a: \(\sin\left(a+b\right)\cdot\sin\left(a-b\right)\)

\(=\frac12\cdot\left\lbrack cos\left(a+b-a+b\right)-cos\left(a+b+a-b\right)\right\rbrack\)

\(=\frac12\left\lbrack cos2b-cos2a\right\rbrack=\frac12\cdot\left\lbrack2\cdot cos^2b-1-\left(2\cdot cos^2a-1\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack2\cdot cos^2b-2\cdot cos^2a\right\rbrack=cos^2b-cos^2a\)

\(=\left(1-\sin^2b\right)-\left(1-\sin^2a\right)=\sin^2a-\sin^2b\)

b: \(4\cdot\sin\left(x+\frac{\pi}{3}\right)\cdot\sin\left(x-\frac{\pi}{3}\right)\)

\(=4\cdot\frac12\cdot\left\lbrack cos\left(x+\frac{\pi}{3}-x+\frac{\pi}{3}\right)-cos\left(x+\frac{\pi}{3}+x+\frac{\pi}{3}\right)\right\rbrack\)

\(=2\cdot\left\lbrack cos\left(\frac23\pi\right)-cos2x\right\rbrack=2\cdot\left\lbrack-\frac12-cos2x\right\rbrack=-1-2\cdot cos2x\)

\(=-1-2\cdot\left(1-2\cdot\sin^2x\right)=-1-2+4\cdot\sin^2x=4\cdot\sin^2x-3\)

c: \(\sin\left(x+\frac{\pi}{4}\right)-\sin\left(x-\frac{\pi}{4}\right)\)

\(=\sin x\cdot cos\left(\frac{\pi}{4}\right)+cosx\cdot\sin\left(\frac{\pi}{4}\right)-\left\lbrack\sin x\cdot cos\left(\frac{\pi}{4}\right)-cosx\cdot\sin\left(\frac{\pi}{4}\right)\right\rbrack\)

\(=\frac{\sqrt2}{2}\cdot\sin x+\frac{\sqrt2}{2}\cdot cosx-\frac{\sqrt2}{2}\cdot\sin x+\frac{\sqrt2}{2}\cdot cosx=\frac{\sqrt2}{2}\cdot2\cdot cosx=\sqrt2\cdot cosx\)

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

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

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

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

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

b: \(\frac{\tan a+\tan b}{\cot a+cotb}\)

\(=\frac{\tan a+\tan b}{\frac{1}{\tan a}+\frac{1}{\tan b}}=\frac{\tan a+\tan b}{\frac{\tan a+\tan b}{\tan a\cdot\tan b}}=\tan a\cdot\tan b\)

c: \(\frac{\tan^2a-\tan^2b}{\tan^2a\cdot\tan^2b}\)

\(=\left(\frac{\sin^2a}{cos^2a}-\frac{\sin^2b}{cos^2b}\right):\left(\frac{\sin^2a}{cos^2a}\cdot\frac{\sin^2b}{cos^2b}\right)\)

\(=\frac{\left(\sin^2a\cdot cos^2b\right)-\sin^2b\cdot cos^2a}{cos^2a\cdot cos^2b}:\frac{\sin^2a\cdot\sin^2b}{cos^2a\cdot cos^2b}\)

\(=\frac{\left(\sin a\cdot cosb-sinb\cdot cosa\right)\left(\sin a\cdot cosb+\sin b\cdot cosa\right)}{\sin^2a\cdot\sin^2b}\)

\(=\frac{\sin\left(a-b\right)\cdot\sin\left(a+b\right)}{\sin^2a\cdot\sin^2b}=\frac{\frac12\cdot\left\lbrack cos\left(a-b-a-b\right)-cos\left(a-b+a+b\right)\right\rbrack}{\sin^2a\cdot sin^2b}\)

\(=\frac{\frac12\cdot\left\lbrack cos\left(-2b\right)-cos2a\right\rbrack}{\sin^2a\cdot\sin^2b}=\frac{\frac12\cdot\left\lbrack cos2b-cos2a\right\rbrack}{\sin^2a\cdot\sin^2b}=\frac{\frac12\cdot\left\lbrack1-2\cdot\sin^2b-1+2\cdot\sin^2a\right\rbrack}{\sin^2a\cdot\sin^2b}\)

\(=\frac{\sin^2a-\sin^2b}{\sin^2a\cdot\sin^2b}\)

MN K BT?

Ta có : A+B+C= 180
=>sin(A+B)/2 = sin(180/2 - C/2) = cosC/2
ttcó: sinC/2 = cos(A+B)/2
=> sA+sB+sC =2cosC/2*cos(A-B)/2 + 2cos(A+B)/2*cosC/2
=2cosC/2
=4cosA/2cosB/2cosC/2

a: \(A=2\left(\sin^6x+cos^6x\right)-3\cdot\left(\sin^4x+cos^4x\right)\)

\(=2\cdot\left\lbrack\left(\sin^2x+cos^2x\right)^3-3\cdot\sin^2x\cdot cos^2x\cdot\left(\sin^2x+cos^2x\right)\right\rbrack-3\cdot\left\lbrack\left(sin^2x+cos^2x\right)^2-2\cdot\sin^2x\cdot cos^2x\right\rbrack\)

\(=2\left\lbrack1-3\cdot sin^2x\cdot cos^2x\right\rbrack-3\cdot\left\lbrack1-2\cdot\sin^2x\cdot cos^2x\right\rbrack\)

\(=2-6\cdot\sin^2x\cdot cos^2x-3+6\cdot\sin^2x\cdot cos^2x\)

=2-3

=-1

c: \(C=\frac{\sin^2x}{1+\cot x}+\frac{cos^2x}{1+\tan x}+\sin x\cdot cosx\)

\(=\frac{\sin^2x}{1+\frac{cosx}{\sin x}}+\frac{cos^2x}{1+\frac{\sin x}{cosx}}+\sin x\cdot cosx=\sin^2x:\frac{\sin x+cosx}{\sin x}+cos^2x:\frac{\sin x+cosx}{cosx}+\sin x\cdot cosx\)

\(=\frac{\sin^3x+cos^3x}{\sin x+cosx}+\sin x\cdot cosx\)

\(=\frac{\left(\sin x+cosx\right)\left(\sin^2x-\sin x\cdot cosx+cos^2x\right)}{\sin x+cosx}+\sin x\cdot cosx\)

\(=\sin^2x-\sin x\cdot cosx+cos^2x+\sin x\cdot cosx\)

\(=\sin^2x+cos^2x=1\)

d: \(D=\frac{\cot^2x-cos^2x}{cot^2x}+\frac{\sin x\cdot cosx}{\cot x}\)

\(=\left(\frac{cos^2x}{\sin^2x}-cos^2x\right):\frac{cos^2x}{sin^2x}+\frac{\sin x\cdot cosx}{\frac{cosx}{\sin x}}\)

\(=cos^2x\left(\frac{1}{\sin^2x}-1\right)\cdot\frac{\sin^2x}{cos^2x}+\frac{\sin x\cdot cosx\cdot\sin x}{cosx}\)

\(=\frac{1-\sin^2x}{\sin^2x}\cdot\sin^2x+\sin^2x=1-\sin^2x+\sin^2x=1\)

\(sin^3A.sin\left(B-C\right)=sin^2A.sinA.sin\left(B-C\right)\)

\(=sin^2A.sin\left(B+C\right).sin\left(B-C\right)=-\frac{1}{2}sin^2A\left(cos2B-cos2C\right)\)

\(=-\frac{1}{2}sin^2A\left(1-2sin^2B-1+2sin^2C\right)=sin^2A.sin^2B-sin^2A.sin^2C\)

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

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

\(=-cos^2\left(a+b\right)+cos\left(a+b\right).cos\left(a-b\right)\)

\(=cos\left(a+b\right)\left[cos\left(a-b\right)-cos\left(a+b\right)\right]\)

\(=2sina.sinb.cos\left(a+b\right)\)