\(\sqrt{\alpha(\alpha-}\)3)^2 với a lớn hơn hoặc bằng 3
Đề bải rút gọn biểu thức
Ảnh nhanh mình tick nha
\(\sqrt{\alpha(\alpha-}\)3)2 với a lớn hơn hoặc bằng 3
Help me
đề bài rút gon biểu thức
=> căn a × căn (a-3)^2
=> căn a ×|a-3|
=> căn a × (a-3)
\(F=\dfrac{\sin\alpha-2\sin\left(2\alpha\right)+\sin\left(3\alpha\right)}{\cos\alpha-3\cos\left(2\alpha\right)+\cos\left(3\alpha\right)}\)
Mn rút gọn giùm mình biểu thức này với. Mình cảm ơn ạ :<
Mẫu số là \(-3cos2a\) hay \(-2cos2a\) vậy bạn? -3 không hợp lý
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 \(\dfrac{\pi}{2}\))
\(\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cos2a+sin3a}{2cos3a.cos2a+cos3a}=\dfrac{sin3a\left(2cos2a+1\right)}{cos3a\left(2cos2a+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)
\(\dfrac{1+sin4a-cos4a}{1+sin4a+cos4a}=\dfrac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin2a.cos2a+2cos^22a-1}=\dfrac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\dfrac{sin2a}{cos2a}=tan2a\)
\(96\sqrt{3}sin\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=48\sqrt{3}sin\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=24\sqrt{3}sin\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=12\sqrt{3}sin\left(\dfrac{\pi}{6}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=6\sqrt{3}sin\left(\dfrac{\pi}{3}\right)=6\sqrt{3}.\dfrac{\sqrt{3}}{2}=9\)
\(A+B+C=\pi\Rightarrow A+B=\pi-C\Rightarrow tan\left(A+B\right)=tan\left(\pi-C\right)\)
\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=-tanC\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)
\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\)
2) Cho số thực alpha <= 1 . Rút gọn biểu thức P= sqrt 15 2 - sqrt 10. (a - 1) ^ 2 3 .
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
Rút gọn các biểu thức sau:
A= \(\dfrac{cos^2\alpha-sin^2\alpha}{cot^2\alpha-tan^2\alpha}-cos^2\alpha\)
B= \(\sqrt{sin^4\alpha+6cos^2\alpha+3cos^4\alpha}+\sqrt{cos^4\alpha+6sin^2\alpha+3sin^4\alpha}\)
\(A=\dfrac{cos^2a-sin^2a}{\dfrac{cos^2a}{sin^2a}-\dfrac{sin^2a}{cos^2a}}-cos^2a=\dfrac{cos^2a.sin^2a\left(cos^2a-sin^2a\right)}{\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)}-cos^2a\)
\(=cos^2a.sin^2a-cos^2a=cos^2a\left(sin^2a-1\right)=-cos^4a\)
\(B=\sqrt{\left(1-cos^2a\right)^2+6cos^2a+3cos^4a}+\sqrt{\left(1-sin^2a\right)^2+6sin^2a+3sin^4a}\)
\(=\sqrt{4cos^4a+4cos^2a+1}+\sqrt{4sin^4a+4sin^2a+1}\)
\(=\sqrt{\left(2cos^2a+1\right)^2}+\sqrt{\left(2sin^2a+1\right)^2}\)
\(=2\left(sin^2a+cos^2a\right)+2=4\)
Rút gọn các biểu thức sau:
a, \(\sqrt 2 \sin \left( {\alpha + \frac{\pi }{4}} \right) - cos\alpha \),
b, \({\left( {cos\alpha + \sin \alpha } \right)^2} - \sin 2\alpha \)
\(a,\sqrt{2}sin\left(\alpha+\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha cos\dfrac{\pi}{4}+cos\alpha sin\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha\cdot\dfrac{\sqrt{2}}{2}+cos\alpha\cdot\dfrac{\sqrt{2}}{2}\right)-cos\alpha\\ =\sqrt{2}\cdot sin\alpha\cdot\dfrac{\sqrt{2}}{2}+\sqrt{2}\cdot cos\alpha\cdot\dfrac{\sqrt{2}}{2}-cos\alpha\\ =sin\alpha+cos\alpha-cos\alpha\\ =sin\alpha\)
\(b,\left(cos\alpha+sin\alpha\right)^2-sin2\alpha\\ =cos^2\alpha+sin^2\alpha=2cos\alpha sin\alpha-2sin\alpha cos\alpha\\ =sin^2\alpha+cos^2\alpha\\ =1\)
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 VS M.N~!!!!!!!!!!!!
Đặt \(\sin^2\alpha=x\Rightarrow\cos^2\alpha=1-\sin^2\alpha\)
\(A=x^3+\left(1-x\right)^3+3x-\left(1-x\right)=x^3+1-3x+3x^2-x^3+3x-1+x=3x^2+x\)
Vậy \(A=3\sin^4\alpha+\sin^2\alpha\). NHỚ NHA!
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) \(\dfrac{tan2\alpha}{tan4\alpha-tan2\alpha}=\dfrac{sin2\alpha}{cos2\alpha}:\left(\dfrac{sin4\alpha}{cos4\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\right)\)
\(=\dfrac{sin2\alpha}{cos2\alpha}:\dfrac{sin4\alpha cos2\alpha-sin2\alpha cos4\alpha}{cos4\alpha cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos2\alpha}.\dfrac{cos4\alpha.cos2\alpha}{sin2\alpha}=cos4\alpha\).
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:
\(\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
\(=sin\dfrac{\alpha}{4}+cos\dfrac{\alpha}{4}-\left(cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\right)=2sin\dfrac{\alpha}{4}\).
c) \(\dfrac{3-4cos2\alpha+cos4\alpha}{3+4cos2\alpha+cos4\alpha}\)\(=\dfrac{4-4cos2\alpha+cos4\alpha-1}{4+4cos2\alpha+cos4\alpha-1}\)
\(=\dfrac{4\left(1-cos2\alpha\right)-2sin^22\alpha}{4\left(1+cos2\alpha\right)-2sin^22\alpha}\)
\(=\dfrac{4cos^2\alpha-2sin^22\alpha}{4sin^2\alpha-2sin^22\alpha}\)
\(=\dfrac{4cos^2\alpha-8sin^2\alpha cos^2\alpha}{4sin^2\alpha-8sin^2\alpha cos^2\alpha}\)
\(=\dfrac{4cos^2\alpha\left(1-2sin^2\alpha\right)}{4sin^2\alpha\left(1-2cos^2\alpha\right)}=cot^2\alpha.\dfrac{cos2\alpha}{-cot2\alpha}\)
\(=-cot^2\alpha\).