Chứng minh \(\frac{\cos3\alpha+\cos\alpha}{\sin3\alpha+\sin\alpha}.\tan2\alpha-8\sin^2\alpha.\cos^2\alpha=\cos4\alpha\) với \(\alpha\ne k\frac{\pi}{4}\left(k\in Z\right)\)
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 \(\dfrac{\pi}{2}\))
\(\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cos2a+sin3a}{2cos3a.cos2a+cos3a}=\dfrac{sin3a\left(2cos2a+1\right)}{cos3a\left(2cos2a+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)
\(\dfrac{1+sin4a-cos4a}{1+sin4a+cos4a}=\dfrac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin2a.cos2a+2cos^22a-1}=\dfrac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\dfrac{sin2a}{cos2a}=tan2a\)
\(96\sqrt{3}sin\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=48\sqrt{3}sin\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=24\sqrt{3}sin\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=12\sqrt{3}sin\left(\dfrac{\pi}{6}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=6\sqrt{3}sin\left(\dfrac{\pi}{3}\right)=6\sqrt{3}.\dfrac{\sqrt{3}}{2}=9\)
\(A+B+C=\pi\Rightarrow A+B=\pi-C\Rightarrow tan\left(A+B\right)=tan\left(\pi-C\right)\)
\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=-tanC\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)
\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\)
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) \(\dfrac{3-4cos2\alpha+cos4\alpha}{3+4cos2\alpha+cos4\alpha}\)\(=\dfrac{4-4cos2\alpha+cos4\alpha-1}{4+4cos2\alpha+cos4\alpha-1}\)
\(=\dfrac{4\left(1-cos2\alpha\right)-2sin^22\alpha}{4\left(1+cos2\alpha\right)-2sin^22\alpha}\)
\(=\dfrac{4cos^2\alpha-2sin^22\alpha}{4sin^2\alpha-2sin^22\alpha}\)
\(=\dfrac{4cos^2\alpha-8sin^2\alpha cos^2\alpha}{4sin^2\alpha-8sin^2\alpha cos^2\alpha}\)
\(=\dfrac{4cos^2\alpha\left(1-2sin^2\alpha\right)}{4sin^2\alpha\left(1-2cos^2\alpha\right)}=cot^2\alpha.\dfrac{cos2\alpha}{-cot2\alpha}\)
\(=-cot^2\alpha\).
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) \(\dfrac{2\sin2\alpha-\sin4\alpha}{2\sin2\alpha+\sin4\alpha}\)
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) \(\dfrac{\sin5\alpha-\sin3\alpha}{2\cos4\alpha}\)
Chứng minh các đẳng thức :
a) \(\tan3\alpha-\tan2\alpha-\tan\alpha=\tan\alpha\tan2\alpha\tan3\alpha\)
b) \(\dfrac{4\tan\alpha\left(1-\tan^2\alpha\right)}{\left(1+\tan^2\alpha\right)^2}=\sin4\alpha\)
c) \(\dfrac{1+\tan^4\alpha}{\tan^2\alpha+\cot^2\alpha}=\tan^2\alpha\)
d) \(\dfrac{\cos\alpha\sin\left(\alpha-3\right)-\sin\alpha\cos\left(\alpha-3\right)}{\cos\left(3-\dfrac{\pi}{6}\right)-\dfrac{1}{2}\sin3}=-\dfrac{2\tan3}{\sqrt{3}}\)
a) \(tan3\alpha-tan2\alpha-tan\alpha=\left(tan3\alpha-tan\alpha\right)-tan2\alpha\)
\(=\left(\dfrac{sin3\alpha}{cos3\alpha}-\dfrac{sin\alpha}{cos\alpha}\right)-\dfrac{sin2\alpha}{cos2\alpha}\)\(=\dfrac{sin3\alpha cos\alpha-cos3\alpha sin\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=sin2\alpha.\left(\dfrac{1}{cos3\alpha cos\alpha}-\dfrac{1}{cos2\alpha}\right)\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos3\alpha cos\alpha}{cos3\alpha cos\alpha cos2\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-\dfrac{1}{2}\left(cos4\alpha+cos2\alpha\right)}{cos3\alpha cos2\alpha cos\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos4\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=\dfrac{sin2\alpha.2sin3\alpha.sin\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=tan3\alpha tan2\alpha tan\alpha\) (Đpcm).
b) \(\dfrac{4tan\alpha\left(1-tan^2\alpha\right)}{\left(1+tan^2\right)^2}=4tan\alpha\left(1-tan^2\alpha\right):\left(\dfrac{1}{cos^2\alpha}\right)^2\)
\(=4tan\alpha\left(1-tan^2\alpha\right)cos^4\alpha\)
\(=4\dfrac{sin\alpha}{cos\alpha}\left(1-\dfrac{sin^2\alpha}{cos^2\alpha}\right)cos^4\alpha\)
\(=4sin\alpha\left(cos^2\alpha-sin^2\alpha\right)cos\alpha\)
\(=4sin\alpha cos\alpha.cos2\alpha\)
\(=2.sin2\alpha.cos2\alpha=sin4\alpha\) (Đpcm).
c) \(\dfrac{1+tan^4\alpha}{tan^2\alpha+cot\alpha}=\left(1+tan^4\alpha\right):\left(tan^2\alpha+cot^2\alpha\right)\)
\(=\left(1+\dfrac{sin^4\alpha}{cos^4\alpha}\right):\left(\dfrac{sin^2\alpha}{cos^2\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}\right)\)
\(=\dfrac{sin^4\alpha+cos^4\alpha}{cos^4\alpha}:\dfrac{sin^4\alpha+cos^4\alpha}{cos^2\alpha sin^2\alpha}\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}=tan^2\alpha\) (Đpcm).
1/ \(\alpha\ne\frac{\pi}{2}+k\pi,k\in Z\) chứng minh rằng: \(\frac{\sin^2\alpha-\cos^2\alpha}{1+2\sin\cos}=\frac{\tan-1}{\tan+1}\)
\(\frac{sin^2a-cos^2a}{1+2sina.cosa}=\frac{\left(sina-cosa\right)\left(sina+cosa\right)}{sin^2a+cos^2a+2sina.cosa}=\frac{\left(sina-cosa\right)\left(sina+cosa\right)}{\left(sina+cosa\right)^2}\)
\(=\frac{sina-cosa}{sina+cosa}=\frac{\frac{sina}{cosa}-\frac{cosa}{cosa}}{\frac{sina}{cosa}+\frac{cosa}{cosa}}=\frac{tana-1}{tana+1}\)
Cho \(\sin\alpha+\cos\alpha=\frac{\sqrt{6}}{2},a\in\left(0;\frac{\pi}{4}\right)\)
Tính giá trị biểu thức: \(P=\cos\left(\alpha+\frac{\pi}{4}\right)+\sqrt{2\left(1-\sin\alpha\cos\alpha+\sin\alpha-\cos\alpha\right)}\)
chứng minh rằng:
\(\frac{1-cos\alpha+cos2\alpha}{sin2\alpha-sin\alpha}\)= cotα ,với α ≠ kπ ( k ∈ Z) và α ≠ \(\pm\) \(\frac{\pi}{3}\) +l2π ( l ∈ Z)
\(\frac{1-cosa+cos2a}{sin2a-sina}=\frac{1-cosa+2cos^2a-1}{2sina.cosa-sina}=\frac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\frac{cosa}{sina}=cota\)
Chứng minh:
a)\(\cos^4\alpha-sin^4\alpha=2cos^2\alpha-1\)
b)\(\frac{cos\alpha}{1-sin\alpha}=\frac{1+sin\alpha}{cos\alpha}\)
c)\(\frac{\left(sin\alpha+cos\alpha\right)^2-\left(sin\alpha-cos\alpha\right)^2}{sin\alpha.cos\alpha}=4\)
Mình cần gấp!!!
a) \(\cos^4\alpha-\sin^4\alpha=\left(\cos^2\alpha+\sin^2\alpha\right)\left(\cos^2\alpha-\sin^2\alpha\right)=\cos^2\alpha-\sin^2\alpha\)
\(2\cos^2\alpha-\left(\sin^2\alpha+\cos^2\alpha\right)=2\cos^2\alpha-1\)
b) \(\frac{\cos\alpha}{1-\sin\alpha}=\frac{1+\sin\alpha}{\cos\alpha}\)\(\Leftrightarrow\)\(\left(1-\sin\alpha\right)\left(1+\sin\alpha\right)=\cos^2\alpha\)
\(\Leftrightarrow\)\(1-\left(\sin^2\alpha+\cos^2\alpha\right)=0\)\(\Leftrightarrow\)\(1-1=0\) ( luôn đúng )
c) \(\frac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha.\cos\alpha}=\frac{2\cos\alpha.2\sin\alpha}{\sin\alpha.\cos\alpha}=4\)
um, hình như câu b) chỗ 1-.... đó hơi sai nếu viết từ bước trên xuống á bạn!
mình nghĩ là: sau dấu bằng đầu tiên, sau đó là:
\(=cos^2\alpha=1-sin^2\alpha\)(luôn đúng)
CẢM ƠN bạn nhiều lắm luôn nha!!!!!