Rút gọn:
B = sin4 α + cos4 α + 2sin2 α*cos2 α
D = sin2 α*sin2 ß + sin2 α*cos2 ß + cos2 α
hì, hồi nãy câu E cho mk/e sửa đề lại nhé
E = cos6 α + sin6 α + 3 sin2 α* cos2 α
(α: an pha, ß: bê ta)
Chứng minh rằng:
a) sin4 α + sin2 α.cos2 α + cos2α = 1
b) (1+tan α).(1+cot α).sin α.cos α=1 + 2.sin α.cos α
c) sin6 α+cos6 α + 3 sin2 α.cos2 α = 1
a: \(=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1^2=1\)
Rút gọn :\(\dfrac{cos2\alpha+cos4\alpha+cos6\alpha}{sin2\alpha+sin4\alpha+sin6\alpha}\)
CMR: α<45* ta có công thức:
a/ \(sin^2\alpha=\frac{1-cos2\text{α}}{2}\)
b/ \(cos^2\text{α}=\frac{1+cos2\text{α}}{2}\)
c/ \(cos2\text{α}=cos^2\text{α}-sin^2\text{α}\)
Tính c o s 2 ( α + x ) + c o s 2 x - 2 cos α . c o s x . c o s ( α + x )
A. 1 2 ( 1 - cos 2 α )
B. c o s 2 α
C. ( 1 - cos 2 α )
D. sin α
Bài 4. a) Tính giá trị biểu thức:
A = cos2 20° + cos2 40° + cos2 50° + cos2 70°.
b) Rút gọn biểu thức:
B = sin6 a + cos6 a + 3 sin2 a. cos2 a
\(a,A=\left(\cos^220^0+\cos^270^0\right)+\left(\cos^240^0+\cos^250^0\right)\\ A=\left(\cos^220^0+\sin^220^0\right)+\left(\cos^240^0+\sin^240^0\right)=1+1=2\\ b,B=\left(\cos^2\alpha\right)^3+\left(\sin^2\alpha\right)^3+3\sin^2\alpha\cdot\cos^2\alpha\cdot\left(\sin^2\alpha+\cos^2\alpha\right)\\ B=\left(\sin^2\alpha+\cos^2\alpha\right)^3=1^3=1\)
a) cos2 (α +x) +cos22 x - 2 cosα cosx.cos(α +x);
b) sin4x.sin10x-sin11x.sin3x-sin7x.sinx
b: =1/2*[cos(10x-4x)-cos(10x+4x)]-1/2*[cos(11x-3x)-cos(11x+3x)]-1/2*[cos(7x-x)-cos(7x+x)]
=1/2*[cos 6x-cos14x-cos8x+cos14x-cos6x+cos8x]
=0
cm 0<=α<=π thì (2cosα-1)^2-4sin^2(α/2-π/4)>(\(\left(\sqrt{2sin\alpha}-2\right)\left(3-cos2\alpha\right)\)
Chung minh rang voi moi goc luong giac α lam cho bieu thuc xac dinh thi
a) \(\dfrac{1-sin2\alpha}{1+sin2\alpha}\)=cot\(^2\)(\(\dfrac{\pi}{4}\)+α) b) \(\dfrac{sin\alpha+sin\beta cos\left(\alpha+\beta\right)}{cos\alpha-sin\beta sin\left(\alpha+\beta\right)}\)=tan\(\left(\alpha+\beta\right)\).
a, \(\dfrac{1-sin2a}{1+sin2a}\)
\(=\dfrac{sin^2a+cos^2a-2sina.cosa}{sin^2a+cos^2a+2sina.cosa}\)
\(=\dfrac{\left(sina-cosa\right)^2}{\left(sina+cosa\right)^2}\)
\(=\dfrac{2sin^2\left(a-\dfrac{\pi}{4}\right)}{2sin^2\left(a+\dfrac{\pi}{4}\right)}\)
\(=\dfrac{sin^2\left(\dfrac{\pi}{4}-a\right)}{sin^2\left(a+\dfrac{\pi}{4}\right)}\)
\(=\dfrac{cos^2\left(\dfrac{\pi}{4}+a\right)}{sin^2\left(\dfrac{\pi}{4}+a\right)}=cot\left(\dfrac{\pi}{4}+a\right)\)
b, \(\dfrac{sina+sinb.cos\left(a+b\right)}{cosa-sinb.sin\left(a+b\right)}\)
\(=\dfrac{sina+sinb.cosa.cosb-sinb.sina.sinb}{cosa-sinb.sina.cosb-sinb.cosa.sinb}\)
\(=\dfrac{sina.\left(1-sin^2b\right)+sinb.cosa.cosb}{cosa.\left(1-sin^2b\right)-sinb.sina.cosb}\)
\(=\dfrac{sina.cos^2b+sinb.cosa.cosb}{cosa.cos^2b-sinb.sina.cosb}\)
\(=\dfrac{\left(sina.cosb+sinb.cosa\right).cosb}{\left(cosa.cosb-sinb.sina\right).cosb}\)
\(=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}=tan\left(a+b\right)\)
Cho góc α thỏa mãn tanα = 5. Tính P= sin4 α - cos4 α
A. P = 2
B. P = 1/2
C. P = 11/13
D. P = 12/13
Chọn D.
Ta có P = ( sin2α - cos2α) ( sin2α + cos2α) = sin2α - cos2α (*)
Chia hai vế của (*) cho cos2 α ta được
Tương đương: P(1 + tan2α) = tan2α - 1