sin^2\left(x\right)}{\cos^2\left(x\right)}\)
Và ta có : \(\cos^2\left(x\right)+\sin^2\left(x\right)=1\)
\(\Leftrightarrow\cos^2\left(x\right)=1-\sin^2\left(x\right)\)
VT: \(\tan^2\left(x\right)-\sin^2\left(x\right)\cdot\tan^2\left(x\right)\)
\(=\tan^2\left(x\right)\cdot\left(1-\sin^2\left(x\right)\right)\)
\(=\frac{\sin^2\left(x\right)}{\cos^2\left(x\right)}\cdot\cos^2\left(x\right)\)
\(=\sin^2\left(x\right)=VP\)(đpcm)
(chúc bạn học tốt
)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 6 2019
\(A=\frac{1-2sina.cosa}{sin^2a-cos^2a}=\frac{sin^2a+cos^2a-2sina.cosa}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{\left(sina-cosa\right)^2}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{sina-cosa}{sina+cosa}\)
b/ \(A=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}=\frac{\frac{1}{3}-1}{\frac{1}{3}+1}=-\frac{1}{2}\)
\(\tan^2\alpha-\sin^2\alpha.\tan^2\alpha=\tan^2\alpha\left(1-\sin^2\alpha\right)=\tan^2\alpha.\cos^2\alpha=\sin^2\alpha\)