# Câu 8 trang 16 SGK Đại số và Giải tích 11 Nâng cao

##### Hướng dẫn giải

Với $$k \in\mathbb Z$$ ta có :

a. $$f(x) = -\sin^2 x$$

$$f(x + kπ) = -\sin^2(x + kπ) = - {\left[ {{{\left( { - 1} \right)}^k}\sin x} \right]^2} = - {\sin ^2}x = f\left( x \right)$$

b.

\eqalign{ & f\left( x \right) = 3{\tan ^2}x + 1 \cr & f\left( {x + k\pi } \right) = 3{\tan ^2}\left( {x + k\pi } \right) + 1 = 3{\tan ^2}x + 1 = f\left( x \right) \cr}

c. $$f(x) = \sin x\cos x$$

\eqalign{ & f\left( {x + k\pi } \right) = \sin \left( {x + k\pi } \right).\cos \left( {x + k\pi } \right) = {\left( { - 1} \right)^k}\sin x.{\left( { - 1} \right)^k}\cos x \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sin x\cos x = f\left( x \right) \cr}

d.

\eqalign{ & f\left( x \right) = \sin x\cos x + {{\sqrt 3 } \over 2}\cos 2x \cr & f\left( {x + k\pi } \right) = \sin \left( {x + k\pi } \right)\cos \left( {x + k\pi } \right) + {{\sqrt 3 } \over 2}\cos \left( {2x + k2\pi } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left( { - 1} \right)^k}\sin x{\left( { - 1} \right)^k}\cos x + {{\sqrt 3 } \over 2}\cos 2x = \sin x\cos x + {{\sqrt 3 } \over 2}\cos 2x = f\left( x \right) \cr}