Question

Chứng minh rằng các biểu thức sau là những hằng số không phụ thuộc \(\alpha,\beta\) :

a) \(\sin6\alpha\cot3\alpha-\cos6\alpha\)

b) \(\left[\tan\left(90^0-\alpha\right)-\cot\left(90^0+\alpha\right)\right]^2-\left[\cot\left(180^0+\alpha\right)+\cot\left(270^0+\alpha\right)\right]^2\)

c) \(\left(\tan\alpha-\tan\beta\right)\cot\left(\alpha-\beta\right)-\tan\alpha\tan\beta\)

d) \(\left(\cot\dfrac{\alpha}{3}-\tan\dfrac{\alpha}{3}\right)\tan\dfrac{2\alpha}{3}\)

Answer 1

a) \(sin6\alpha cot3\alpha cos6\alpha=2.sin3\alpha.cos3\alpha\dfrac{cos3\alpha}{sin3\alpha}-cos6\alpha\)
\(=2cos^23\alpha-\left(2cos^23\alpha-1\right)=1\) (Không phụ thuộc vào x).

Answer 2

b) \(\left[tan\left(90^o-\alpha\right)-cot\left(90^o+\alpha\right)\right]^2\)\(-\left[cot\left(180^o+\alpha\right)+cot\left(270^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+cot\left(90^o-\alpha\right)\right]^2\)\(-\left[cot\alpha+cot\left(90^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+tan\alpha\right]^2-\left[cot\alpha-tan\alpha\right]^2\)
\(=4tan\alpha cot\alpha=4\). (Không phụ thuộc vào \(\alpha\)).

Answer 3

c) \(\left(tan\alpha-tan\beta\right)cot\left(\alpha-\beta\right)-tan\alpha tan\beta\)
\(=\left(\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\beta}{cos\beta}\right).\dfrac{cos\left(\alpha-\beta\right)}{sin\left(\alpha-\beta\right)}-tan\alpha tan\beta\)
\(=\left(\dfrac{sin\alpha cos\beta-cos\alpha sin\beta}{cos\alpha cos\beta}\right).\dfrac{cos\left(\alpha-\beta\right)}{sin\left(\alpha-\beta\right)}\)\(-\dfrac{sin\alpha sin\beta}{cos\alpha cos\beta}\)
\(=\dfrac{sin\left(\alpha-\beta\right)}{cos\alpha cos\beta}.\dfrac{cos\left(\alpha-\beta\right)}{sin\left(\alpha-\beta\right)}-\dfrac{sin\alpha sin\beta}{cos\alpha cos\beta}\)
\(=\dfrac{cos\left(\alpha-\beta\right)}{cos\alpha cos\beta}-\dfrac{sin\alpha sin\beta}{cos\alpha cos\beta}\)
\(=\dfrac{cos\alpha cos\beta+sin\alpha sin\beta-sin\alpha sin\beta}{cos\alpha cos\beta}=\dfrac{cos\alpha cos\beta}{cos\alpha cos\beta}=1\).

Answer 4

d)Áp dụng công thức:

\(cotx-tanx=\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=\dfrac{cos^2x-sin^2x}{sinx.cosx}\)
\(=\dfrac{2cos2x}{sin2x}=2cotx.\)
Ta được:
\(\left(cot\dfrac{\pi}{3}-tan\dfrac{\pi}{3}\right).tan\dfrac{2\pi}{3}\)\(=\dfrac{2cos\dfrac{2\pi}{3}}{sin\dfrac{2\pi}{3}}.tan\dfrac{2\pi}{3}\)
\(=2.cot\dfrac{2\pi}{3}tan\dfrac{2\pi}{3}=2\).