Chứng minh các đẳng thức sau
a) 2( sin^6 a + cos^6 a)+1=3(sin^4 a +cos^4 a)
b) tan a – tan b / cot a – cot b = tan a tan b
Chứng minh các đẳng thức sau
a) 2( sin^6 a + cos^6 a)+1=3(sin^4 a +cos^4 a)
b) tan a – tan b / cot a – cot b = tan a tan b
a,
$VT= 2(\sin^6a+\cos^6a)+1$
$=2(\sin^a+\cos^2a)(\sin^4a-\sin^2a.\cos^2a+\cos^4a)+1$
$= 2\sin^4a-2\sin^2a\cos^2a+2\cos^4a+1$
$= 2\sin^4a+2\cos^4a+(1-2\sin^2a\cos^2a)$
$VP=3(\sin^4a+\cos^4a)$
$=2\sin^4a+2\cos^4a+\sin^4a+\cos^4a$
$= 2\sin^4a+2\cos^4a+(\sin^2a+\cos^2a)-2\sin^2a.\cos^2a$
$= 2\sin^4a+2\cos^4a+1-2\sin^2a\cos^2a= VT$ (đpcm)
b,
$VT=\dfrac{\tan a-\tan b}{\cot a-\cot b}$
$=\dfrac{\frac{\sin a.\cos b -\cos a.\sin b}{\cos a.\cos b}}{\frac{\cos a.\sin b-\sin a.\cos b}{\sin a.\sin b}}$
$=\dfrac{\sin(a-b)}{\cos a.\cos b} : \dfrac{\sin(b-a)}{\sin a.\sin b}$
$=\dfrac{\sin(a-b)}{\cos a.\cos b}. \dfrac{\sin a.\sin b}{-\sin(a-b)}$
$=-\tan a.\tan b$
Giải thích các bước giải:
$a)VT=2(\sin^6a+\cos^6a)+1\\
=2\left [(\sin^2+\cos^2)^3-3\sin^4x\cos^2-3\sin^2x\cos^4x \right ]+1\\
=2\left [1-3\sin^2x\cos^2(\sin^2x+\cos^2x) \right ]+1\\
=2-6\sin^2x\cos^2+1\\
=3-6\sin^2x\cos^2\\
=3(1-2\sin^2x\cos^2)\\
=3(\sin^4x+\cos^4x)=VP\Rightarrow ĐPCM$
$b)VT=\dfrac{\tan a-\tan b}{\cot a-\cot b}\\
=\dfrac{\dfrac{\sin a}{\cos a}-\dfrac{\sin b}{\cos b}}{\dfrac{\cos a}{\sin a}-\dfrac{\cos b}{\sin b}}\\
=\dfrac{\dfrac{\sin a\cos b}{\cos a\cos b}-\dfrac{\sin b\cos a}{\cos a\cos b}}{\dfrac{\cos a\sin b}{\sin a\sin b}-\dfrac{\sin a\cos b}{\sin a\sin b}}\\
=\dfrac{\sin a\sin b.(\sin a\cos b-\sin b\cos a)}{\cos a\cos b.\left (\cos a\sin b-\sin a\cos b \right )}\\
=\dfrac{\sin a\sin b.\sin (a-b)}{\cos a\cos b.\sin (b-a)}\\
=-\tan a\tan b=VP\Rightarrow đpcm$