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10 tháng 5 2017

a) \(sin6\alpha cot3\alpha cos6\alpha=2.sin3\alpha.cos3\alpha\dfrac{cos3\alpha}{sin3\alpha}-cos6\alpha\)
\(=2cos^23\alpha-\left(2cos^23\alpha-1\right)=1\) (Không phụ thuộc vào x).

10 tháng 5 2017

b) \(\left[tan\left(90^o-\alpha\right)-cot\left(90^o+\alpha\right)\right]^2\)\(-\left[cot\left(180^o+\alpha\right)+cot\left(270^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+cot\left(90^o-\alpha\right)\right]^2\)\(-\left[cot\alpha+cot\left(90^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+tan\alpha\right]^2-\left[cot\alpha-tan\alpha\right]^2\)
\(=4tan\alpha cot\alpha=4\). (Không phụ thuộc vào \(\alpha\)).

10 tháng 5 2017

a) \(\dfrac{tan\alpha-tan\beta}{cot\beta-cot\alpha}=\dfrac{\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\beta}{cos\beta}}{\dfrac{cos\beta}{sin\beta}-\dfrac{cos\alpha}{sin\alpha}}\)
\(=\dfrac{\dfrac{sin\alpha cos\beta-cos\alpha sin\beta}{cos\alpha cos\beta}}{\dfrac{cos\beta sin\alpha-cos\alpha sin\beta}{sin\beta sin\alpha}}\)
\(=\dfrac{sin\beta sin\alpha}{cos\beta cos\alpha}=tan\alpha tan\beta\).

10 tháng 5 2017

b) \(tan100^o+\dfrac{sin530^o}{1+sin640^o}=tan100^o+\dfrac{sin170^o}{1+sin280^o}\)
\(=-cot10^o+\dfrac{sin10^o}{1-sin80^o}\)\(=\dfrac{-cos10^o}{sin10^o}+\dfrac{sin10^o}{1-cos10^o}\)
\(=\dfrac{-cos10^o+cos^210^o+sin^210^o}{sin10^o\left(1-cos10^o\right)}\) \(=\dfrac{1-cos10^o}{sin10^o\left(1-cos10^o\right)}=\dfrac{1}{sin10^o}\) .

11 tháng 5 2017

a) \(tan3\alpha-tan2\alpha-tan\alpha=\left(tan3\alpha-tan\alpha\right)-tan2\alpha\)
\(=\left(\dfrac{sin3\alpha}{cos3\alpha}-\dfrac{sin\alpha}{cos\alpha}\right)-\dfrac{sin2\alpha}{cos2\alpha}\)\(=\dfrac{sin3\alpha cos\alpha-cos3\alpha sin\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=sin2\alpha.\left(\dfrac{1}{cos3\alpha cos\alpha}-\dfrac{1}{cos2\alpha}\right)\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos3\alpha cos\alpha}{cos3\alpha cos\alpha cos2\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-\dfrac{1}{2}\left(cos4\alpha+cos2\alpha\right)}{cos3\alpha cos2\alpha cos\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos4\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=\dfrac{sin2\alpha.2sin3\alpha.sin\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=tan3\alpha tan2\alpha tan\alpha\) (Đpcm).

11 tháng 5 2017

b) \(\dfrac{4tan\alpha\left(1-tan^2\alpha\right)}{\left(1+tan^2\right)^2}=4tan\alpha\left(1-tan^2\alpha\right):\left(\dfrac{1}{cos^2\alpha}\right)^2\)
\(=4tan\alpha\left(1-tan^2\alpha\right)cos^4\alpha\)
\(=4\dfrac{sin\alpha}{cos\alpha}\left(1-\dfrac{sin^2\alpha}{cos^2\alpha}\right)cos^4\alpha\)
\(=4sin\alpha\left(cos^2\alpha-sin^2\alpha\right)cos\alpha\)
\(=4sin\alpha cos\alpha.cos2\alpha\)
\(=2.sin2\alpha.cos2\alpha=sin4\alpha\) (Đpcm).

25 tháng 7 2018

bài 1) ta có : \(G=cos\left(\alpha-5\pi\right)+sin\left(\dfrac{-3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)

\(G=cos\left(\alpha-\pi\right)+sin\left(\dfrac{\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\)

\(G=cos\left(\pi-\alpha\right)+sin\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)-tan\left(\pi+\alpha-\dfrac{\pi}{2}\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\) \(G=cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\dfrac{\pi}{2}-\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)=2cos\alpha+1\) bài 2) ta có : \(H=cot\left(\alpha\right).cos\left(\alpha+\dfrac{\pi}{2}\right)+cos\left(\alpha\right)-2sin\left(\alpha-\pi\right)\) \(H=cot\left(\alpha\right).cos\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)+cos\left(\alpha\right)+2sin\left(\pi-\alpha\right)\) \(H=-cot\left(\alpha\right).sin\left(\alpha\right)+cos\left(\alpha\right)+2sin\left(\alpha\right)\) \(H=-cos\alpha+cos\alpha+2sin\alpha=2sin\alpha\)

11 tháng 5 2017

a)\(sin^2\left(180^o-\alpha\right)+tan^2\left(180-\alpha\right).tan^2\left(270^o+\alpha\right)\)\(+sin\left(90^o+\alpha\right)cos\left(\alpha-360^o\right)\)
\(=sin^2\alpha+tan^2\alpha.cot^2\alpha+cos\alpha cos\alpha\)
\(=sin^2\alpha+cos^2\alpha+\left(tan\alpha cot\alpha\right)^2=1+1=2\).

11 tháng 5 2017

\(\dfrac{cos\left(\alpha-180^o\right)}{sin\left(180^o-\alpha\right)}+\dfrac{tan\left(\alpha-180^o\right)cos\left(180^o+\alpha\right)sin\left(270^o+\alpha\right)}{tan\left(270^o+\alpha\right)}\)
\(=\dfrac{cos\left(180^o-\alpha\right)}{sin\left(180^o-\alpha\right)}+\dfrac{-tan\left(180^o-\alpha\right).cos\alpha.sin\left(90^o+\alpha\right)}{-tan\left(90^o+\alpha\right)}\)
\(=tan\left(180^o-\alpha\right)+\dfrac{tan\alpha.cos\alpha.cos\alpha}{cot\alpha}\)
\(=-tan\alpha+tan^2\alpha cos^2\alpha\)
\(=tan\alpha\left(-1+tan\alpha cos^2\alpha\right)\)
\(=tan\alpha\left(sin\alpha cos\alpha-1\right)\).

AH
Akai Haruma
Giáo viên
1 tháng 10 2018

a)

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

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

\(=2\sin ^2a\)

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

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

\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)

c)

\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)

\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)

\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)

AH
Akai Haruma
Giáo viên
1 tháng 10 2018

d)

\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)

\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)

f)

\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)

\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)

\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)

11 tháng 5 2017

a)\(sin\left(\alpha+\dfrac{\pi}{2}\right)=cos\left[\dfrac{\pi}{2}-\left(\alpha+\dfrac{\pi}{2}\right)\right]=cos\left(-\alpha\right)=cos\alpha\).
b) \(cos\left(x+\dfrac{\pi}{2}\right)=sin\left[\dfrac{\pi}{2}-\left(x+\dfrac{\pi}{2}\right)\right]=sin\left(-x\right)=-sinx\).
c) \(tan\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{sin\left(\alpha+\dfrac{\pi}{2}\right)}{cos\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{cos\alpha}{-sin\alpha}=-cot\alpha\).
d) \(cot\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{cos\left(\alpha+\dfrac{\pi}{2}\right)}{sin\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{-sin\alpha}{cos\alpha}=-tan\alpha\).

AH
Akai Haruma
Giáo viên
21 tháng 2 2019

Bạn xem lại biểu thức A. Biểu thức $A$ sau khi rút gọn thì \(A=\frac{-2\sin ^2a}{3\cos 2a}\) vẫn phụ thuộc vào $a$

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Sử dụng công thức: \(\sin (90-a)=\cos a; \cot (90-a)=\tan a\), ta có:

\(B=\tan ^260(\sin ^8a-\cos ^8a)+4\cos 60(\cos ^6a-\sin ^6a)-\cos ^6a(\tan ^2a-1)^3\)

\(=3(\sin ^8a-\cos ^8a)+2(\cos ^6a-\sin ^6a)-\cos ^6a\left(\frac{\sin ^2a}{\cos ^2a}-1\right)^3\)

\(=3(\sin ^8a-\cos ^8a)+2(\cos ^6a-\sin ^6a)-(\sin ^2a-\cos ^2a)^3\)

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

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

\(=(\sin ^2a-\cos ^2a)[3(\sin ^4a+\cos ^4a)-2(\cos ^4a+\sin ^2a\cos ^2a+\sin ^4a)-(\sin ^2a-\cos ^2a)^2]\)

\(=(\sin ^2a-\cos ^2a).0=0\). Do đó giá trị của biểu thức không phụ thuộc vào $a$

8 tháng 7 2023

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