Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)
\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)
Với 0 < α < :
a) sin(α - π) < 0; b) cos( - α) < 0;
c) tan(α + π) > 0; d) cot(α + ) < 0
Vì \(0< \alpha< \dfrac{\pi}{2}\) nên các giá trị lượng giác của \(\alpha\) đều dương.
a) \(sin\left(\alpha-\pi\right)=-sin\left(\pi-\alpha\right)=-sin\alpha< 0\).
b) \(cos\left(\dfrac{3\pi}{2}-\alpha\right)=cos\left(\dfrac{3\pi}{2}-\alpha-2\pi\right)=cos\left(-\dfrac{\pi}{2}-\alpha\right)\)
\(=cos\left(\dfrac{\pi}{2}+\alpha\right)=-sin\alpha< 0\).
c) \(tan\left(\alpha+\pi\right)=tan\alpha>0\).
d) \(cot\left(\alpha+\dfrac{\pi}{2}\right)=-tan\alpha< 0\).
bài 1) ta có : \(G=cos\left(\alpha-5\pi\right)+sin\left(\dfrac{-3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)
\(G=cos\left(\alpha-\pi\right)+sin\left(\dfrac{\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\)
\(G=cos\left(\pi-\alpha\right)+sin\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)-tan\left(\pi+\alpha-\dfrac{\pi}{2}\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\) \(G=cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\dfrac{\pi}{2}-\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)=2cos\alpha+1\) bài 2) ta có : \(H=cot\left(\alpha\right).cos\left(\alpha+\dfrac{\pi}{2}\right)+cos\left(\alpha\right)-2sin\left(\alpha-\pi\right)\) \(H=cot\left(\alpha\right).cos\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)+cos\left(\alpha\right)+2sin\left(\pi-\alpha\right)\) \(H=-cot\left(\alpha\right).sin\left(\alpha\right)+cos\left(\alpha\right)+2sin\left(\alpha\right)\)
\(H=-cos\alpha+cos\alpha+2sin\alpha=2sin\alpha\)
\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)
\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x+sinx\)
\(=1+sinx\)
\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)
\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)
\(=1+cosx\)
G = \(cos\left(a+\pi-6\text{}\text{}\pi\right)+sin\left(-2\pi+\dfrac{\pi}{2}+a\right)-tan\left(\dfrac{\pi}{2}+a\right)\cdot cot\left(\pi+\dfrac{\pi}{2}-a\right)\)
= \(cos\left(a+\pi\right)+sin\left(\dfrac{\pi}{2}+a\right)-tan\left(\dfrac{\pi}{2}+a\right)\cdot cot\left(\dfrac{\pi}{2}-a\right)\)
= \(-cosa+cosa-\left(-cota\cdot tana\right)=1\)
Lời giải:
Theo công thức lượng giác:
\(F=\sin (\pi +a)-\cos (\frac{\pi}{2}-a)+\cot (2\pi -a)+\tan (\frac{3\pi}{2}-a)\)
\(=-\sin a-\sin a+\cot (\pi -a)+\tan (\frac{\pi}{2}-a)\)
\(=-2\sin a-\cot a+\cot a=-2\sin a\)
1.
\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)
\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)
\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)
\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)
\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)
\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)
\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)
\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)
2.
\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)
\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)
\(=cos3x\)
\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)
\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)
\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)
\(=tan^2a.cot^2b-2\)
Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)
Do \(\pi< \alpha< \dfrac{3\pi}{2}\) nên \(sin\alpha,cos\alpha< 0;tan\alpha,cot\alpha< 0\).
\(cos\left(\alpha-\dfrac{\pi}{2}\right)=cos\left(\dfrac{\pi}{2}-\alpha\right)=sin\alpha< 0\).
\(sin\left(\dfrac{\pi}{2}+\alpha\right)=cos\alpha< 0\).
\(tan\left(\dfrac{3\pi}{2}-\alpha\right)=tan\left(\dfrac{3\pi}{2}-\alpha-2\pi\right)\)\(=tan\left(-\dfrac{\pi}{2}-\alpha\right)\)\(=-tan\left(\dfrac{\pi}{2}+\alpha\right)=cot\left(\alpha\right)>0\).
\(cot\left(\alpha+\pi\right)=cot\left(\alpha\right)>0\).