Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{cos^2x-sin^2y}{sin^2x.sin^2y}-\frac{cos^2x.cos^2y}{sin^2x.sin^2y}=\frac{cos^2x-sin^2y-cos^2x.cos^2y}{sin^2x.sin^2y}=\frac{cos^2x\left(1-cos^2y\right)-sin^2y}{sin^2x.sin^2y}\)
\(=\frac{cos^2x.sin^2y-sin^2y}{sin^2x.sin^2y}=\frac{-sin^2y\left(1-cos^2x\right)}{sin^2x.sin^2y}=\frac{-sin^2x.sin^2y}{sin^2x.sin^2y}=-1\)
a) \(A=sin\left(\dfrac{\pi}{4}+x\right)-cos\left(\dfrac{\pi}{4}-x\right)\)
\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-\left(cos\dfrac{\pi}{4}.cosx+sin\dfrac{\pi}{4}.sinx\right)\)
\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-cos\dfrac{\pi}{4}.cosx-sin\dfrac{\pi}{4}.sinx\)
\(\Leftrightarrow A=\dfrac{\sqrt{2}}{2}.cosx+\dfrac{\sqrt{2}}{2}.sinx-\dfrac{\sqrt{2}}{2}.cosx-\dfrac{\sqrt{2}}{2}.sinx\)
\(\Leftrightarrow A=0\)
b) \(B=cos\left(\dfrac{\pi}{6}-x\right)-sin\left(\dfrac{\pi}{3}+x\right)\)
\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-\left(sin\dfrac{\pi}{3}.cosx+cos\dfrac{\pi}{3}.sinx\right)\)
\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-sin\dfrac{\pi}{3}.cosx-cos\dfrac{\pi}{3}.sinx\)
\(\Leftrightarrow B=\dfrac{\sqrt{3}}{2}.cosx+\dfrac{1}{2}.sinx-\dfrac{\sqrt{3}}{2}.cosx-\dfrac{1}{2}.sinx\)
\(\Leftrightarrow B=0\)
c) \(C=sin^2x+cos\left(\dfrac{\pi}{3}-x\right).cos\left(\dfrac{\pi}{3}+x\right)\)
\(\Leftrightarrow C=sin^2x+\left(cos\dfrac{\pi}{3}.cosx+sin\dfrac{\pi}{3}.sinx\right).\left(cos\dfrac{\pi}{3}.cosx-sin\dfrac{\pi}{3}.sinx\right)\)
\(\Leftrightarrow C=sin^2x+\left(\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right).\left(\dfrac{1}{2}.cosx-\dfrac{\sqrt{3}}{2}.sinx\right)\)
\(\Leftrightarrow C=sin^2x+\dfrac{1}{4}.cos^2x-\dfrac{3}{4}.sin^2x\)
\(\Leftrightarrow C=\dfrac{1}{4}.sin^2x+\dfrac{1}{4}.cos^2x\)
\(\Leftrightarrow C=\dfrac{1}{4}\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow C=\dfrac{1}{4}\)
d) \(D=\dfrac{1-cos2x+sin2x}{1+cos2x+sin2x}.cotx\)
\(\Leftrightarrow D=\dfrac{1-\left(1-2sin^2x\right)+2sinx.cosx}{1+2cos^2a-1+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sin^2x+2sinx.cosx}{2cos^2x+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sinx\left(sinx+cosx\right)}{2cosx\left(cosx+sinx\right)}.cotx\)
\(\Leftrightarrow D=\dfrac{sinx}{cosx}.cotx\)
\(\Leftrightarrow D=tanx.cotx\)
\(\Leftrightarrow D=1\)
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)
\(B=sin^4\alpha-cos^4\alpha-2sin^2\alpha+1\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-cos^2\alpha\right)-2sin^2\alpha+1\)
\(=sin^2\alpha-cos^2\alpha-2sin^2\alpha+1\)
\(=-sin^2\alpha-cos^2\alpha+1\)
\(=-\left(sin^2\alpha+cos^2\alpha\right)+1=-1+1=0\).
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 $a$
------------
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 $a$
Sửa đề: sin^4x+cos^4x+1
\(A=\dfrac{\left(sin^2x+cos^2x\right)^3-3sin^2xcos^2x+2}{\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x+1}\)
\(=\dfrac{3\left(1-sin^2xcos^2x\right)}{2\left(1-sin^2xcos^2x\right)}=\dfrac{3}{2}\)
\(A=2(\sin ^6x+\cos ^6x)-3(\sin ^4x+\cos ^4x)\)
\(=2(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-3(\sin ^4x+\cos ^4x)\)
\(=2(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-3(\sin ^4x+\cos ^4x)\)
\(=-(\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x)=-(\sin ^2x+\cos ^2x)^2=-1^2=-1\)
là giá trị không phụ thuộc vào biến (đpcm)
-----------------------
\(B=\sin ^6x+\cos ^6x-2\sin ^4x-\cos ^4x+\sin ^2x\)
\(=(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)-2\sin ^4x-\cos ^4x+\sin ^2x\)
\(=\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x-2\sin ^4x-\cos ^4x+\sin ^2x\)
\(=-\sin ^4x-\sin ^2x\cos ^2x+\sin ^2x=-\sin ^2x(\sin ^2x+\cos ^2x)+\sin ^2x\)
\(=-\sin ^2x+\sin ^2x=0\)
là giá trị không phụ thuộc vào biến (đpcm)
\(C=(\sin ^4x+\cos ^4x-1)(\tan ^2x+\cot ^2x+2)=(\sin ^4x+\cos ^4x-1)(\frac{\sin ^2x}{\cos ^2x}+\frac{\cos ^2x}{\sin ^2x}+2)\)
\(=(\sin ^4x+\cos ^4x-1).\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin ^2x\cos ^2x}=(\sin ^4x+\cos ^4x-1).\frac{(\sin ^2x+\cos ^2x)^2}{\sin ^2x\cos ^2x}\)
\(=(\sin ^4x+\cos ^4x-1).\frac{1}{\sin ^2x\cos ^2x}=\frac{(\sin ^2x)^2+(\cos ^2x)^2+2\sin ^2x\cos ^2x-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}\)
\(=\frac{(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}=\frac{1-2\sin ^2x\cos ^2x-1}{\sin ^2x\cos ^2x}=\frac{-2\sin ^2x\cos ^2x}{\sin ^2x\cos ^2x}=-2\)
là giá trị không phụ thuộc vào biến $x$
--------------------
\(D=\frac{1}{\cos ^6x}-\tan ^6x-\frac{\tan ^2x}{\cos ^2x}=\frac{1}{\cos ^6x}-\frac{\sin ^6x}{\cos ^6x}-\frac{\sin ^2x}{\cos ^4x}\)
\(=\frac{1-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{(\sin ^2x+\cos ^2x)^3-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)
\(=\frac{\sin ^6x+\cos ^6x+3\sin ^2x\cos ^2x(\sin ^2x+\cos ^2x)-\sin ^6x-\sin ^2x\cos ^2x}{\cos ^6x}\)
\(=\frac{\cos ^6x+3\sin ^2x\cos ^2x-\sin ^2x\cos ^2x}{\cos ^6x}=\frac{\cos ^4x+2\sin ^2x}{\cos ^4x}\)
\(=1+\frac{2\sin ^2x}{\cos ^4x}\)
Giá trị biểu thức này vẫn phụ thuộc vào $x$. Bạn xem lại đề.
Giả sử các biểu thức đều có nghĩa
\(A=2\left(\left(sin^2x\right)^3+\left(cos^2x\right)^3\right)-3\left(sin^4x+cos^4x+2sin^2xcos^2x-2sin^2xcos^2x\right)\)
\(A=2\left(sin^2x+cos^2x\right)\left(\left(sin^2x+cos^2x\right)^2-3sin^2xcos^2x\right)-3\left(\left(sin^2x+cos^2x\right)^2-2sin^2xcos^2x\right)\)
\(A=2\left(1-3sin^2xcos^2x\right)-3\left(1-2sin^2xcos^2x\right)\)
\(A=2-6sin^2xcos^2x-3+6sin^2xcos^2x=-1\)
b/ \(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{tanx-1}=\dfrac{1+cotx}{1-cotx}-\dfrac{2}{\dfrac{1}{cotx}-1}\)
\(B=\dfrac{1+cotx}{1-cotx}-\dfrac{2cotx}{1-cotx}=\dfrac{1+cotx-2cotx}{1-cotx}=\dfrac{1-cotx}{1-cotx}=1\)
c/ \(C=cos^4x-sin^4x+cos^4x+sin^2xcos^2x+3sin^2x\)
\(C=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)
\(C=cos^2x-sin^2x+cos^2x+3sin^2x\)
\(C=2cos^2x+2sin^2x=2\left(cos^2x+sin^2x\right)=2\)
\(E=\dfrac{\left(cosx-siny\right)\left(cosx+siny\right)}{sin^2x\cdot sin^2y}-\dfrac{cos^2x}{sin^2x}\cdot\dfrac{cos^2y}{sin^2y}\)
\(=\dfrac{cos^2x\left(1-cos^2y\right)-sin^2y}{sin^2x\cdot sin^2y}\)
\(=\dfrac{sin^2y\left(cos^2x-1\right)}{sin^2x\cdot sin^2y}=-1\)