Tính:
A = 5 . sin2 151π/6 + 3 . cos2 . 85π/3 - 4 tan2 . 193π/6 + 7 cot2 37π/3.
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NL
Nguyễn Lê Phước Thịnh
CTVHS
23 tháng 8 2021
\(\cot\alpha=\dfrac{1}{2}\)
\(\sin\alpha=\dfrac{kề}{\sqrt{5}kề}=\dfrac{\sqrt{5}}{5}\)
\(\cos\alpha=\sqrt{1-\dfrac{5}{25}}=\dfrac{2\sqrt{5}}{5}\)
NC
1
20 tháng 4 2019
\(sin^6a+cos^6a=\left(sin^2x\right)^3+\left(cos^2x\right)^3\)
\(=\left(sin^2x+cos^2x\right)\left(sin^4x+cos^4x-sin^2x.cos^2x\right)\)
\(=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)^2-\frac{3}{4}.\left(2sinx.cosx\right)^2\)
\(=1-\frac{3}{4}sin^22x=1-\frac{3}{4}\left(\frac{1}{2}-\frac{1}{2}cos4x\right)=\frac{5}{8}+\frac{3}{8}cos4x\)
2/
\(\frac{1+sin2a-cos2a}{1+cos2a}=\frac{1+2sina.cosa-\left(1-2sin^2a\right)}{1+2cos^2a-1}=\frac{2sina.cosa+2sin^2a}{2cos^2a}\)
\(=\frac{2sina.cosa}{2cos^2a}+\frac{2sin^2a}{2cos^2a}=tana+tan^2a\)
HM
1
NL
Nguyễn Lê Phước Thịnh
CTVHS
25 tháng 4
\(\tan\alpha+\cot\alpha=3\)
=>\(\frac{\sin\alpha}{cos\alpha}+\frac{cos\alpha}{\sin\alpha}=3\)
=>\(\frac{\sin^2\alpha+cos^2\alpha}{\sin\alpha\cdot cos\alpha}=3\)
=>\(\frac{1}{\sin\alpha\cdot cos\alpha}=3\)
=>\(\sin\alpha\cdot cos\alpha=\frac13\)
\(\tan^2\alpha+\cot^2\alpha=\left(\tan\alpha+\cot\alpha\right)^2-2\cdot tan\alpha\cdot\cot a\)
\(=3^2-2\)
=9-2
=7
\(A=5\cdot\sin^2\left(\frac{151}{6}\pi\right)+3\cdot cos^2\left(\frac{85}{3}\pi\right)-4\cdot\tan^2\left(\frac{193}{6}\pi\right)+7\cdot\cot^2\left(\frac{37\pi}{3}\right)\)
\(=5\cdot\sin^2\left(24\pi+\pi+\frac{\pi}{6}\right)+3\cdot cos^2\left(28\pi+\frac{\pi}{3}\right)-4\cdot\tan^2\left(32\pi+\frac{\pi}{6}\right)+7\cdot\cot^2\left(12\pi+\frac{\pi}{3}\right)\)
\(=5\cdot\sin^2\left(\pi+\frac{\pi}{6}\right)+3\cdot cos^2\left(\frac{\pi}{3}\right)-4\cdot\tan^2\left(\frac{\pi}{6}\right)+7\cdot\cot^2\left(\frac{\pi}{3}\right)\)
\(=5\cdot\frac14+3\cdot\frac14-4\cdot\frac13+7\cdot\frac13\)
=5/4+3/4-4/3+7/3
=2+1
=3