Tính:cos4a+sin2a.cos2a+sin4a
a là alpha nhé
cho \(sin\alpha=\frac{1}{4}\). tính giá trị biểu thức A=(sin4a+2sin2a).cosa
\(A=\left(2sin2a.cos2a+2sin2a\right)cosa\)
\(=2sin2a.cosa\left(cos2a+1\right)=4sina.cosa.cosa\left(1-2sin^2a+1\right)\)
\(=4sina.cos^2a\left(2-2sin^2a\right)=8sina\left(1-sin^2a\right)\left(1-sin^2a\right)\)
\(=8sina.\left(1-sin^2a\right)^2=8.\frac{1}{4}\left(1-\frac{1}{16}\right)^2=...\)
cho a là góc nhọn.Rút gọn biểu thức
A=sin6a +cos6a +3.sin2a.cos2a
\(A=\left(sin^2a+cos^2a\right)^3-3\cdot sin^2a\cdot cos^2a\left(sin^2a+cos^2a\right)+3\cdot sin^2a\cdot cos^2a\)
\(=1-3\cdot sin^2a\cdot cos^2a+3\cdot sin^2a\cdot cos^2a\)
=1
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) \(\sin^2\left(45^0+\alpha\right)-\sin^2\left(30^0-\alpha\right)-\sin15^0\cos\left(15^0+2\alpha\right)=\sin2\alpha\)
Cho \(\alpha\) là góc nhọn. Rút gọn biểu thức:
\(A=sin^6\alpha+cos^6\alpha+3sin^2\alpha-cos^2\alpha\)
cảm ơn các bạn trước nhé
\(A=sin^6\alpha+cos^6\alpha+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha\right)^3+\left(cos^2\alpha\right)^3+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha\right)+3sin^2\alpha-cos^2\alpha\)
\(=sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha\)
\(=\left(sin^2\alpha+cos^2\alpha\right)^2-2sin^2\alpha.cos^2\alpha-sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha\)
\(1-3sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha=3sin^2\alpha\left(1-cos^2\alpha\right)+\left(1-cos^2\alpha\right)\)
\(=\left(3sin^2\alpha+1\right).sin^2\alpha=0\)
Cho \(\alpha\)là góc nhọn. Rút gọn biểu thức
A = \(\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha-\cos^2\alpha\)
( Giúp mình bài này với nhé , mình đang cần gấp ) :D
a) cos4a - sin4a +1 = 2cos2a
b) cos6a + sin6a + 3sin2a.cos2a = 1
b: \(=\left(\cos^2\alpha+\sin^2\alpha\right)^3-3\cos^2\alpha\sin^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)+3\cdot\sin^2\alpha\cdot\cos^2\alpha\)
=1
\(cos^4a-sin^4a+1=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1\)
\(=cos^2a-sin^2a+1=cos^2a-sin^2a+sin^2a+cos^2a\)
\(=2cos^2a\)
\(cos^6a+sin^6a+3sin^2a.cos^2a\)
\(=\left(cos^2a+sin^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a.cos^2a\)
\(=1-3sin^2a.cos^2a.1+3sin^2a.cos^2a\)
\(=1\)
tìm a biết a là góc tù và sin4a + cos4a = 5/8
\(sin^4a+cos^4a=\dfrac{5}{8}\)
\(\Leftrightarrow\left(sin^2a+cos^2a\right)^2-2sin^2a.cos^2a=\dfrac{5}{8}\)
\(\Leftrightarrow1-2sin^2a\left(1-sin^2a\right)=\dfrac{5}{8}\)
\(\Leftrightarrow2sin^4a-2sin^2a+\dfrac{3}{8}=0\Rightarrow\left[{}\begin{matrix}sin^2a=\dfrac{3}{4}\\sin^2a=\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sina=\dfrac{\sqrt{3}}{2}\\sina=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=150^0\\a=120^0\end{matrix}\right.\)
Rút gọn biểu thức
A=\(\frac{2Sina-Sin4a}{2Sina+Sin4a}\)
B=\(\frac{Sin\left(\frac{\pi}{4}-a\right)+Cos\left(\frac{\pi}{4}-a\right)}{Sin\left(\frac{\pi}{4}-a\right)-Cos\left(\frac{\pi}{4}-a\right)}\)
bài 3 Rút gọn các biểu thức sau
a) A= sin4a - cos4a +2sin2a . cos2a
$\sin^4 a-cos^4 a+2\sin^2 a.\cos^2 a\\=(\sin^4 a-\cos^4 a)+2\sin^2 a.\cos^2 a\\=(\sin^2 a+\cos^2 a)(\sin^2-\cos ^2 )+2\sin^2 a.\cos^2 a\\=\sin^2 a-\cos^2 a+2\sin^2 a.\cos^2 a$
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) \(\dfrac{\cos2x-\sin4x-\cos6x}{\cos2x+\sin4x-\cos6x}\)