Rút gọn: Sin3a+Sin3a Cos2a
Giúp mình vs. Cảm ơn
Biểu thức rút gọn của biểu thức \(A=\dfrac{\sin2a+\sin5a-\sin3a}{1+\cos a-2\sin^22a}\) là : ?
\(A=\dfrac{sin2\alpha+sin5\alpha-sin3\alpha}{1+cos\alpha-2sin^22\alpha}\)
\(=\dfrac{2sin\alpha.cos\alpha+2.cos4\alpha.sin\alpha}{cos4\alpha+cos\alpha}\)
\(=\dfrac{2sin\alpha.\left(cos\alpha+cos4\alpha\right)}{cos4\alpha+cos\alpha}=2sin\alpha\)
rút gọn biểu thức \(\frac{sina+sin3a}{2cos4a}\)
Rút gọn A=\(\dfrac{\sin a+\sin3a+\sin5a}{\cos a+\cos3a+\cos5a}\)
A = \(\dfrac{2\sin3a.\cos2a+\sin3a}{2\cos3a.\cos2a+\cos3a}=\dfrac{\sin3a.\left(2\cos2a+1\right)}{\cos3a.\left(2\cos2a+1\right)} =\dfrac{\sin3a}{\cos3a}=\tan3a\)
Rút gọn biểu thức
A= sin2a +sin5a- sin3a/1+ cos- 2sin22a
\(A=\frac{sin2a+sin5a-sin3a}{1+cosa-2sin^22a}=\frac{2sina.cosa+2cos4a.sina}{cos4a+cosa}=\frac{2sina\left(cos4a+cosa\right)}{cos4a+cosa}=2sina\)
Rút gọn biểu thức
\(A=\frac{sin2a+sin5a-sin3a}{1+cosa-2sin^22a}\)
\(A=\frac{2sina.cosa+2cos4a.sina}{cos4a+cosa}=\frac{2sina\left(cos4a+cosa\right)}{cos4a+cosa}=2sina\)
a. \(\dfrac{sina+sin3a+sin5a}{cosa+cos3a+cos5a}\)= tan3a
b. \(\dfrac{1+cosa}{1-cosa}tan^2\dfrac{a}{2}-cos^2a=sin^2a\)
giúp mk vs ạ
a.
\(\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cosa+sin3a}{2cos3a.cosa+cos3a}=\dfrac{sin3a\left(2cosa+1\right)}{cos3a\left(2cosa+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)
b.
\(\dfrac{1+cosa}{1-cosa}.\dfrac{sin^2\dfrac{a}{2}}{cos^2\dfrac{a}{1}}-cos^2a=\dfrac{1+cosa}{1-cosa}.\dfrac{\dfrac{1-cosa}{2}}{\dfrac{1+cosa}{2}}-cos^2a\)
\(=\dfrac{1+cosa}{1-cosa}.\dfrac{1-cosa}{1+cosa}-cos^2a=1-cos^2a=sin^2a\)
Cho tan a = 2. Tính giá trị biểu thức: E= 8 cos 3 a - 2 sin 3 a + cos a 2 c o s a - s i n 3 a
A. - 3 2
B. 2
C. 4
D. 5 2
Cho tan a= 2. Tính giá trị biểu thức E = 8 cos 3 a - 2 sin 3 a + cos a 2 cos a - sin 3 a
A.2
B. - 3 2
C.4
D. 5 2
rút gọn:
a, A=\(\frac{sina+sin2a+sin3a}{cosa+cos2a+cos3a}\)
b, B=\(\frac{sin^2a+sin^2a.tan^2a}{cos^2a+cos^2a.cot^2a}\)
\(A=\frac{sina+sin3a+sin2a}{cosa+cos3a+cos2a}=\frac{2sin2a.cosa+sin2a}{2cos2a.cosa+cos2a}=\frac{sin2a\left(2cosa+1\right)}{cos2a\left(2cosa+1\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(B=\frac{sin^2a\left(1+tan^2a\right)}{cos^2a\left(1+cot^2a\right)}=\frac{sin^2a.\frac{1}{cos^2a}}{cos^2a.\frac{1}{sin^2a}}=\frac{sin^4a}{cos^4a}=tan^4a\)