Lời giải:

Ta có:

$\sin (2a+b)=3\sin b$

$\Leftrightarrow \sin (a+b+a)=3\sin (a+b-a)$

$\Leftrightarrow \sin (a+b)\cos a+\cos (a+b)\sin a=3\sin (a+b)\cos a-3\cos (a+b)\sin a$

$\Leftrightarrow 4\cos (a+b)\sin a=2\sin (a+b)\cos a$

$\Leftrightarrow 2\cos (a+b)\sin a=\sin (a+b)\cos a$

$\Rightarrow \frac{2\sin a}{\cos a}=\frac{\sin (a+b)}{\cos (a+b)}$

$\Rightarrow 2\tan a=\tan (a+b)$

Ta có đpcm.

Em học lớp 9 nên giúp được câu 2 thôi nha :)

\(pt:x^2-mx+m+8=0\)

\(\Delta=\left(-m\right)^2-4\left(m+8\right)=m^2-4m+32=\left(m-2\right)^2+28>0\forall m\)

⇒ pt luôn có 2 nghiệm phân biệt

Theo hệ thức Vi-et: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=m+8\end{matrix}\right.\)

Để pt có 2 nghiệm phân biệt cùng âm thì:

\(\left\{{}\begin{matrix}\Delta>0\left(TM\right)\\P>0\\S< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 0\\m+8>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 0\\m>-8\end{matrix}\right.\Leftrightarrow-8< m< 0\)

3. a, P = 2 sinx ( cos x + cos 3x + cos 5x)

= 2 sinx . [ 2.cos3x.cos (-2x) + cos 3x]

= 2 sinx . [ cos 3x ( cos 2x + 1)]

= 2 sinx cos 3x . (2 cos x - 1 + 1)

= 4 sinx . cos x .cos 3x = 2 . sin2x .cos 3x

#mã mã#

Lời giải:

Đặt $a-b=x; b-c=y, c-a=z$ thì $x+y+z=0$.

ĐKĐB tương đương với:

$x^2+y^2+z^2=(y-z)^2+(z-x)^2+(x-y)^2$

$\Leftrightarrow x^2+y^2+z^2=2(x^2+y^2+z^2)-2(xy+yz+xz)$

$\Leftrightarrow x^2+y^2+z^2=2(xy+yz+xz)$
$\Leftrightarrow 2(x^2+y^2+z^2)=x^2+y^2+z^2+2(xy+yz+xz)$

$\Leftrightarrow 2(x^2+y^2+z^2)=(x+y+z)^2=0$

$\Rightarrow x=y=z=0$

$\Leftrightarrow a-b=b-c=c-a=0$

$\Leftrightarrow a=b=c$ (ta có đpcm)

Đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\c-a=z\end{matrix}\right.\) thì ta có \(x+y+z=0\). Điều kiện đã cho tương đương \(x^2+y^2+z^2=\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\)

\(\Leftrightarrow x^2+y^2+z^2=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2+y^2+z^2=2\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=4\left(xy+yz+zx\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2=4\left(xy+yz+zx\right)\)

\(\Leftrightarrow4\left(xy+yz+zx\right)=0\)

\(\Leftrightarrow xy+yz+zx=0\)

\(\Leftrightarrow x^2+y^2+z^2=0\)

\(\Leftrightarrow x=y=z=0\)

\(\Leftrightarrow a-b=b-c=c-a=0\)

\(\Leftrightarrow a=b=c\)

Ta có đpcm

Lời giải:

a)

\(\frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{(\sin ^2a+\cos ^2a)+\cos ^2a-1}{\cot ^2a}=\frac{1+\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{(\frac{\cos a}{\sin a})^2}=\sin ^2a\)

b)

\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)

\(=\frac{\sin ^2a+\cos ^2a}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\tan ^2a+1-1=\tan ^2a\)

c)

\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}=\frac{\sin ^4a(\cos ^2a-1)}{\cos ^4a(\sin ^2a-1)}\)

\(=\frac{\sin ^4a(-\sin ^2a)}{\cos ^4a(-\cos ^2a)}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)

\(\dfrac{1-\sin^2a\cos^2a}{\sin^2a}-\sin^2a\)

\(=\dfrac{1-\sin^2a\cos^2a-\sin^2a\sin^2a}{\sin^2a}\)

\(=\dfrac{1-\sin^2a\left(\cos^2a+\sin^2a\right)}{\sin^2a}\)

\(=\dfrac{\cos^2a+\sin^2a-\sin^2a}{\sin^2a}\)

\(=\dfrac{\cos^2a}{\sin^2a}=\cot^2a\)

\(\frac{cos\left(a-b\right)}{sin\left(a+b\right)}=\frac{cosa.cosb+sina.sinb}{sina.cosb+cosa.sinb}=\frac{\frac{cosa.cosb}{sina.sinb}+1}{\frac{sina.cosb}{sina.sinb}+\frac{cosa.sinb}{sina.sinb}}=\frac{cota.cotb+1}{cota+cotb}\)

Bạn ghi đề ko đúng

\(sin\left(a+b\right)sin\left(a-b\right)=\frac{1}{2}\left[cos2b-cos2a\right]\)

\(=\frac{1}{2}\left[1-2sin^2b-1+2sin^2a\right]\)

\(=sin^2a-sin^2b\)

\(=1-cos^2a-1+cos^2b=cos^2b-cos^2a\)

Câu này bạn cũng ghi đề ko đúng

\(cos\left(a+b\right)cos\left(a-b\right)=\frac{1}{2}\left[cos2a+cos2b\right]\)

\(=\frac{1}{2}\left[2cos^2a-1+1-2sin^2b\right]=cos^2a-sin^2b\)

\(=1-sin^2a-1+cos^2b=cos^2b-sin^2a\)