Chứng minh các đẳng thứ sau:
\(1,sin^8x-cos^8x=-(\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x) \)
2\(sin^2x×cos^4x=\dfrac{1}{16}+\dfrac{1}{32}cos2x-\dfrac{1}{16}cos4x-\dfrac{1}{32}cos6x\)
Chứng minh đẳng thức:
1 ,\(tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)+cot\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)=\dfrac{2}{cosx}\)
2 ,\(sin^8x-cos^8x=-\left(\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x\right)\)
3 ,\(3-4cos2x+cos4x=8sin^4x\)
4 ,\(sin\left(2x+\dfrac{\pi}{3}\right).cos\left(x-\dfrac{\pi}{6}\right)-cos\left(2x+\dfrac{\pi}{3}\right).cos\left(\dfrac{2\pi}{3}-x\right)=cosx\)
5 ,\(\sqrt{3}cos2x+sin2x+sin\left(4x-\dfrac{\pi}{3}\right)=4cos\left(2x-\dfrac{\pi}{6}\right).sin^2\left(x+\dfrac{\pi}{6}\right)\)
1,\(VT=\dfrac{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}+\dfrac{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}\)\(=\dfrac{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)^2+cos^2\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}\)
\(=\dfrac{1}{\dfrac{1}{2}.sin\left(\dfrac{\pi}{2}+x\right)}=\dfrac{2}{cosx}=VP\)
2,\(VT=\left(sin^4x-cos^4x\right)\left(sin^4x+cos^4x\right)=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(=\left(sin^2-cos^2x\right)\left(1-2sin^2x.cos^2x\right)\)\(=-cos2x\left(1-\dfrac{1}{2}sin^22x\right)\)\(=-\dfrac{cos2x\left(2-sin^22x\right)}{2}=-\dfrac{cos2x\left(1+cos^22x\right)}{2}\)
\(VP=-\left(\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x\right)=-\dfrac{7}{8}cos2x-\dfrac{1}{8}\left[4cos^32x-3cos2x\right]=-\dfrac{7}{8}.cos2x-\dfrac{1}{2}cos^32x+\dfrac{3}{8}cos2x\)
\(=-\dfrac{1}{2}cos2x-\dfrac{1}{2}cos^32x=\dfrac{-cos2x\left(1+cos^22x\right)}{2}\)
\(\Rightarrow VT=VP\)(đpcm)
3, \(VT=3-4\left(1-2sin^2x\right)+1-2sin^22x=8sin^2x-2sin^22x=8sin^2x-8.sin^2x.cos^2x=8sin^2x\left(1-cos^2x\right)=8sin^4x=VP\)
4,\(VP=\dfrac{1}{2}\left[sin\left(x+\dfrac{\pi}{2}\right)+sin\left(3x+\dfrac{\pi}{6}\right)\right]-\dfrac{1}{2}\left[cos\left(3x-\dfrac{\pi}{3}\right)+cos\left(x+\pi\right)\right]\)
\(=\dfrac{1}{2}\left(cosx+sin3x.\dfrac{\sqrt{3}}{2}+\dfrac{cos3x}{2}\right)-\dfrac{1}{2}\left(\dfrac{cos3x}{2}+sin3x.\dfrac{\sqrt{3}}{2}-cosx\right)\)
\(=\dfrac{1}{2}.2cosx=cosx=VP\)
5, \(VP=4cos\left(2x-\dfrac{\pi}{6}\right).\left(sinx.\dfrac{\sqrt{3}}{2}+\dfrac{cosx}{2}\right)^2\)\(=cos\left(2x-\dfrac{\pi}{6}\right).\left(sinx.\sqrt{3}+cosx\right)^2\)
\(VT=2.cos\left(2x-\dfrac{\pi}{6}\right)+2.sin\left(2x-\dfrac{\pi}{6}\right).cos\left(2x-\dfrac{\pi}{6}\right)=2cos\left(2x-\dfrac{\pi}{6}\right)\left[1+sin\left(2x-\dfrac{\pi}{6}\right)\right]\)
\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(1+\dfrac{sin2x.\sqrt{3}}{2}-\dfrac{cos2x}{2}\right)\)\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x+cos^2x+sinx.cosx.\sqrt{3}-\dfrac{cos^2x-sin^2x}{2}\right)\)
\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x.\dfrac{3}{2}+sinx.cosx.\sqrt{3}+\dfrac{cos^2x}{2}\right)\)\(=cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x.3+2sinx.cosx.\sqrt{3}+cos^2x\right)\)
\(=cos\left(2x-\dfrac{\pi}{6}\right)\left(sinx.\sqrt{3}+cosx\right)^2\)
\(\Rightarrow VT=VP\) (dpcm)
Giải PT
a1) \(3.\cos4x-2^{ }\cos^23x=1\)
a2) \(2\cos2x-8\cos x+7=\dfrac{1}{\cos x}\)
a3) \(\dfrac{\left(1+\sin x+\cos2x\right)\sin\left(x+\dfrac{\pi}{4}\right)}{1+\tan x}=\dfrac{1}{\sqrt{2}}\cos x\)
a4) \(9\sin x+6\cos x-3\sin2x+\cos2x=8\)
a) Pt \(\Leftrightarrow3.cos4x-\left(cos6x+1\right)=1\)
\(\Leftrightarrow3cos4x-cos6x-2=0\)
Đặt \(t=2x\)
Pttt:\(3cos2t-cos3t-2=0\)
\(\Leftrightarrow3\left(2cos^2t-1\right)-\left(4cos^3t-3cost\right)-2=0\)
\(\Leftrightarrow-4cos^3t+6cos^2t+3cost-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{1+\sqrt{21}}{4}\left(vn\right)\\cost=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=k2\pi\\t=\pm arc.cos\left(\dfrac{1-\sqrt{21}}{4}\right)+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\dfrac{1}{2}.arccos\left(\dfrac{1-\sqrt{21}}{4}\right)+k\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a2) \(2cos2x-8cosx+7=\dfrac{1}{cosx}\) (ĐK: \(x\ne\dfrac{\pi}{2}+k\pi\))
\(\Leftrightarrow2.\left(2cos^2x-1\right)-8cosx+7=\dfrac{1}{cosx}\)
\(\Leftrightarrow2.\left(2cos^2x-1\right)cosx-8cos^2x+7cosx=1\)
\(\Leftrightarrow4cos^3x-8cos^2x+5cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\) (tm) (\(k\in Z\))
Vậy...
a3) Đk: \(x\ne-\dfrac{\pi}{4}+k\pi;x\ne\dfrac{\pi}{2}+k\pi\)
Pt \(\Leftrightarrow\dfrac{\left(1+sinx+1-2sin^2x\right).\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)}{1+\dfrac{sinx}{cosx}}=\dfrac{1}{\sqrt{2}}cosx\)
\(\Leftrightarrow\dfrac{\left(-2sin^2x+sinx+2\right).\left(sinx+cosx\right)cosx}{cosx+sinx}=cosx\)
\(\Leftrightarrow\left(2+sinx-2sin^2x\right).cosx=cosx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\2+sinx-2sin^2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=0\left(ktm\right)\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\) (\(k\in Z\))
Vậy...
a4) Pt \(\Leftrightarrow9sinx+6cosx-6sinx.cosx+1-2sin^2x=8\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sin^2x-9sinx+7\right)=0\)
\(\Leftrightarrow6cosx\left(1-sinx\right)-\left(2sinx-7\right)\left(sinx-1\right)=0\)
\(\Leftrightarrow\left(1-sinx\right)\left(6cosx+2sinx+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\6cosx+2sinx=7\left(vn\right)\end{matrix}\right.\) (\(6cosx+2sinx=7\) vô nghiệm do \(6^2+2^2< 7^2\))
\(\Rightarrow sinx=1\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi;k\in Z\)
Vậy...
Chứng minh
a) \(\dfrac{1+\cos x+\cos2x+\cos3x}{2\cos^2x+\cos x-1}=2\cos x\)
b) \(\cos\dfrac{5x}{2}.\cos\dfrac{3x}{2}+\sin\dfrac{7x}{2}.\sin\dfrac{x}{2}=\cos x.\cos2x\)
a, \(\dfrac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)
\(=\dfrac{1+cos2x+cosx+cos3x}{2cos^2x+cosx-1}\)
\(=\dfrac{2cos^2x+2cos2x.cosx}{cos2x+cosx}\)
\(=\dfrac{2cosx\left(cos2x+cosx\right)}{cos2x+cosx}=2cosx\)
b) \(cos\dfrac{5x}{2}.cos\dfrac{3x}{2}+sin\dfrac{7x}{2}.sin\dfrac{x}{2}\)
\(=cos\dfrac{4x+x}{2}.cos\dfrac{4x-x}{2}+sin\dfrac{4x+3x}{2}.sin\dfrac{4x-3x}{2}\)
\(=\dfrac{1}{2}\left(cos4x+cosx\right)-\dfrac{1}{2}\left(cos4x-cos3x\right)\)
\(=\dfrac{1}{2}\left(cosx+cos3x\right)=\dfrac{1}{2}.2cos2x.cos\left(-x\right)\)\(=cosx.cos2x\)
1. Rút gọn biểu thức \(P=cos^4x-sin^4x\)
\(A.P=cos2x\) \(B.P=\dfrac{3}{4}+\dfrac{1}{4}cos4x\) \(C.P=\dfrac{1}{4}+\dfrac{3}{4}cos4x\) \(D.P=\dfrac{3}{4}-\dfrac{1}{4}cos4x\)
2.Đơn giản biểu thức \(D=sin\left(\dfrac{5\pi}{2}-\alpha\right)+cos\left(13\pi+\alpha\right)-3sin\left(\alpha-5\pi\right)\)
\(A.3sina-2cosa\) \(B.3sina\) \(C.-3sina\) \(D.2cosa+3sina\)
Trắc nghiệm nhưng mong mn trình bày bài làm giúp em để tham khảo với ạ. Em cảm ơn
1.Ý A
\(P=cos^4x-sin^4x=\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=cos2x\)
2. Ý B
\(D=sin\left(\dfrac{5\pi}{2}-\alpha\right)+cos\left(13\pi+\alpha\right)-3sin\left(\alpha-5\pi\right)\)
\(=sin\left(2\pi+\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha+\pi-6\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha+\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\alpha=3sin\alpha\)
Giải các phương trình sau
a) \(sin^6x+cos^6x=cos2x+\dfrac{1}{16}\)
b) \(sin^4\dfrac{x}{2}+cos^4\dfrac{x}{2}=\dfrac{5}{2}-2sinx\)
c) \(cos5xcosx=cos4xcos2x+4-3sin^2x\)
d) \(2cosxcos2x=1+cos2x+cos3x\)
e) \(sin3x+cos2x=2\left(sin2xcosx-1\right)\)
a.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos2x+\dfrac{1}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=cos2x+\dfrac{1}{16}\)
\(\Leftrightarrow\dfrac{15}{16}-\dfrac{3}{4}\left(1-cos^22x\right)=cos2x\)
\(\Leftrightarrow\dfrac{3}{4}cos^22x-cos2x+\dfrac{3}{16}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{4-\sqrt{7}}{6}\\cos2x=\dfrac{4+\sqrt{7}}{6}>1\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(\dfrac{4-\sqrt{7}}{6}\right)+k\pi\)
b.
\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{5}{2}-2sinx\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^2x=\dfrac{5}{2}-2sinx\)
\(\Leftrightarrow\dfrac{1}{2}sin^2x-2sinx+\dfrac{3}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=3\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)
c.
\(\Leftrightarrow\dfrac{1}{2}cos6x+\dfrac{1}{2}cos4x=\dfrac{1}{2}cos6x+\dfrac{1}{2}cos2x+4-3\left(\dfrac{1}{2}-\dfrac{1}{2}cos2x\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(2cos^22x-1\right)=\dfrac{1}{2}cos2x+\dfrac{5}{2}+\dfrac{3}{2}cos2x\)
\(\Leftrightarrow cos^22x-2cos2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow2x=\pi+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\)
Rút gọn các biểu thức sau :
a)\(\dfrac{1+\sin4a-\cos4a}{1+\cos4a+\sin4a}\)
b) \(\dfrac{1+\cos a}{1-\cos a}\tan^2\dfrac{a}{2}-\cos^2a\)
c) \(\dfrac{\cos2x-\sin4x-\cos6x}{\cos2x+\sin4x-\cos6x}\)
Tìm nghiệm của các phương trinh:
1,\(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
2,\(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}\left(1+cot2xcotx\right)=0\)
3,\(cos^4x+sin^4x+cos\left(x-\dfrac{\pi}{4}\right)sin\left(3x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)
4,\(cos5x+cos2x+2sin3xsin2x=0\) trên \(\left[0;2\pi\right]\)
5,\(\dfrac{cos\left(cosx+2sinx\right)+3sinx\left(sinx+\sqrt{2}\right)}{sin2x-1}=1\)
6,\(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
7,\(cos\left(2x+\dfrac{\pi}{4}\right)+cos\left(2x-\dfrac{\pi}{4}\right)+4sinx=2+\sqrt{2}\left(1-sinx\right)\)
1, \(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx+sin3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{2sin2x.cosx+cosx}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{cosx\left(2sin2x+1\right)}{1+2sin2x}=\dfrac{2+2cos^2x}{5}\)
⇒ cosx = \(\dfrac{2+2cos^2x}{5}\)
⇔ 2cos2x - 5cosx + 2 = 0
⇔ \(\left[{}\begin{matrix}cosx=2\\cosx=\dfrac{1}{2}\end{matrix}\right.\)
⇔ \(x=\pm\dfrac{\pi}{3}+k.2\pi\) , k là số nguyên
2, \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\left(1+cot2x.cotx\right)=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cos2x.cosx+sin2x.sinx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cosx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2cosx}{2cosx.sin^4x}=0\)
⇒ \(48-\dfrac{1}{cos^4x}-\dfrac{1}{sin^4x}=0\). ĐKXĐ : sin2x ≠ 0
⇔ \(\dfrac{1}{cos^4x}+\dfrac{1}{sin^4x}=48\)
⇒ sin4x + cos4x = 48.sin4x . cos4x
⇔ (sin2x + cos2x)2 - 2sin2x. cos2x = 3 . (2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}\) . (2sinx . cosx)2 = 3(2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}sin^22x\) = 3sin42x
⇔ \(sin^22x=\dfrac{1}{2}\) (thỏa mãn ĐKXĐ)
⇔ 1 - 2sin22x = 0
⇔ cos4x = 0
⇔ \(x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)
3, \(sin^4x+cos^4x+sin\left(3x-\dfrac{\pi}{4}\right).cos\left(x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)
⇔ \(\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
⇔ \(1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{3}{2}=0\)
⇔ \(\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}-\dfrac{1}{2}sin^22x=0\)
⇔ sin2x - sin22x - (1 + cos4x) = 0
⇔ sin2x - sin22x - 2cos22x = 0
⇔ sin2x - 2 (cos22x + sin22x) + sin22x = 0
⇔ sin22x + sin2x - 2 = 0
⇔ \(\left[{}\begin{matrix}sin2x=1\\sin2x=-2\end{matrix}\right.\)
⇔ sin2x = 1
⇔ \(2x=\dfrac{\pi}{2}+k.2\pi\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
4, cos5x + cos2x + 2sin3x . sin2x = 0
⇔ cos5x + cos2x + cosx - cos5x = 0
⇔ cos2x + cosx = 0
⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}=0\)
⇔ \(cos\dfrac{3x}{2}=0\)
⇔ \(\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\)
⇔ x = \(\dfrac{\pi}{3}+k.\dfrac{2\pi}{3}\)
Do x ∈ [0 ; 2π] nên ta có \(0\le\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\le2\pi\)
⇔ \(-\dfrac{1}{2}\le k\le\dfrac{5}{2}\). Do k là số nguyên nên k ∈ {0 ; 1 ; 2}
Vậy các nghiệm thỏa mãn là các phần tử của tập hợp
\(S=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\)
5, \(\dfrac{cos^2x+sin2x+3sin^2x+3\sqrt{2}sinx}{sin2x-1}=1\)
⇒ \(cos^2x+sin2x+3sin^2x+3\sqrt{2}sinx=sin2x-1\)
⇒ cos2x + 3sin2x + 3\(\sqrt{2}\)sin2x + 1 = 0
⇔ 2 + 2sin2x + 3\(\sqrt{2}\)sin2x = 0
⇔ 2 + 1 - cos2x + 3\(\sqrt{2}\) sin2x = 0
⇔ \(3\sqrt{2}sin2x-cos2x=-1\)
Còn lại tự giải
7, \(cos\left(2x+\dfrac{\pi}{4}\right)+cos\left(2x-\dfrac{\pi}{4}\right)+4sinx=2+\sqrt{2}\left(1-sinx\right)\)
⇔ \(2cos2x.cos\dfrac{\pi}{4}+4sinx=2+\sqrt{2}\left(1-sinx\right)\)
⇔ \(\sqrt{2}cos2x+4sinx=2+\sqrt{2}-\sqrt{2}sinx\)
Dùng công thức : cos2x = 1 - 2sin2x đưa về phương trình bậc 2 ẩn sinx
Chứng minh các đẳng thức:
\(cos^3xsinx-sin^3xcosx=\dfrac{1}{4}sin4x\)
\(sin^4x+cos^4x=\dfrac{1}{4}\left(3+cos4x\right)\)
\(cos^3xsinx-sin^3xcosx=sinx.cosx\left(cos^2x-sin^2x\right)=\dfrac{1}{2}sin2x.cos2x=\dfrac{1}{4}sin4x\)
\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=1-\dfrac{1}{2}\left(2sinx.cosx\right)^2=1-\dfrac{1}{2}sin^22x\)
\(=1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{1}{4}\left(3+cos4x\right)\)
chứng minh đẳng thức lượng giác
a) \(\dfrac{1-cos^2\left(\dfrac{\pi}{2}-x\right)}{1-sin^2\left(\dfrac{\pi}{2}-x\right)}\) - cot\(\left(\dfrac{\pi}{2}-x\right)\) . tan\(\left(\dfrac{\pi}{2}-x\right)\) = \(\dfrac{1}{sin^2x}\)
b) \(\left(\dfrac{1}{cos2x}+1\right)\).tan\(x\) = \(tan2x\)
Để chứng minh các định lượng đẳng cấp, ta sẽ sử dụng các công thức định lượng giác cơ bản và các quy tắc biến đổi đẳng thức. a) Bắt đầu với phương trình ban đầu: 1 - cos^2(π/2 - x) / (1 - sin^2(π/2 - x)) = -cot(π/2 - x) * tan( π/2 - x) Ta biết rằng: cos^2(π/2 - x) = sin^2(x) (công thức lượng giác) sin^2(π/2 - x) = cos^2(x) (công thức lượng giác) Thay vào phương trình ban đầu, ta có: 1 - sin^2(x) / (1 - cos^2(x)) = -cot(π/2 - x) * tan(π/ 2 - x) Tiếp theo, ta sẽ tính toán một số lượng giác: cot(π/2 - x) = cos(π/2 - x) / sin(π/2 - x) = sin(x) / cos(x) = tan(x) (công thức lượng giác) tan(π/2 - x) = sin(π/2 - x) / cos(π/2 - x) = cos(x) / sin(x) = 1 / tan(x) (công thức lượng giác) Thay vào phương trình, ta có: 1 - sin^2(x) / (1 - cos^2(x)) = -tan(x) * (1/tan(x)) = -1 Vì vậy, ta đã chứng minh là đúng. b) Bắt đầu với phương thức ban đầu: (1/cos^2(x) + 1) * tan(x) = tan^2(x) Tiếp tục chuyển đổi phép tính: 1/cos^2(x) + 1 = tan^2(x) / tan(x) = tan(x) Tiếp theo, ta sẽ tính toán một số giá trị lượng giác: 1/cos^2(x) = sec^2(x) (công thức) lượng giác) sec^2(x) + 1 = tan^2(x) + 1 = sin^2(x)/cos^2(x) + 1 = (sin^2(x) + cos^2(x) ))/cos^2(x) = 1/cos^2(x) Thay thế vào phương trình ban đầu, ta có: 1/cos^2(x) + 1 = 1/cos^2(x) Do đó, ta đã chứng minh được b)đúng.