Giải Bài 1.5 trang 17 SGK Toán 11

Lời giải

a) Chứng minh $\cos^4\alpha - \sin^4\alpha = 2\cos^2\alpha - 1$.

Biến đổi vế trái, coi $\cos^4\alpha - \sin^4\alpha$ là hiệu hai bình phương:

$$\cos^4\alpha - \sin^4\alpha = \left(\cos^2\alpha\right)^2 - \left(\sin^2\alpha\right)^2 = \left(\cos^2\alpha - \sin^2\alpha\right)\left(\cos^2\alpha + \sin^2\alpha\right).$$

Vì $\cos^2\alpha + \sin^2\alpha = 1$ nên:

$$\cos^4\alpha - \sin^4\alpha = \cos^2\alpha - \sin^2\alpha.$$

Thay $\sin^2\alpha = 1 - \cos^2\alpha$:

$$\cos^2\alpha - \sin^2\alpha = \cos^2\alpha - \left(1 - \cos^2\alpha\right) = 2\cos^2\alpha - 1.$$

Vậy $\cos^4\alpha - \sin^4\alpha = 2\cos^2\alpha - 1$ (đpcm).

b) Chứng minh $\dfrac{\cos^2\alpha + \tan^2\alpha - 1}{\sin^2\alpha} = \tan^2\alpha$ (điều kiện $\cos\alpha \neq 0$, $\sin\alpha \neq 0$).

Biến đổi tử thức của vế trái. Nhóm $\cos^2\alpha - 1 = -\sin^2\alpha$:

$$\cos^2\alpha + \tan^2\alpha - 1 = \left(\cos^2\alpha - 1\right) + \tan^2\alpha = -\sin^2\alpha + \tan^2\alpha.$$

Thay $\tan^2\alpha = \dfrac{\sin^2\alpha}{\cos^2\alpha}$:

$$-\sin^2\alpha + \dfrac{\sin^2\alpha}{\cos^2\alpha} = \sin^2\alpha\left(\dfrac{1}{\cos^2\alpha} - 1\right) = \sin^2\alpha \cdot \dfrac{1 - \cos^2\alpha}{\cos^2\alpha} = \sin^2\alpha \cdot \dfrac{\sin^2\alpha}{\cos^2\alpha} = \sin^2\alpha \cdot \tan^2\alpha.$$

Do đó:

$$\dfrac{\cos^2\alpha + \tan^2\alpha - 1}{\sin^2\alpha} = \dfrac{\sin^2\alpha \cdot \tan^2\alpha}{\sin^2\alpha} = \tan^2\alpha.$$

Vậy $\dfrac{\cos^2\alpha + \tan^2\alpha - 1}{\sin^2\alpha} = \tan^2\alpha$ (đpcm).

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