Thực hiện phép tính: A = cos2 520 + cos2 380 + sin2 300
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\(a,A=\left(\cos^220^0+\cos^270^0\right)+\left(\cos^240^0+\cos^250^0\right)\\ A=\left(\cos^220^0+\sin^220^0\right)+\left(\cos^240^0+\sin^240^0\right)=1+1=2\\ b,B=\left(\cos^2\alpha\right)^3+\left(\sin^2\alpha\right)^3+3\sin^2\alpha\cdot\cos^2\alpha\cdot\left(\sin^2\alpha+\cos^2\alpha\right)\\ B=\left(\sin^2\alpha+\cos^2\alpha\right)^3=1^3=1\)
\(\dfrac{1+cos2a-sin2a}{1+cos2a+sin2a}=\dfrac{2cos^2a-2sina.cosa}{2cos^2a+2sinacosa}\)
\(=\dfrac{2cosa\left(cosa-sina\right)}{2cosa\left(cosa+sina\right)}=\dfrac{cosa-sina}{cosa+sina}=\dfrac{\sqrt{2}sin\left(\dfrac{\pi}{4}-a\right)}{\sqrt{2}cos\left(\dfrac{\pi}{4}-a\right)}=tan\left(\dfrac{\pi}{4}-a\right)\)
\(\dfrac{1+cos2a-cosa}{sin2a-sina}=\dfrac{2cos^2a-cosa}{2sina.cosa-sina}=\dfrac{cosa\left(2cosa-1\right)}{sina\left(2cosa-1\right)}=\dfrac{cosa}{sina}=cota\)
\(\cot\alpha=\dfrac{1}{2}\)
\(\sin\alpha=\dfrac{kề}{\sqrt{5}kề}=\dfrac{\sqrt{5}}{5}\)
\(\cos\alpha=\sqrt{1-\dfrac{5}{25}}=\dfrac{2\sqrt{5}}{5}\)
\(\cos\alpha=\sqrt{1-\dfrac{9}{25}}=\dfrac{4}{5}\)
a: \(A=\cos\alpha\cdot\sin^3\alpha+\cos^3\alpha\cdot\sin\alpha\)
\(=\dfrac{4}{5}\cdot\dfrac{27}{125}+\dfrac{64}{125}\cdot\dfrac{3}{5}\)
\(=\dfrac{4\cdot27+64\cdot3}{625}\)
\(=\dfrac{300}{625}=\dfrac{12}{25}\)
\(cos^2\left(a-b\right)-sin^2\left(a+b\right)\)
\(=\left(cosa.cosb+sina.sinb\right)^2-\left(sina.cosb+cosa.sinb\right)^2\)
\(=cos^2a.cos^2b+sin^2a.sin^2b-sin^2a.cos^2b-cos^2a.sin^2b\)
\(=cos^2b\left(cos^2a-sin^2a\right)-sin^2b\left(cos^2a-sin^2a\right)\)
\(=\left(cos^2b-sin^2b\right)\left(cos^2a-sin^2a\right)\)
\(=cos2a.cos2b\left(dpcm\right)\)
a) sin230 độ - sin240 độ - sin250 độ + sin2 60 độ
= cos260o - cos250o - sin250o + sin260o
= (cos260o + sin260o) - (cos250o + sin250o)
= 1 - 1 = 0
b) cos225 độ - cos235độ + cos245 độ -cos2 55 độ + cos2 65 độ
= sin265o - sin255o + cos245o - cos255o + cos265o
= (sin265o + cos265o) - (sin255o + cos255o) + cos245o
= 1 - 1 +1/2
= 1/2