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Lời giải:
a)
\(\frac{\sin a}{1+\cos a}+\cot a=\frac{\sin a}{1+\cos a}+\frac{\cos a}{\sin a}=\frac{\sin ^2a+\cos^2a+\cos a}{\sin a(1+\cos a)}\)
\(=\frac{1+\cos a}{\sin a(1+\cos a)}=\frac{1}{\sin a}\) (đpcm)
b)
\(\frac{1}{\cos a}-\frac{\cos a}{1+\sin a}=\frac{1+\sin a-\cos ^2a}{\cos a(1+\sin a)}=\frac{(1-\cos ^2a)+\sin a}{\cos a(\sin a+1)}\)
\(=\frac{\sin^2a+\sin a}{\cos a(\sin a+1)}=\frac{\sin a(\sin a+1)}{\cos a(\sin a+1)}=\frac{\sin a}{\cos a}=\tan a\) (đpcm)
c)
\(\frac{\tan a-\sin a}{\sin ^3a}=\frac{\frac{\sin a}{\cos a}-\sin a}{\sin ^3a}=\frac{\frac{1}{\cos a}-1}{\sin ^2a}=\frac{1-\cos a}{\cos a\sin ^2a}=\frac{1-\cos a}{\cos a(1-\cos ^2a)}=\frac{1}{\cos a(1+\cos a)}\)
d)
\(\frac{\sin a+\cos a-1}{\sin a-\cos a+1}=\frac{(\sin a+\cos a-1)(\sin a+\cos a+1)}{(\sin a-\cos a+1)(\sin a+\cos a+1)}=\frac{(\sin a+\cos a)^2-1}{(\sin a+1)^2-\cos ^2a}\)
\(=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-\cos ^2a}=\frac{1+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-(1-\sin ^2a)}\)
\(=\frac{2\sin a\cos a}{2\sin ^2a+2\sin a}=\frac{2\sin a\cos a}{2\sin a(\sin a+1)}=\frac{\cos a}{1+\sin a}\) (đpcm)
Mấu chốt trong các bài này là việc sử dụng công thức $\sin ^2a+\cos ^2a=1$
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\(P=\sin^2x+cos\left(\frac{\pi}{3}-x\right)cos\left(\frac{\pi}{3}+x\right)\)
\(=\sin^2x+cos^2\left(\frac{\pi}{3}\right)-sin^2x\)
\(=\cos^2\left(\frac{\pi}{3}\right)=\frac{1}{4}\)
=> P không phụ thuộc vào x
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\(\frac{sin2a-cos2a}{sin2a+cos2a}=\frac{\left(sin2a-cos2a\right)^2}{\left(sin2a+cos2a\right)\left(sin2a-cos2a\right)}\)
\(=\frac{sin^22a+cos^22a-2sin2a.cos2a}{sin^22a-cos^22a}=\frac{1-sin4a}{-cos4a}\)
\(=-\frac{1}{cos4a}+\frac{sin4a}{cos4a}=tan4a-\frac{1}{cos4a}\)
\(c\text{os}3a=4cosa.c\text{os}\left(\frac{\pi}{3}-a\right).c\text{os}\left(\frac{\pi}{3}+a\right)\)
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Sử dụng công thức \(cosx.cosy=\frac{1}{2}\left(cos\left(x+y\right)+cos\left(x-y\right)\right)\) với 2 cái cos cuối cùng
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\(\frac{1+cosx-sinx}{1-cosx-sinx}=\frac{1+2cos^2\frac{x}{2}-1-2sin\frac{x}{2}.cos\frac{x}{2}}{1-1+2sin^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}=\frac{2cos^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}{2sin^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}\)
\(=\frac{-2cos\frac{x}{2}\left(sin\frac{x}{2}-cos\frac{x}{2}\right)}{2sin\frac{x}{2}\left(sin\frac{x}{2}-cos\frac{x}{2}\right)}=\frac{-cos\frac{x}{2}}{sin\frac{x}{2}}=-cot\frac{x}{2}\)
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\(A=\frac{sin2x+sin6x+cos7x+cos3x}{sin3x-sinx}=\frac{2sin4x.cos2x+2cos5x.cos2x}{2cos2x.sinx}=\frac{2cos2x\left(sin4x+cos5x\right)}{2cos2x.sinx}\)
\(=\frac{sin4x+cos5x}{sinx}\)
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\(3sin^4x-\left(1-sin^2x\right)^2=\frac{1}{2}\Leftrightarrow3sin^4x-\left(sin^4x-2sin^2x+1\right)=\frac{1}{2}\)
\(\Leftrightarrow2sin^4x+2sin^2x-\frac{3}{2}=0\) \(\Rightarrow\left[{}\begin{matrix}sin^2x=\frac{1}{2}\\sin^2x=-\frac{3}{2}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow cos^2x=1-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow B=\left(\frac{1}{2}\right)^2+3\left(\frac{1}{2}\right)^2=1\)
\(4sin^4x+3\left(1-sin^2x\right)^2=\frac{7}{4}\Leftrightarrow4sin^4x+3\left(sin^4x-2sin^2x+1\right)=\frac{7}{4}\)
\(\Leftrightarrow7sin^4x-6sin^2x+\frac{5}{4}=0\Rightarrow\left[{}\begin{matrix}sin^2x=\frac{1}{2}\Rightarrow cos^2x=\frac{1}{2}\\sin^2x=\frac{5}{14}\Rightarrow cos^2x=\frac{9}{14}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}C=3\left(\frac{1}{2}\right)^2+4\left(\frac{1}{2}\right)^2=\frac{7}{4}\\C=3\left(\frac{5}{14}\right)^2+4\left(\frac{9}{14}\right)^2=\frac{57}{28}\end{matrix}\right.\)
\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)
\(=1-\frac{1}{2}\left(2sinx.cosx\right)^2=1-\frac{1}{2}sin^22x\)
\(=1-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2}cos4x\right)=\frac{3}{4}+\frac{1}{4}cos4x\)