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\(A=\frac{\cos57}{\cos57}+\frac{\cot58}{\cot58}-2\left(1+1\right)\)\()\)
=1+1-4
=-2
\(\cos a\times\sin b=-\dfrac{1}{2}\left[\sin\left(a-b\right)-\sin\left(a+b\right)\right]\)
\(=-\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{2}{3}\right)=\dfrac{-1}{2}\times1=-\dfrac{1}{2}\)
Vì A+B+C=180^{\circ}A+B+C=180∘ nên V T=\dfrac{\sin ^{3} \dfrac{B}{2}}{\cos \left(\dfrac{180^{\circ}-B}{2}\right)}+\dfrac{\cos ^{3} \dfrac{B}{2}}{\sin \left(\dfrac{180^{\circ}-B}{2}\right)}-\dfrac{\cos \left(180^{\circ}-B\right)}{\sin B} \cdot \tan BVT=cos(2180∘−B)sin32B+sin(2180∘−B)cos32B−sinBcos(180∘−B)⋅tanB.
V T=\dfrac{\sin ^{3} \dfrac{B}{2}}{\cos \left(\dfrac{180^{\circ}-B}{2}\right)}+\dfrac{\cos ^{3} \dfrac{B}{2}}{\sin \left(\dfrac{180^{\circ}-B}{2}\right)}-\dfrac{\cos \left(180^{\circ}-B\right)}{\sin B} \cdot \tan BVT=cos(2180∘−B)sin32B+sin(2180∘−B)cos32B−sinBcos(180∘−B)⋅tanB =\dfrac{\sin ^{3} \dfrac{B}{2}}{\sin \dfrac{B}{2}}+\dfrac{\cos ^{3} \dfrac{B}{2}}{\cos \dfrac{B}{2}}-\dfrac{-\cos B}{\sin B} \cdot \tan B=\sin ^{2} \dfrac{B}{2}+\cos ^{2} \dfrac{B}{2}+1=2=V P=sin2Bsin32B+cos2Bcos32B−sinB−cosB⋅tanB=sin22B+cos22B+1=2=VP
Suy ra điều phải chứng minh.
Lời giải:
$-1=\cos (a-b)=\cos a\cos b+\sin a\sin b$
$\Rightarrow -2=2\cos a\cos b+2\sin a\sin b$
Mà: $2=\cos ^2a+\sin ^2a+\cos ^2b+\sin ^2b$
Cộng theo vế 2 đẳng thức trên lại suy ra:
$0=(\cos a+\cos b)^2+(\sin a+\sin b)^2$
$\Rightarrow \cos a=-\cos b; \sin a=-\sin b$
$\frac{1}{2}=\sin (a+b)=\sin a\cos b-\cos a\sin b$
$=(-\sin b)(-\cos a)-\cos a\sin b=0$ (vô lý)
DO đó không tính được $\cos a\cos b$
a, \(sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2cos^2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2\cdot\left[1+cos2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right]=0\)
\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-1-cos\left(\dfrac{\pi}{2}-x\right)=0\)
\(\Leftrightarrow sin\dfrac{s}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x-sinx=0\)
\(\Leftrightarrow sinx\cdot\left(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\text{ (1) }\\sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\text{ (2) }\end{matrix}\right.\)
(1) : \(sinx=0\Leftrightarrow x=k\pi\left(k\in Z\right)\)
(2) : \(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-cos\dfrac{x}{2}\cdot2sin\dfrac{x}{2}\cdot cos\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot cos^2\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot\left(1-sin^2\dfrac{x}{2}\right)-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}+2sin^3\dfrac{x}{2}-1=0\)
\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\pi+k4\pi\left(k\in Z\right)\)
b, \(tanx-3cotx=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cos}{sinx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{sinx-cosx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow sin^2x-3cos^2x=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)
\(\Leftrightarrow\left(sinx-\sqrt{3}\cdot cosx\right)\cdot\left(sinx+\sqrt{3}\cdot cosx\right)=4\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)
\(\Leftrightarrow\left(sinx+\sqrt{3}\cdot cosx\right)\cdot\left[\left(sinx-\sqrt{3}\cdot cosx\right)-4sinx\cdot cosx\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}\cdot cosx=0\text{ (1) }\\sinx-\sqrt{3}\cdot cosx-4sinx\cdot cosx=0\text{ (2) }\end{matrix}\right.\)
(1) : \(sinx+\sqrt{3}\cdot cosx=0\)
\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=0\)
\(\Leftrightarrow cos\dfrac{\pi}{3}\cdot sinx+sin\dfrac{\pi}{3}\cdot cosx=0\)
\(\Leftrightarrow sin\cdot\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\left(k\in Z\right)\)
(2) : \(sinx-\sqrt{3}cosx-4sinx\cdot cosx=0\)
\(\Leftrightarrow sinx-\sqrt{3}cos=2sin2x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cos2=sin2x\)
\(\Leftrightarrow cos\dfrac{\pi}{3}-sinx-sin\dfrac{\pi}{3}\cdot cosx=sin2x\)
\(\Leftrightarrow sin\cdot\left(x-\dfrac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=2x+k2\pi\\x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\left(k\in Z\right)\end{matrix}\right.\)
a)\(sin^2\left(180^o-\alpha\right)+tan^2\left(180-\alpha\right).tan^2\left(270^o+\alpha\right)\)\(+sin\left(90^o+\alpha\right)cos\left(\alpha-360^o\right)\)
\(=sin^2\alpha+tan^2\alpha.cot^2\alpha+cos\alpha cos\alpha\)
\(=sin^2\alpha+cos^2\alpha+\left(tan\alpha cot\alpha\right)^2=1+1=2\).
\(\dfrac{cos\left(\alpha-180^o\right)}{sin\left(180^o-\alpha\right)}+\dfrac{tan\left(\alpha-180^o\right)cos\left(180^o+\alpha\right)sin\left(270^o+\alpha\right)}{tan\left(270^o+\alpha\right)}\)
\(=\dfrac{cos\left(180^o-\alpha\right)}{sin\left(180^o-\alpha\right)}+\dfrac{-tan\left(180^o-\alpha\right).cos\alpha.sin\left(90^o+\alpha\right)}{-tan\left(90^o+\alpha\right)}\)
\(=tan\left(180^o-\alpha\right)+\dfrac{tan\alpha.cos\alpha.cos\alpha}{cot\alpha}\)
\(=-tan\alpha+tan^2\alpha cos^2\alpha\)
\(=tan\alpha\left(-1+tan\alpha cos^2\alpha\right)\)
\(=tan\alpha\left(sin\alpha cos\alpha-1\right)\).
a) ta có : \(A=\dfrac{sin33}{cos57}+\dfrac{tan32}{cot58}-2\left(sin20.cos70+cos20.sin70\right)\)
\(\Leftrightarrow A=\dfrac{sin33}{cos\left(90-33\right)}+\dfrac{tan32}{cot\left(90-32\right)}-2\left(sin20.cos\left(90-20\right)+cos20.sin\left(90-20\right)\right)\)
\(\Leftrightarrow A=\dfrac{sin33}{sin33}+\dfrac{tan32}{tan32}-2\left(sin20.sin20+cos20.cos20\right)\)\(\Leftrightarrow A=1+1-2\left(sin^220+cos^220\right)=1+1-2=0\)
b) sữa đề chút nha
ta có : \(B=\dfrac{sin^215+sin^275-sin^212-sin^278}{cos^213+cos^277+cos^21+cos^289}+\dfrac{2tan55}{cot35}\)
\(\Leftrightarrow B=\dfrac{sin^215+sin^2\left(90-15\right)-sin^212-sin^2\left(90-12\right)}{cos^213+cos^2\left(90-13\right)+cos^21+cos^2\left(90-1\right)}+\dfrac{2tan\left(90-35\right)}{cot35}\)
\(\Leftrightarrow B=\dfrac{sin^215+cos^215-sin^212-cos^212}{cos^213+sin^213+cos^21+sin^21}+\dfrac{2cot35}{cot35}\) \(\Leftrightarrow B=\dfrac{sin^215+cos^215-\left(sin^212+cos^212\right)}{cos^213+sin^213+cos^21+sin^21}+\dfrac{2cot35}{cot35}\)
\(\Leftrightarrow B=\dfrac{1-1}{cos^213+sin^213+cos^21+sin^21}+2=0+2=2\)