\(a)\;sin(\alpha  + \beta ).sin(\alpha  - \beta ) = \;\frac{1}{2}.\left[ {cos\left( {\alpha  + \beta  - \alpha  + \beta } \right) - cos\left( {\alpha  + \beta  + \alpha  - \beta } \right)} \right]\)

\(\begin{array}{l} = \;\frac{1}{2}.(cos2\beta  - cos2\alpha ) = \;\frac{1}{2}.(1 - 2si{n^2}\beta  - 1 + 2si{n^2}\alpha )\\ = si{n^2}\alpha  - si{n^2}\beta \end{array}\)

\(\begin{array}{l}b)\;co{s^4}\alpha  - co{s^4}\left( {\alpha  - \frac{\pi }{2}} \right) = \;co{s^4}\alpha  - si{n^4}\alpha \\ = \;(co{s^2}\alpha  + si{n^2}\alpha )(co{s^2}\alpha  - si{n^2}\alpha )\\ = \;co{s^2}\alpha -si{n^2}\alpha  = cos2\alpha .\end{array}\)

Chọn A

+) Xét \(\beta  =  - \alpha \), khi đó:

\(\begin{array}{l}cos\beta  = cos\left( {-{\rm{ }}\alpha } \right) = cos\alpha ;\\sin\beta  = sin\left( {-{\rm{ }}\alpha } \right) = -sin\alpha  \Leftrightarrow sin\alpha  = -sin\beta .\end{array}\)

Do đó A thỏa mãn.

Đáp án: A

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

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

\(K=\frac{sin\left(\frac{a+b}{2}\right)}{cos\left(\frac{a+b}{2}\right)}=tan\left(\frac{a+b}{2}\right)\)

a,

\(\begin{array}{l}\cos \left( {\alpha  - b} \right) + \cos \left( {\alpha  + \beta } \right)\\ = \cos \alpha \cos \beta  + \sin \alpha sin\beta  + \cos \alpha \cos \beta  - \sin \alpha sin\beta \\ = 2\cos \alpha \cos \beta \end{array}\)

\(\begin{array}{l}\cos \left( {\alpha  - b} \right) - \cos \left( {\alpha  + \beta } \right)\\ = \cos \alpha \cos \beta  + \sin \alpha sin\beta  - \cos \alpha \cos \beta  + \sin \alpha sin\beta \\ = 2\sin \alpha sin\beta \end{array}\)

b,

\(\begin{array}{l}\sin \left( {\alpha  - \beta } \right) - \sin \left( {\alpha  + \beta } \right)\\ = \sin \alpha \cos \beta  - \cos \alpha sin\beta  - \sin \alpha \cos \beta  - \cos \alpha sin\beta \\ =  - 2\cos \alpha sin\beta \end{array}\)

\(\begin{array}{l}\sin \left( {\alpha  - \beta } \right) + \sin \left( {\alpha  + \beta } \right)\\ = \sin \alpha \cos \beta  - \cos \alpha sin\beta  + \sin \alpha \cos \beta  + \cos \alpha sin\beta \\ = 2\sin \alpha \cos \beta \end{array}\)

\(\dfrac{sin\left(a-b\right)}{sina.sinb}+\dfrac{sin\left(b-c\right)}{sinb.sinc}+\dfrac{sin\left(c-a\right)}{sinc.sina}\)

\(=\dfrac{sina.cosb-cosa.sinb}{sina.sinb}+\dfrac{sinb.cosc-cosb.sinc}{sinb.sinc}+\dfrac{sinc.cosa-cosc.sina}{sina.sinc}\)

\(=\dfrac{cosb}{sinb}-\dfrac{cosa}{sina}+\dfrac{cosc}{sincc}-\dfrac{cosb}{sinb}+\dfrac{cosa}{sina}-\dfrac{cosc}{sincc}\)

\(=0\)

\(sina+sinb=2sin\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=\frac{\sqrt{2}}{2}\)

\(\Rightarrow sin\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=\frac{\sqrt{2}}{4}\) (1)

\(cosa+cosb=2cos\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=\frac{\sqrt{6}}{2}\)

\(\Rightarrow cos\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=\frac{\sqrt{6}}{4}\) (2)

(1); (2) \(\Rightarrow tan\left(\frac{a+b}{2}\right)=\frac{\sqrt{3}}{3}\) \(\Rightarrow tan\left(a+b\right)=\sqrt{3}\) \(\Rightarrow a+b=60^0\)

\(\Rightarrow sin\left(a+b\right)=sin\left(60^0\right)=\frac{\sqrt{3}}{2}\)

1.

Ý tưởng thế này: nhìn vế trái phần đáp án có \(tan\left(a+b\right)\) nên cần biến đổi giả thiết xuất hiện \(sin\left(a+b\right)\) , vậy ta làm như sau:

\(sina.cos\left(a+b\right)=sin\left(a+b-a\right)\)

\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b\right).cosa-cos\left(a+b\right).sina\)

\(\Leftrightarrow2sina.cos\left(a+b\right)=sin\left(a+b\right).cosa\)

\(\Rightarrow2tana=tan\left(a+b\right)\)

2.

Đây là 1 dạng cơ bản, nhìn vào lập tức cần ghép x với 3x (đơn giản vì \(\frac{x+3x}{2}=2x\))

\(A=\frac{sin3x-sinx+cos2x}{cosx-cos3x+sin2x}=\frac{2cos2x.sinx+cos2x}{2sin2x.sinx+sin2x}=\frac{cos2x\left(2sinx+1\right)}{sin2x\left(2sinx+1\right)}\)

\(=\frac{cos2x}{sin2x}=cot2x\)