\(A=\dfrac{sin2\alpha+sin5\alpha-sin3\alpha}{1+cos\alpha-2sin^22\alpha}\)

\(=\dfrac{2sin\alpha.cos\alpha+2.cos4\alpha.sin\alpha}{cos4\alpha+cos\alpha}\)

\(=\dfrac{2sin\alpha.\left(cos\alpha+cos4\alpha\right)}{cos4\alpha+cos\alpha}=2sin\alpha\)

Bạn kiểm tra lại đề bài câu 1, câu này chỉ có thể rút gọn đến \(2cot^2x+2cotx+1\) nên biểu thức ko hợp lý

Đồng thời kiểm tra luôn đề câu 2, trong cả 2 căn thức đều xuất hiện \(6sin^2x\) rất không hợp lý, chắc chắn phải có 1 cái là \(6cos^2x\)

Câu 1 đề vẫn có vấn đề:

\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2\left(1+cot^2x\right)cot^2x}{\left(tanx-1\right)\left(tan^2x+1\right)cot^2x}=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^2x}{tanx-1}\)

\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^3x}{1-cotx}=\dfrac{1+cotx-2cot^3x}{1-cotx}\)

\(=\dfrac{\left(1-cotx\right)\left(1+2cotx+2cot^2x\right)}{1-cotx}=1+2cotx+2cot^2x\)

Có thể coi như ko thể rút gọn tiếp

2.

\(\sqrt{\left(1-cos^2x\right)^2+6cos^2x+3cos^4x}+\sqrt{\left(1-sin^2x\right)^2+6sin^2x+3sin^4x}\)

\(=\sqrt{4cos^4x+4cos^2x+1}+\sqrt{4sin^4x+4sin^2x+1}\)

\(=\sqrt{\left(2cos^2x+1\right)^2}+\sqrt{\left(2sin^2x+1\right)^2}\)

\(=2\left(cos^2x+sin^2x\right)+2=4\)

Đề bài ko chính xác, biểu thức này không rút gọn được (có thể coi việc biến đổi khả dĩ duy nhất \(1+2sina.cosa=\left(sina+cosa\right)^2\) không phải là hành động rút gọn)

chỉnh lại đề 1 chút: \(A=\dfrac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{cos^2\alpha+sin^2\alpha+2sin\alpha.cos\alpha}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}\)

\(=\dfrac{\left(cos\alpha+sin\alpha\right)^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{cos\alpha+sin\alpha}{cos\alpha-sin\alpha}\)

rút gọn biểu thức:

E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))

a/\(sina-1=2sin\dfrac{a}{2}.cos\dfrac{a}{2}-sin^2\dfrac{a}{2}-cos^2\dfrac{a}{2}=-\left(sin\dfrac{a}{2}-cos\dfrac{a}{2}\right)^2\)

b/\(P=\dfrac{cosa+cos5a+2cos3a}{sina+sin5a+2sin3a}=\dfrac{2cos3a.cos2a+2cos3a}{2sin3a.cos2a+2sin3a}=\dfrac{2cos3a\left(cos2a+1\right)}{2sin3a\left(cos2a+1\right)}=cot3a\)

c/\(P=sin\left(30+60\right)=sin90=1\)

d/

\(A=cos\dfrac{2\pi}{7}+cos\dfrac{6\pi}{7}+cos\dfrac{4\pi}{7}\Rightarrow A.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)

\(=\dfrac{1}{2}sin\dfrac{3\pi}{7}-\dfrac{1}{2}sin\dfrac{\pi}{7}+\dfrac{1}{2}sin\dfrac{5\pi}{7}-\dfrac{1}{2}sin\dfrac{3\pi}{7}+\dfrac{1}{2}sin\dfrac{7\pi}{7}-\dfrac{1}{2}sin\dfrac{5\pi}{7}\)

\(=-\dfrac{1}{2}sin\dfrac{\pi}{7}\Rightarrow A=-\dfrac{1}{2}\)

e/

\(tan\dfrac{\pi}{24}+tan\dfrac{7\pi}{24}=\dfrac{sin\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}}+\dfrac{sin\dfrac{7\pi}{24}}{cos\dfrac{7\pi}{24}}=\dfrac{sin\dfrac{\pi}{24}cos\dfrac{7\pi}{24}+sin\dfrac{7\pi}{24}cos\dfrac{\pi}{24}}{cos\dfrac{\pi}{24}.cos\dfrac{7\pi}{24}}\)

\(=\dfrac{sin\left(\dfrac{\pi}{24}+\dfrac{7\pi}{24}\right)}{\dfrac{1}{2}cos\dfrac{\pi}{4}+\dfrac{1}{2}cos\dfrac{\pi}{3}}=\dfrac{2sin\dfrac{\pi}{3}}{cos\dfrac{\pi}{4}+cos\dfrac{\pi}{3}}=\dfrac{\sqrt{3}}{\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}+1}\)

sina - 1 = sina - sin\(\dfrac{\pi}{2}\)

a) \(\dfrac{\sin2\text{a}+\cos a}{1+\cos2\text{a}+\cos a}=2\tan a\)

a) \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos\alpha}=\dfrac{2sin\alpha cos\alpha+sin\alpha}{2cos^2\alpha+cos\alpha}\)\(=\dfrac{sin\alpha\left(2cos\alpha+1\right)}{cos\alpha\left(2cos\alpha+1\right)}=\dfrac{sin\alpha}{cos\alpha}=tan\alpha\).

b) \(\dfrac{4sin^2\alpha}{1-cos^2\dfrac{\alpha}{2}}=\dfrac{4sin^2\alpha}{sin^2\dfrac{\alpha}{2}}=\dfrac{4.sin^2\dfrac{\alpha}{2}.cos^2\dfrac{\alpha}{2}}{sin^2\dfrac{\alpha}{2}}=4sin^2\dfrac{\alpha}{2}\).

\(\cos \left( {a + b} \right)\cos \left( {a - b} \right) - \sin \left( {a + b} \right)\sin \left( {a - b} \right)\)

\( = \frac{1}{2}\left[ {\cos \left( {a + b - a + b} \right) + \cos \left( {a + b + a - b} \right)} \right] - \frac{1}{2}\left[ {\cos \left( {a + b - a + b} \right) - \cos \left( {a + b + a - b} \right)} \right]\)

\( = \frac{1}{2}\left( {\cos 2b + \cos 2a - \cos 2b + \cos 2a} \right) = \frac{1}{2}.2\cos 2a = \cos 2a = 1 - 2{\sin ^2}a\)

Vậy chọn đáp án C

A= cos⁴ ∝ + cos² ∝ . sin² ∝ + sin² ∝

=cos⁴ ∝+(cos² ∝+1).sin² ∝

=cos⁴ ∝+(1+cos⁴ ∝)(1-cos⁴ ∝)

=cos⁴ ∝+1-cos⁴ ∝=1

\(A=\dfrac{sinx+sin3x+sin2x}{cosx+cos3x+cos2x}=\dfrac{2sin2x.cosx+sin2x}{2cos2x.cosx+cos2x}=\dfrac{sin2x\left(2cosx+1\right)}{cos2x\left(2cosx+1\right)}=tan2x\)