Chứng Minh
\(tan^2a+cot^2a=\dfrac{2\left(3+cos4a\right)}{1-cos4a}\)
Những câu hỏi liên quan
Cho \(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\).
Tính cos4a
\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)
=>\(\dfrac{sin^2a+1}{cos^2a}+\dfrac{cos^2a+1}{sin^2a}=7\)
=>\(\dfrac{sin^4a+sin^2a+cos^4a+cos^2a}{sin^2a\cdot cos^2a}=7\)
=>\(sin^4a+cos^4a+1=7\cdot sin^2a\cdot cos^2a\)
=>\(\left(sin^2a+cos^2a\right)^2-2\cdot sin^2a\cdot cos^2a+1=7\cdot sin^2a\cdot cos^2a\)
=>\(2=9\cdot sin^2a\cdot cos^2a\)
=>\(8=9\cdot sin^22a\)
=>16=9(1-cos4a)
=>1-cos4a=16/9
=>cos4a=-7/9
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Chứng minh đẳng thức sau :
\(\frac{6+2cos4a}{1-cos4a}=tan^2a+cot^2a\)
Lời giải:
Áp dụng công thức: $\cos 2x=\cos ^2x-\sin ^2x=1-2\sin ^2x=2\cos ^2x-1$ ta có:
\(\frac{6+2\cos 4a}{1-\cos 4a}=\frac{6+2(2\cos ^22a-1)}{2\sin ^22a}=\frac{2+2\cos ^22a}{\sin ^22a}=\frac{2+2(\cos ^2a-\sin ^2a)^2}{4\sin ^2a\cos ^2a}\)
\(=\frac{1+(\sin ^2a-\cos ^2a)^2}{2\sin ^2a\cos ^2a}=\frac{(\sin ^2a+\cos ^2a)^2+(\sin ^2a-\cos ^2a)^2}{2\sin ^2a\cos ^2a}=\frac{2(\sin ^4a+\cos ^4a)}{2\sin ^2a\cos ^2a}=\frac{\sin ^4a+\cos ^4a}{\sin ^2a\cos ^2a}\)
\(=\frac{\sin ^2a}{\cos ^2a}+\frac{\cos ^2a}{\sin ^2a}=\tan ^2a+\cot ^2a\) (đpcm)
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Chứng minh rằng: cot2a + tan2a= \(\frac{2\cos4a+6}{1-\cos4a}\)
\(cot^2a+tan^2a=\frac{cos^2a}{sin^2a}+\frac{sin^2a}{cos^2a}=\frac{cos^4a+sin^4a}{sin^2a.cos^2a}=\frac{8\left(\frac{1+cos2a}{2}\right)^2+8\left(\frac{1-cos2a}{2}\right)^2}{2\left(2sina.cosa\right)^2}\)
\(=\frac{2+4cos2a+2cos^22a+2-4cos2a+2cos^22a}{2sin^22a}=\frac{4+4cos^22a}{2sin^22a}\)
\(=\frac{4+4\left(\frac{1+cos4a}{2}\right)}{2\left(\frac{1-cos4a}{2}\right)}=\frac{6+2cos4a}{1-cos4a}\)
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Biến đổi thành tích :
a) \(1+\cos\left(\dfrac{\pi}{2}+3\alpha\right)-\sin\left(\dfrac{3}{2}\pi-3\alpha\right)+\cot\left(\dfrac{5}{2}\pi+3\alpha\right)\)
b) \(\dfrac{\cos7\alpha-\cos8\alpha-\cos9\alpha+\cos10\alpha}{\sin7\alpha-\sin8\alpha-\sin9\alpha+\sin10\alpha}\)
c) \(-\cos5a.\cos4a-\cos4a\cos3a+2\cos^22a\cos a\)
Rút gọn các biểu thức sau :
a)\(\dfrac{1+\sin4a-\cos4a}{1+\cos4a+\sin4a}\)
b) \(\dfrac{1+\cos a}{1-\cos a}\tan^2\dfrac{a}{2}-\cos^2a\)
c) \(\dfrac{\cos2x-\sin4x-\cos6x}{\cos2x+\sin4x-\cos6x}\)
Chứng minh
\(\frac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)=2cosx
\(\frac{cos4a\cdot tan2a-sin4a}{cos4a\cdot cot2a+sin4a}\)=-tan22a
Giúp mình vs! Mình đang cần gấp :((
CHỨNG MINH:
\(\frac{1+\cos4a}{\cot a-\tan a}=\frac{1}{4}\sin4a\)\(\frac{\cot^22a-1}{2\cot a}-\cos8a.\cot4a=\sin8a\)
chứng minh rằng
a) \(cos^4a+sin^4a-6sin^2a.cos^2a=cos4a\)
b) \(tan\frac{3\pi}{5}-tan\frac{2\pi}{5}-tan\frac{\pi}{5}=tan\frac{\pi}{5}.tan\frac{2\pi}{5}.tan\frac{3\pi}{5}\)
\(cos^4a+sin^4a-6sin^2a.cos^2a\)
\(=cos^4a+sin^4a-2sin^2a.cos^2a-4sin^2a.cos^2a\)
\(=\left(cos^2a-sin^2a\right)^2-\left(2sina.cosa\right)^2\)
\(=cos^22a-sin^22a\)
\(=cos4a\)
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Chứng minh các hệ thức sau :
a) sinalpha+sinleft(alpha+dfrac{14}{3}piright)+sinleft(alpha-dfrac{8}{3}piright)0
b) dfrac{sin4a}{1+cos4a}.dfrac{cos2a}{1+cos2a}cotleft(dfrac{3}{2}pi-aright)
c) left(cos a-cos bright)^2-left(sin a-sin bright)^2-4sin^2dfrac{a-b}{2}cosleft(a+bright)
d) sin^2left(45^0+alpharight)-sin^2left(30^0-alpharight)-sin15^0cosleft(15^0+2alpharight)sin2alpha
Đọc tiếp
Chứng minh các hệ thức sau :
a) \(\sin\alpha+\sin\left(\alpha+\dfrac{14}{3}\pi\right)+\sin\left(\alpha-\dfrac{8}{3}\pi\right)=0\)
b) \(\dfrac{\sin4a}{1+\cos4a}.\dfrac{\cos2a}{1+\cos2a}=\cot\left(\dfrac{3}{2}\pi-a\right)\)
c) \(\left(\cos a-\cos b\right)^2-\left(\sin a-\sin b\right)^2=-4\sin^2\dfrac{a-b}{2}\cos\left(a+b\right)\)
d) \(\sin^2\left(45^0+\alpha\right)-\sin^2\left(30^0-\alpha\right)-\sin15^0\cos\left(15^0+2\alpha\right)=\sin2\alpha\)