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\(A=\dfrac{\dfrac{3sina}{sina}-\dfrac{cosa}{sina}}{\dfrac{2sina}{sina}+\dfrac{cosa}{sina}}=\dfrac{3-cota}{2+cota}=\dfrac{3-3}{2+3}=0\)
\(B=\dfrac{\dfrac{sin^2a}{sin^2a}-\dfrac{3sina.cosa}{sin^2a}+\dfrac{2}{sin^2a}}{\dfrac{2sin^2a}{sin^2a}+\dfrac{sina.cosa}{sin^2a}+\dfrac{cos^2a}{sin^2a}}=\dfrac{1-3cota+2\left(1+cot^2a\right)}{2+cota+cot^2a}=\dfrac{1-3.3+2\left(1+3^2\right)}{2+3+3^2}=...\)
a. \(A=\dfrac{3sin\alpha-cos\alpha}{2sin\alpha+cos\alpha}=\dfrac{3\dfrac{sin\alpha}{cos\alpha}-1}{2\dfrac{sin\alpha}{cos\alpha}+1}=\dfrac{3.\dfrac{1}{3}-1}{2.\dfrac{1}{3}+1}=0\)
b.\(B=\dfrac{sin^2\alpha-3sin\alpha.cos\alpha+2}{2sin^2\alpha+sin\alpha.cos\alpha+cos^2\alpha}\)\(=\dfrac{1-\dfrac{3cos\alpha}{sin\alpha}+\dfrac{2}{sin^2\alpha}}{2+\dfrac{cos\alpha}{sin\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}}=\dfrac{1-3.3+\dfrac{2}{sin^2\alpha}}{2+3+3^2}\)
Mà \(\dfrac{cos\alpha}{sin\alpha}=3,cos^2\alpha+sin^2\alpha=1\Rightarrow sin^2\alpha=\dfrac{1}{10}\)
\(B=\dfrac{1-3.3+\dfrac{2}{\dfrac{1}{10}}}{2+3+3^2}=\dfrac{6}{7}\)
1,\(A=3\left(sin^4x+cos^4x\right)-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)\)
\(=3\left(sin^4x+cos^4x\right)-2\left(sin^4x-sin^2x.cos^4x+cos^4x\right)\)
\(=sin^4x+2sin^2x.cos^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\)
Vậy...
2,\(B=cos^6x+2sin^4x\left(1-sin^2x\right)+3\left(1-cos^2x\right)cos^4x+sin^4x\)
\(=-2cos^6x+3sin^4x-2sin^6x+3cos^4x\)
\(=-2\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)
\(=-2\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3\left(cos^4x+sin^4x\right)\)\(=cos^4x+sin^4x+2sin^2x.cos^2x=1\)
Vậy...
3,\(C=\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}\right)\right]+\dfrac{1}{2}\left[cos\left(-\dfrac{7\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)
\(=cos\left(-\dfrac{7\pi}{12}\right)+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x+\dfrac{11\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)+cos\left(2x-\dfrac{\pi}{12}+\pi\right)\right]\)
\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}+\dfrac{1}{2}\left[cos\left(2x-\dfrac{\pi}{12}\right)-cos\left(2x-\dfrac{\pi}{12}\right)\right]\)\(=\dfrac{-\sqrt{6}+\sqrt{2}}{4}\)
Vậy...
4, \(D=cos^2x+\left(-\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx\right)^2+\left(-\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right)^2\)
\(=cos^2x+\dfrac{1}{4}cos^2x+\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x+\dfrac{1}{4}cos^2x-\dfrac{\sqrt{3}}{4}cosx.sinx+\dfrac{3}{4}sin^2x\)
\(=\dfrac{3}{2}\left(cos^2x+sin^2x\right)=\dfrac{3}{2}\)
Vậy...
5, Xem lại đề
6,\(F=-cosx+cosx-tan\left(\dfrac{\pi}{2}+x\right).cot\left(\pi+\dfrac{\pi}{2}-x\right)\)
\(=tan\left(\pi-\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=tan\left(\dfrac{\pi}{2}-x\right).cot\left(\dfrac{\pi}{2}-x\right)\)\(=cotx.tanx=1\)
Vậy...
a)
\((\sin x+\cos x)^2=\sin ^2x+2\sin x\cos x+\cos ^2x\)
\(=(\sin ^2x+\cos ^2x)+2\sin x\cos x=1+2\sin x\cos x\)
b)
\(\sin ^4x+\cos ^4x=\sin ^4x+2\sin ^2x\cos ^2x+\cos ^4x-2\sin ^2\cos ^2x\)
\(=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x\)
\(=1-2\sin ^2x\cos ^2x\)
c)
\(\tan ^2x-\sin ^2x=(\frac{\sin x}{\cos x})^2-\sin ^2x\)
\(=\sin ^2x\left(\frac{1}{\cos ^2x}-1\right)=\sin ^2x. \frac{1-\cos ^2x}{\cos ^2x}=\sin ^2x.\frac{\sin ^2x}{\cos ^2x}\)
\(=\sin ^2x\left(\frac{\sin x}{\cos x}\right)^2=\sin ^2x\tan ^2x\)
d)
\(\sin ^6x+\cos ^6x=(\sin ^2x)^3+(\cos ^2x)^3\)
\(=(\sin ^2x+\cos ^2x)(\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x)\)
\(=\sin ^4x-\sin ^2x\cos ^2x+\cos ^4x\)
\(=(\sin ^4x+\cos ^4x)-\sin ^2x\cos ^2x=1-2\sin ^2x\cos ^2x-\sin ^2x\cos ^2x\)
\(=1-3\sin ^2x\cos ^2x\) (theo kq phần b)
e)
\(\sin x\cos x(1+\tan x)(1+\cot x)=\sin x\cos x(1+\frac{\sin x}{\cos x})(1+\frac{\cos x}{\sin x})\)
\(=\sin x\cos x.\frac{\cos x+\sin x}{\cos x}.\frac{\sin x+\cos x}{\sin x}\)
\(=(\sin x+\cos x)^2=\sin ^2x+\cos ^2x+2\sin x\cos x\)
\(=1+2\sin x\cos x\)
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P/s: Nói chung cứ bám vào công thức \(\sin ^2x+\cos ^2x=1\)
\(A=sin^6x+sin^4x.cos^2x+2\left(sin^2x.cos^4x+sin^4x.cos^2x\right)+cos^4x\)
\(=sin^4x\left(sin^2x+cos^2x\right)+2sin^2x.cos^2x\left(sin^2x+cos^2x\right)+cos^4x\)
\(=sin^4x+2sin^2x.cos^2x+cos^4x\)
\(=\left(sin^2x+cos^2x\right)^2=1\)
d.
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^4x\)
\(tan^4x-3tan^2x-4tanx-3=0\)
\(\Leftrightarrow\left(tan^2x+tanx+1\right)\left(tan^2x-tanx-3\right)=0\)
\(\Leftrightarrow tan^2x-tanx-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1-\sqrt{13}}{2}\\tanx=\frac{1+\sqrt{13}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(\frac{1-\sqrt{13}}{2}\right)+k\pi\\x=arctan\left(\frac{1+\sqrt{13}}{2}\right)+k\pi\end{matrix}\right.\)
\(B=cos^2x.cot^2x+cos^2x-cot^2x+2\left(sin^2x+cos^2x\right)\)
\(=cos^2x\left(cot^2x+1\right)-cot^2x+2\)
\(=\frac{cos^2x}{sin^2x}-cot^2x+1=cot^2x-cot^2x+1=1\)
\(M=cos^4x-sin^4x+cos^4x+sin^2x.cos^2x+3sin^2x\)
\(=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+cos^2x\left(cos^2x+sin^2x\right)+3sin^2x\)
\(=cos^2x-sin^2x+cos^2x+3sin^2x\)
\(=2\left(sin^2x+cos^2x\right)=2\)