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sin4 x + cos4 x + 2 sin2 x cos2 x
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29 tháng 8 2021
\(\cos\alpha=\sqrt{1-\dfrac{9}{25}}=\dfrac{4}{5}\)
a: \(A=\cos\alpha\cdot\sin^3\alpha+\cos^3\alpha\cdot\sin\alpha\)
\(=\dfrac{4}{5}\cdot\dfrac{27}{125}+\dfrac{64}{125}\cdot\dfrac{3}{5}\)
\(=\dfrac{4\cdot27+64\cdot3}{625}\)
\(=\dfrac{300}{625}=\dfrac{12}{25}\)
NA
Ngoc Anh Thai
Giáo viên
15 tháng 5 2021
\(sin^6\left(\pi+x\right)=sin^6x,cos^6\left(x-\pi\right)=cos^6\pi\\ sin^4\left(x+2\pi\right)=sin^4x,sin^4\left(x-\dfrac{3\pi}{2}\right)=cos^4x,cos^2\left(x-\dfrac{\pi}{2}\right)=sin^2x.\)
Khi đó \(A=sin^6x+cos^6x-2sin^4x-cos^4x+sin^2x\\ =\left(sin^2x+cos^2x\right)^2-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)-\left(sin^4x+cos^4x\right)-sin^4x+sin^2x\\ =1-3sin^2x.cos^2x-\left[1-2sin^2x.cos^2x\right]-sin^2x.\left(sin^2x-1\right)\\ =1-3sin^2x.cos^2x-1+2sin^2x.cos^2x+sin^2x.cos^2x\\ =0\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 4 2019
Nhân cả tử và mẫu của phân số chứa tan với \(sina.cosa\)
\(A=\frac{sin^2x-cos^2x}{sin^2x+cos^2x}+cos2x=sin^2x-cos^2x+cos2x=-cos2x+cos2x=0\)
\(B=\frac{1+sin4a-cos4a}{1+sin4a+cos4a}=\frac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin4a.cos4a+2cos^22a-1}\)
\(B=\frac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\frac{sin2a}{cos2a}=tan2a\)
\(C=\frac{3-4cos2a+2cos^22a-1}{3+4cos2a+2cos^22a-1}=\frac{2\left(cos^22a-2cos2a-1\right)}{2\left(cos^22a+2cos2a+1\right)}\)
\(C=\frac{\left(cos2a-1\right)^2}{\left(cos2a+1\right)^2}=\frac{\left(1-2sin^2a-1\right)^2}{\left(2cos^2a-1+1\right)^2}=\frac{sin^4a}{cos^4a}=tan^4a\)
\(D=\frac{sin^22a+4sin^4a-\left(2sina.cosa\right)^2}{4-4sin^2a-sin^22a}=\frac{sin^22a+4sin^4a-sin^22a}{4\left(1-sin^2a\right)-\left(2sina.cosa\right)^2}=\frac{4sin^4a}{4cos^2a-4sin^2a.cos^2a}\)
\(=\frac{sin^4a}{cos^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^2a.cos^2a}=\frac{sin^4a}{cos^4a}=tan^4a\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 5 2021
\(\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cos2a+sin3a}{2cos3a.cos2a+cos3a}=\dfrac{sin3a\left(2cos2a+1\right)}{cos3a\left(2cos2a+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)
\(\dfrac{1+sin4a-cos4a}{1+sin4a+cos4a}=\dfrac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin2a.cos2a+2cos^22a-1}=\dfrac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\dfrac{sin2a}{cos2a}=tan2a\)
\(96\sqrt{3}sin\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=48\sqrt{3}sin\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=24\sqrt{3}sin\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=12\sqrt{3}sin\left(\dfrac{\pi}{6}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=6\sqrt{3}sin\left(\dfrac{\pi}{3}\right)=6\sqrt{3}.\dfrac{\sqrt{3}}{2}=9\)
\(A+B+C=\pi\Rightarrow A+B=\pi-C\Rightarrow tan\left(A+B\right)=tan\left(\pi-C\right)\)
\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=-tanC\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)
\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\)
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