Question

Giúp em mấy câu khó với ạ

Answer 1

Chính xác là em cần những câu nào nhỉ? Cứ làm đi, thấy câu nào vướng mắc thì hỏi

Answer 2

2.

\(A_2=\dfrac{3}{2}.\left(2sin15^0cos15^0\right)+\dfrac{sin60^0}{\left(sin^215^0+cos^215^0\right)\left(sin^215^0-cos^215^0\right)}\)

\(=\dfrac{3}{2}sin\left(2.15^0\right)+\dfrac{sin60^0}{-\left(cos^215^0-sin^215^0\right)}\)

\(=\dfrac{3}{2}sin30^0-\dfrac{sin60^0}{cos\left(2.15^0\right)}\)

\(=\dfrac{3}{2}sin30^0-\dfrac{sin60^0}{cos30^0}\)

\(=\dfrac{3}{2}.\dfrac{1}{2}-\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{\sqrt{3}}{2}}=\dfrac{3}{4}-1=-\dfrac{1}{4}\)

Answer 3

3.

\(A_3=tan9^0+tan81^0-\left(tan27^0+tan63^0\right)\)

\(=\dfrac{sin9^0}{cos9^0}+\dfrac{sin81^0}{cos81^0}-\left(\dfrac{sin27^0}{cos27^0}+\dfrac{sin63^0}{cos63^0}\right)\)

\(=\dfrac{sin9^0.cos81^0+cos9^0.sin81^0}{cos9^0.cos81^0}-\left(\dfrac{sin27^0.cos63^0+cos27^0.sin63^0}{cos27^0.cos63^0}\right)\)

\(=\dfrac{sin\left(9^0+81^0\right)}{cos9^0.cos81^0}-\dfrac{sin\left(27^0+63^0\right)}{cos27^0.cos63^0}\)

\(=\dfrac{2}{2cos9^0.cos81^0}-\dfrac{2}{2cos27^0.cos63^0}\)

\(=\dfrac{2}{cos90^0+cos72^0}-\dfrac{2}{cos93^0+cos36^0}\)

\(=\dfrac{2}{cos72^0}-\dfrac{2}{cos36^0}=\dfrac{2\left(cos36^0-cos72^0\right)}{cos72^0.cos36^0}\)

\(=\dfrac{4sin54^0.sin18^0}{sin\left(90^0-72^0\right).sin\left(90^0-36^0\right)}\)

\(=\dfrac{4sin54^0.sin18^0}{sin18^0.sin54^0}=4\)

Answer 4

3.

\(A_4=4tan\dfrac{\pi}{8}+2tan\dfrac{\pi}{16}+\dfrac{sin\dfrac{\pi}{32}}{cos\dfrac{\pi}{32}}-\dfrac{cos\dfrac{\pi}{32}}{sin\dfrac{\pi}{32}}\)

\(=4tan\dfrac{\pi}{8}+2tan\dfrac{\pi}{16}+\dfrac{sin^2\dfrac{\pi}{32}-cos^2\dfrac{\pi}{32}}{sin\dfrac{\pi}{32}.cos\dfrac{\pi}{32}}\)

\(=4tan\dfrac{\pi}{8}+2tan\dfrac{\pi}{16}-\dfrac{cos\dfrac{\pi}{16}}{\dfrac{1}{2}sin\dfrac{\pi}{16}}\)

\(=4tan\dfrac{\pi}{8}+\dfrac{2sin\dfrac{\pi}{16}}{cos\dfrac{\pi}{16}}-\dfrac{2cos\dfrac{\pi}{16}}{sin\dfrac{\pi}{16}}\)

\(=4tan\dfrac{\pi}{8}+\dfrac{2\left(sin^2\dfrac{\pi}{16}-cos^2\dfrac{\pi}{16}\right)}{sin\dfrac{\pi}{16}.cos\dfrac{\pi}{16}}\)

\(=\dfrac{4sin\dfrac{\pi}{8}}{cos\dfrac{\pi}{8}}-\dfrac{2cos\dfrac{\pi}{8}}{\dfrac{1}{2}sin\dfrac{\pi}{8}}\)

\(=\dfrac{4\left(sin^2\dfrac{\pi}{8}-cos^2\dfrac{\pi}{8}\right)}{sin\dfrac{\pi}{8}.cos\dfrac{\pi}{8}}=\dfrac{-8.cos\dfrac{\pi}{4}}{sin\dfrac{\pi}{4}}\)

\(=-8cot\dfrac{\pi}{4}=-8\)