Rút gọn \(\dfrac{sin2\alpha+sin5\alpha-sin3\alpha}{1+cos\alpha-2sin^22\alpha}\)
H24
Những câu hỏi liên quan
với điều kiện xác định của các giá trị lượng giác , cho P= \(\frac{sin2\alpha+sin5\alpha-sin3\alpha}{1+cos\alpha-2sin^22\alpha}\), đơn giản biểu thức P bằng??
\(P=\frac{sin2a+2cos4a.sina}{cos4a+cosa}=\frac{2sina.cosa+2sina.cos4a}{cos4a+cosa}=\frac{2sina\left(cosa+cos4a\right)}{cos4a+cosa}=2sina\)
Đúng 0
Bình luận (0)
Rút gọn các biểu thức :
a) dfrac{2sin2alpha-sin4alpha}{2sin2alpha+sin4alpha}
b) tanalphaleft(dfrac{1+cos^2alpha}{sinalpha}-sinalpharight)
c) dfrac{sinleft(dfrac{pi}{4}-alpharight)+cosleft(dfrac{pi}{4}-alpharight)}{sinleft(dfrac{pi}{4}-alpharight)-cosleft(dfrac{pi}{4}-alpharight)}
d) dfrac{sin5alpha-sin3alpha}{2cos4alpha}
Đọc tiếp
Rút gọn các biểu thức :
a) \(\dfrac{2\sin2\alpha-\sin4\alpha}{2\sin2\alpha+\sin4\alpha}\)
b) \(\tan\alpha\left(\dfrac{1+\cos^2\alpha}{\sin\alpha}-\sin\alpha\right)\)
c) \(\dfrac{\sin\left(\dfrac{\pi}{4}-\alpha\right)+\cos\left(\dfrac{\pi}{4}-\alpha\right)}{\sin\left(\dfrac{\pi}{4}-\alpha\right)-\cos\left(\dfrac{\pi}{4}-\alpha\right)}\)
d) \(\dfrac{\sin5\alpha-\sin3\alpha}{2\cos4\alpha}\)
1.chứng minh hệ thức: dfrac{sinalpha+sin3alpha+sin5alpha}{cosalpha+cos3alpha+cos5alpha}tan3alpha2.rút gọn biểu thức: dfrac{1+sin4alpha-cos4alpha}{1+cos4alpha+sin4alpha}3. Tính 96sqrt{3}sindfrac{pi}{48}cosdfrac{pi}{48}cosdfrac{pi}{24}cosdfrac{pi}{12}cosdfrac{pi}{6}4. chứng minh rằng trong một △ABC ta có: tanA + tanB + tanC tanA tanB tanC (A,B,C cùng khác dfrac{pi}{2})
Đọc tiếp
1.\(\)chứng minh hệ thức: \(\dfrac{sin\alpha+sin3\alpha+sin5\alpha}{cos\alpha+cos3\alpha+cos5\alpha}=tan3\alpha\)
2.rút gọn biểu thức: \(\dfrac{1+sin4\alpha-cos4\alpha}{1+cos4\alpha+sin4\alpha}\)
3. Tính \(96\sqrt{3}sin\dfrac{\pi}{48}cos\dfrac{\pi}{48}cos\dfrac{\pi}{24}cos\dfrac{\pi}{12}cos\dfrac{\pi}{6}\)
4. chứng minh rằng trong một △ABC ta có:
tanA + tanB + tanC = tanA tanB tanC (A,B,C cùng khác \(\dfrac{\pi}{2}\))
\(\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cos2a+sin3a}{2cos3a.cos2a+cos3a}=\dfrac{sin3a\left(2cos2a+1\right)}{cos3a\left(2cos2a+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)
\(\dfrac{1+sin4a-cos4a}{1+sin4a+cos4a}=\dfrac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin2a.cos2a+2cos^22a-1}=\dfrac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\dfrac{sin2a}{cos2a}=tan2a\)
\(96\sqrt{3}sin\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=48\sqrt{3}sin\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=24\sqrt{3}sin\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=12\sqrt{3}sin\left(\dfrac{\pi}{6}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=6\sqrt{3}sin\left(\dfrac{\pi}{3}\right)=6\sqrt{3}.\dfrac{\sqrt{3}}{2}=9\)
\(A+B+C=\pi\Rightarrow A+B=\pi-C\Rightarrow tan\left(A+B\right)=tan\left(\pi-C\right)\)
\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=-tanC\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)
\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\)
Đúng 0
Bình luận (0)
Rút gọn biểu thức \(P=\dfrac{1}{\sin\alpha.\sin2\alpha}+\dfrac{1}{\sin2\alpha.\sin3\alpha}+.....+\dfrac{1}{\sin n\alpha.\sin\left(n+1\right)\alpha}\)
(Giúp mik với !!!)
\(P.sina=\dfrac{sina}{sin2a.sina}+\dfrac{sina}{sin3a.sin2a}+...+\dfrac{sina}{sin\left(n+1\right)a.sinna}\)
\(=\dfrac{sin\left(2a-a\right)}{sin2a.sina}+\dfrac{sin\left(3a-2a\right)}{sin3a.sin2a}+...+\dfrac{sin\left[\left(n+1\right)a-na\right]}{sin\left(n+1\right)a.sinna}\)
\(=\dfrac{sin2a.cosa-cos2a.sina}{sin2a.sina}+\dfrac{sin3a.cos2a-cos3a.sin2a}{sin3a.sin2a}+\dfrac{sin\left(n+1\right)a.cosna-cos\left(n+1\right)a.sinna}{sin\left(n+1\right)a.sinna}\)
\(=cota-cot2a+cot2a-cot3a+...+cot\left(na\right)-cot\left(n+1\right)a\)
\(=cota-cot\left(n+1\right)a\)
\(\Rightarrow P=\dfrac{cota-cot\left(n+1\right)a}{sina}\)
Đúng 1
Bình luận (0)
BT
9 tháng 5 2016 lúc 19:09
rút gọn hệ thức :
a) A = \(\frac{\sin2\alpha+\sin3\alpha+\sin4\alpha}{\cos2\alpha+\cos3\alpha+\cos4\alpha}\)
b) B = \(\frac{\sin\alpha+2\sin2\alpha+\sin3\alpha}{\cos\alpha+2\cos2\alpha+\cos3\alpha}\)
Rút gọn các biểu thức :
a) dfrac{tan2alpha}{tan4alpha-tan2alpha}
b) sqrt{1+sinalpha}-sqrt{1-sinalpha}, với 0 alpha dfrac{pi}{2}
c) dfrac{3-4cos2alpha+cos4alpha}{3+4cos2alpha+cos4alpha}
d) dfrac{sinalpha+sin3alpha+sin5alpha}{cosalpha+cos3alpha+cos5alpha}
Đọc tiếp
Rút gọn các biểu thức :
a) \(\dfrac{\tan2\alpha}{\tan4\alpha-\tan2\alpha}\)
b) \(\sqrt{1+\sin\alpha}-\sqrt{1-\sin\alpha}\), với \(0< \alpha< \dfrac{\pi}{2}\)
c) \(\dfrac{3-4\cos2\alpha+\cos4\alpha}{3+4\cos2\alpha+\cos4\alpha}\)
d) \(\dfrac{\sin\alpha+\sin3\alpha+\sin5\alpha}{\cos\alpha+\cos3\alpha+\cos5\alpha}\)
a) \(\dfrac{tan2\alpha}{tan4\alpha-tan2\alpha}=\dfrac{sin2\alpha}{cos2\alpha}:\left(\dfrac{sin4\alpha}{cos4\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\right)\)
\(=\dfrac{sin2\alpha}{cos2\alpha}:\dfrac{sin4\alpha cos2\alpha-sin2\alpha cos4\alpha}{cos4\alpha cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos2\alpha}.\dfrac{cos4\alpha.cos2\alpha}{sin2\alpha}=cos4\alpha\).
Đúng 0
Bình luận (0)
b) \(\sqrt{1+sin\alpha}-\sqrt{1-sin\alpha}=\sqrt{sin^2\dfrac{\alpha}{2}+2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)\(-\sqrt{sin^2\dfrac{\alpha}{2}-2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)
\(=\sqrt{\left(sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right)^2}-\sqrt{\left(sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right)^2}\)
\(=\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
Vì \(0< \alpha< \dfrac{\pi}{2}\) nên \(0< \alpha< \dfrac{\pi}{4}\).
Trong \(\left(0;\dfrac{\pi}{4}\right)\) thì \(sin\dfrac{\alpha}{2}\) tăng dần từ 0 tới \(\dfrac{\sqrt{2}}{2}\) và \(cos\dfrac{\alpha}{2}\) giảm dần từ 1 tới \(\dfrac{\sqrt{2}}{2}\) nên \(\left|sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right|=-\left(sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right)=cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\).
Vì vậy:
\(\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
\(=sin\dfrac{\alpha}{4}+cos\dfrac{\alpha}{4}-\left(cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\right)=2sin\dfrac{\alpha}{4}\).
Đúng 0
Bình luận (0)
c) \(\dfrac{3-4cos2\alpha+cos4\alpha}{3+4cos2\alpha+cos4\alpha}\)\(=\dfrac{4-4cos2\alpha+cos4\alpha-1}{4+4cos2\alpha+cos4\alpha-1}\)
\(=\dfrac{4\left(1-cos2\alpha\right)-2sin^22\alpha}{4\left(1+cos2\alpha\right)-2sin^22\alpha}\)
\(=\dfrac{4cos^2\alpha-2sin^22\alpha}{4sin^2\alpha-2sin^22\alpha}\)
\(=\dfrac{4cos^2\alpha-8sin^2\alpha cos^2\alpha}{4sin^2\alpha-8sin^2\alpha cos^2\alpha}\)
\(=\dfrac{4cos^2\alpha\left(1-2sin^2\alpha\right)}{4sin^2\alpha\left(1-2cos^2\alpha\right)}=cot^2\alpha.\dfrac{cos2\alpha}{-cot2\alpha}\)
\(=-cot^2\alpha\).
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Chứng minh đẳng thức
a) \(\dfrac{cos\alpha+cos3\alpha+cos5\alpha}{sin\alpha+sin3\alpha+sin5\alpha}=cot3\alpha\)
b) \(\left(\dfrac{cosa}{sinb}+\dfrac{sina}{cosb}\right)\dfrac{1-cos4b}{cos\left(a-b\right)}=4sin2b\)
\(\dfrac{cosa+cos5a+cos3a}{sina+sin5a+sin3a}=\dfrac{2cos3a.cos2a+cos3a}{2sin3a.cos2a+sin3a}\)
\(=\dfrac{cos3a\left(2cos2a+1\right)}{sin3a\left(2cos2a+1\right)}=\dfrac{cos3a}{sin3a}=cot3a\)
\(\left(\dfrac{cosa}{sinb}+\dfrac{sina}{cosb}\right)\left(\dfrac{1-cos4b}{cos\left(a-b\right)}\right)=\dfrac{\left(cosa.cosb+sina.sinb\right)}{sinb.cosb}.\dfrac{2sin^22b}{cos\left(a-b\right)}\)
\(=\dfrac{cos\left(a-b\right)}{\dfrac{1}{2}sin2b}.\dfrac{2sin^22b}{cos\left(a-b\right)}=4sin2b\)
Đúng 0
Bình luận (2)
BT
30 tháng 4 2016 lúc 12:35
rút gọn biểu thức : a) A = \(\frac{sin2\alpha+sin3\alpha+sin4\alpha}{cos2\alpha+cos3\alpha+cos4\alpha}\) ; b) B = \(\frac{sin\alpha+2sin2\alpha+sin3\alpha}{cosa+2cos2\alpha+cos3a}\)
Bài 1: Rút gọn:
A= \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos2\alpha}\)
B= \(\dfrac{4sin^2\alpha}{1-cos^2\dfrac{\alpha}{2}}\)
C= \(\dfrac{1+cos\alpha-sin\alpha}{1-cos\alpha-sin\alpha}\)
