Question

 

Cho góc nhọn α. Tính tanα biết cosα = \(\dfrac{3}{4}\).

Answer 1

Có sin²a + cos²a = 1

Mà cos a = \(\dfrac{3}{4}\)

=>  sin²a + (\(\dfrac{3}{4}\))² = 1

=> sin²a + \(\dfrac{3^2}{4^2}\) = 1

=> sin²a + \(\dfrac{9}{16}\)= 1

=> sin²a = \(\dfrac{7}{16}\)

=> sin a = \(\dfrac{\sqrt{7}}{4}\)

Có tan a = \(\dfrac{\text{sin a}}{\text{cos a}}\)

Mà \(\left\{{}\begin{matrix}\text{cos a = }\dfrac{3}{4}\\\text{sin a = }\dfrac{\sqrt{7}}{4}\end{matrix}\right.\)

=> tan a = \(\dfrac{\dfrac{\sqrt{7}}{4}}{\dfrac{3}{4}}\) = \(\dfrac{\sqrt{7}}{4}\): \(\dfrac{3}{4}\) = ​\(\dfrac{\sqrt{7}}{4}\).\(\dfrac{4}{3}\) =​\(\dfrac{\sqrt{7}}{3}\)