Cho cos $\alpha$ = $\frac{5}{13}$ ( 0 < $\alpha$ < $\frac{\alpha}{2}$ ) . Tính a, cos(10$\pi$ + $\alpha$ ), cot($\frac{17\alpha}{2}$ + $\alpha$) b,

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1. Giải thích các bước giải:

   Cho $\cos\alpha=\dfrac5{13} , (0<\alpha<\dfrac{\pi}{2})$

   Tính:

   a.$\cos(10\pi+\alpha), \cot (\dfrac{17\pi}{2}+\alpha)$

   Ta có:

   $0<\alpha<\dfrac{\pi}{2}\to \sin\alpha>0$

   Mà $\sin^2\alpha=1-\cos^2\alpha=\dfrac{144}{169}\to \sin\alpha=\dfrac{12}{13}$

   $\to\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{12}{5}$

   Ta có:

   $\cos(10\pi+\alpha)=\cos(\alpha)=\dfrac5{13}$

   $\cot(\dfrac{17\pi}{2}+\alpha)$

   $=\cot(8\pi+\dfrac{\pi}{2}+\alpha)$

   $=\cot(\dfrac{\pi}{2}+\alpha)$

   $=\tan(-\alpha)$

   $=-\tan(\alpha)$

   $=-\dfrac{12}5$

   b.Ta có:

   $\sin2\alpha=2\sin\alpha\cos\alpha=\dfrac{60}{169}$

   $\cos2\alpha=2\cos^2\alpha-1=-\dfrac{119}{169}$

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