Tính B= \(\sin^6x+\cos^6x+3\sin^2x.cos^2x\)( với x là góc nhọn tùy ý)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
5 tháng 7 2021
1,\(A=3\left(sin^4x+cos^4x\right)-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)
\(=3\left(sin^4x+cos^4x\right)-2\left(sin^4x-sin^2x.cos^4x+cos^4x\right)\)
\(=sin^4x+2sin^2x.cos^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\)
Vậy...
2,\(B=cos^6x+2sin^4x\left(1-sin^2x\right)+3\left(1-cos^2x\right)cos^4x+sin^4x\)
\(=-2cos^6x+3sin^4x-2sin^6x+3cos^4x\)
\(=-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)
\(=-2\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)\(=cos^4x+sin^4x+2sin^2x.cos^2x=1\)
Vậy...
3,\(C=\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}\right)\right]+\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)
\(=cos\left(-\dfrac{7\pi}{12}\right)+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}+\pi\right)\right]\)
\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)-cos\left(2x-\dfrac{\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)
Vậy...
4, \(D=cos^2x+\left(-\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)^2+\left(-\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right)^2\)
\(=cos^2x+\dfrac{1}{4}cos^2x+\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x+\dfrac{1}{4}cos^2x-\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x\)
\(=\dfrac{3}{2}\left(cos^2x+sin^2x\right)=\dfrac{3}{2}\)
Vậy...
5, Xem lại đề
6,\(F=-cosx+cosx-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\pi+\dfrac{\pi}{2}-x\right)\)
\(=tan\left(\pi-\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=tan\left(\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=cotx.tanx=1\)
Vậy...
AG
1
Y
29 tháng 5 2019
\(=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x\cdot cos^2x+cos^4x\right)\)
\(+\left(sin^2x+cos^2x\right)^2-2sin^2x\cdot cos^2x+5\cdot sin^2x\cdot cos^2x\)
\(=sin^4x+cos^4x-sin^2x\cdot cos^2x+1-2\cdot sin^2x\cdot cos^2x+5\cdot sin^2x\cdot cos^2x\)
\(=1-2\cdot sin^2x\cdot cos^2x-sin^2x\cdot cos^2x+1-2\cdot sin^2x\cdot cos^2x+5\cdot sin^2x\cdot cos^2x\)
\(=2\)
TN
3 tháng 6 2017
= (sin^2x + cos^2x)^2 - 3sin^4x.cos^2x - 3sin^2x.cos^4x
= 1 - 3/4sin^2 (2x).sin^2x - 3/4sin^2(2x).cos^2x
= 1 - 3/4sin^2(2x)
AH
Akai Haruma
Giáo viên
1 tháng 8 2019
Lời giải:
* $x$ là biến chứ không phải tham số bạn nhé*
\(A=2[(\cos ^2x)^3+(\sin ^2x)^3]-3(\cos ^4x+\sin ^4x)\)
\(=2(\cos ^2x+\sin ^2x)(\cos ^4x-\cos ^2x\sin ^2x+\sin ^4x)-3(\cos ^4x+\sin ^4x)\)
\(=2(\cos ^4x-\cos ^2x\sin ^2x+\sin ^4x)-3(\cos ^4x+\sin ^4x)\)
\(=-(\cos ^4x+2\cos ^2x\sin ^2x+\sin ^4x)=-(\cos ^2x+\sin ^2x)^2=-1^2=-1\)
là giá trị không phụ thuộc vào biến (đpcm)
--------------------------
\(B=\frac{\tan ^2x}{\sin ^2x\cos ^2x}-(1+\tan ^2x)^2=\frac{\sin ^2x}{\cos ^2x.\sin ^2x\cos ^2x}-(1+\frac{\sin ^2x}{\cos ^2x})^2\)
\(=\frac{1}{\cos ^4x}-(\frac{\cos ^2x+\sin ^2x}{\cos ^2x})^2=\frac{1}{\cos ^4x}-(\frac{1}{\cos ^2x})^2=\frac{1}{\cos ^4x}-\frac{1}{\cos ^4x}=0\)
là giá trị không phụ thuộc vào biến $x$ (đpcm)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 2 2021
\(sinx+cosx=m\Leftrightarrow\left(sinx+cosx\right)^2=m^2\)
\(\Leftrightarrow1+2sinx.cosx=m^2\Rightarrow sinx.cosx=\dfrac{m^2-1}{2}\)
\(A=sin^2x+cos^2x=1\)
\(B=sin^3x+cos^3x=\left(sinx+cosx\right)^3-3sinx.cosx\left(sinx+cosx\right)\)
\(=m^3-\dfrac{3m\left(m^2-1\right)}{2}=\dfrac{2m^3-3m^3+3m}{2}=\dfrac{3m-m^3}{2}\)
\(C=\left(sin^2+cos^2x\right)^2-2\left(sinx.cosx\right)^2=1-2\left(\dfrac{m^2-1}{2}\right)^2\)
\(D=\left(sin^2x\right)^3+\left(cos^2x\right)^3=\left(sin^2x+cos^2x\right)^3-3\left(sin^2x+cos^2x\right)\left(sinx.cosx\right)^2\)
\(=1-3\left(\dfrac{m^2-1}{2}\right)^2\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2020
6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2020
5.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 3 2019
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/ \(C=cos^4x-sin^4x+cos^4x+sin^2xcos^2x+3sin^2x\)
\(C=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)
\(C=cos^2x-sin^2x+cos^2x+3sin^2x\)
\(C=2cos^2x+2sin^2x=2\left(cos^2x+sin^2x\right)=2\)
15 tháng 12 2023
\(sin^6x+cos^6x+3\cdot sin^2x\cdot cos^2x\)
\(=\left(sin^2x+cos^2x\right)^3-3\cdot sin^2x\cdot cos^2x\left(sin^2x+cos^2x\right)+3\cdot sin^2x\cdot cos^2x\)
\(=1^3-3\cdot sin^2x\cdot cos^2x+3\cdot sin^2x\cdot cos^2x\)
=1