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Question

1, (PT vô tỷ) : Giải PT : $\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+2-2\sqrt{x+1}}=\frac{x+5}{2}$

2, Rút gọn Bt:

a/ $(x.sina-y.cosa)^2+(x.cosa+y.sina)^2$

b/ $\sqrt{\frac{1+cosx}{1-cosx}}-\sqrt{\frac{1-cosx}{1+cosx}}, x\in (0;\pi )$

c/ $cosx-\sqrt{tan^2x-sin^2x}; x\in (\frac{\pi }{2};\frac{3\pi }{2})$

score 1 · Answer 1

2c) vì $x \in (\dfrac{\pi}{2};\ \dfrac{3\pi}{2}) \Rightarrow | \cos x | = -\cos x$

Ta có $\tan^2 x - \sin^2 x = \sin^2 x(\dfrac{1}{\cos^2 x} - 1) = \sin^2 x (1 + \tan^2 x - 1) = \sin^2 x \tan^2 x = \dfrac{\sin^4 x}{\cos^2 x}$

Vậy KQ $= \cos x - \sqrt{\dfrac{\sin^4 x}{\cos^2 x}} = \cos x + \dfrac{\sin^2 x}{\cos x} = \dfrac{\cos^2 x + \sin^2 x}{\cos x} = \dfrac{1}{\cos x}$

score 1 · Answer 2

2b) $\dfrac{\sqrt{1+\cos x}}{\sqrt{1-\cos x}} - \dfrac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}} = \dfrac{1+\cos x -(1-\cos x)}{\sqrt{1 -\cos^2 x}}$

$= \dfrac{2\cos x}{| \sin x |}$ vì $x \in (0; \ \pi) \Rightarrow | \sin x | = \sin x$

Vậy KQ $= 2\tan x$

score 1 · Answer 3

2a) Phá bình phương tung tóe ta được =))

$x^2\sin^2 a + y^2\cos^2 a + x^2\cos^2 a + y^2\sin^2 a = x^2 + y^2$

score 1 · Answer 4

1) Gợi ý $\sqrt{(\sqrt{x+1} + 1)^2} + \sqrt{(\sqrt{x+1} -1)^2} = \dfrac{x+5}{2}$

$\Leftrightarrow \sqrt{x+1} + 1 + | \sqrt{x+1} - 1 | = \dfrac{x+5}{2}$

Tự phá trị tuyệt đối làm nốt nhé

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