Bài 2: 

\(\cos\alpha=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}\)

\(\tan\alpha=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)

\(\cot\alpha=\dfrac{\sqrt{5}}{2}\)

Vì sin(\(\alpha\) ) = cos (\(90-\alpha\)) nên \(sin^2\alpha=cos^2\left(90-\alpha\right)\)

a/ \(sin^230-sin^240-sin^250+sin^260=\left(cos^260+sin^260\right)-\left(cos^250+sin^250\right)=1-1=0\)

b/ \(cos^225-cos^235+cos^245-cos^255+cos^265=\left(sin^265+cos^265\right)-\left(sin^255+cos^255\right)+cos^245=1-1+cos^245=cos^245=\dfrac{1}{2}\)

1: BC=4cm

2: M=2

4: AH=2,4cm

1: BC=4cm

2: M=2

4: AH=2,4cm

a)

\(\cos225^0=\cos\left(180^0+45^0\right)=-\cos45^0=-\dfrac{\sqrt{2}}{2}\)

\(\sin240^0=\sin\left(180^0+60^0\right)=-\sin60^0=-\dfrac{\sqrt{3}}{2}\)

\(\cos\left(-15^0\right)=-\cot15^0=-\tan75^0=-\tan\left(30^0+45^0\right)\)

\(=\dfrac{-\tan30^0-\tan45^0}{1-\tan30^0\tan45^0}=\dfrac{-\dfrac{1}{\sqrt{3}}-1}{1-\dfrac{1}{\sqrt{3}}}=-\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)

\(=-\dfrac{\left(\sqrt{3}+1\right)^2}{2}=-2-\sqrt{3}\)

\(\tan75^0=\cot15^0=2+\sqrt{3}\)

b)

\(\sin\dfrac{7\pi}{12}=\sin\left(\dfrac{\pi}{3}+\dfrac{\pi}{4}\right)=\sin\dfrac{\pi}{3}\cos\dfrac{\pi}{4}+\cos\dfrac{\pi}{3}\sin\dfrac{\pi}{4}\)

\(=\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)

\(\cos\left(-\dfrac{\pi}{12}\right)=\cos\left(\dfrac{\pi}{4}-\dfrac{\pi}{3}\right)=\cos\dfrac{\pi}{4}\cos\dfrac{\pi}{3}+\sin\dfrac{\pi}{3}\sin\dfrac{\pi}{4}\)

\(=\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=0,9659\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=0,9659\)

\(\tan\dfrac{13\pi}{12}=\tan\left(\pi+\dfrac{\pi}{12}\right)=\tan\dfrac{\pi}{12}=\tan\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)\)

\(=\dfrac{\tan\dfrac{\pi}{3}-\tan\dfrac{\pi}{4}}{1+\tan\dfrac{\pi}{3}\tan\dfrac{\pi}{4}}=\dfrac{\sqrt{3}-1}{1+\sqrt{3}}=2-\sqrt{3}\)

Có

A=\(\left(sin^215^o+sin^275^o\right)+\left(sin^240^o+sin^250^o\right)+\left(sin^260^o+sin^230^o\right)\)

\(=\left(sin^215^o+cos^215^o\right)+...\)

\(=1\cdot3=3\)

Câu c tương tự mà mk nghĩ đề sai dấu - trước cos^245độ

Nói chung nếu: a+b=90 độ

thì: \(sin^2a+sin^2b=1\)

b) thì áp dụng nếu a+b=90 độ:

\(tana=cotb\) và ngược lại

Mà \(tana\cdot cota=1\)

Nói chung là công thức......

a) \(\sqrt{\frac{1+\cos x}{1-\cos x}}-\sqrt{\frac{1-\cos x}{1+\cos x}}=\frac{\sqrt{\left(1+\cos x\right)^2}-\sqrt{\left(1-\cos x\right)^2}}{\sqrt{\left(1-\cos x\right)\left(1+\cos x\right)}}\)

\(=\frac{1+\cos x-1+\cos x}{\sqrt{1-\cos^2x}}=\frac{2\cos x}{\sqrt{\sin^2x}}=\frac{2\cos x}{\sin x}=2\cot x\)

b) \(\frac{1}{\tan x+1}+\frac{1}{\cot x+1}=\frac{\tan x+1+\cot x+1}{\left(\tan x+1\right)\left(\cot x+1\right)}\)

\(=\frac{\tan x+\cot x+2}{\tan x+\cot x+\tan x.\cot x+1}=\frac{\tan x+\cot x+2}{\tan x+\cot x+2}=1\)

c) (ko bt có sai đề ko, làm mãi ko ra) 

d) \(\sin^21^0+\sin^22^0+\sin^23^0+...+\sin^289^0\)

\(=\left(\sin^21^0+\sin^289^0\right)+\left(\sin^22^0+\sin^288^0\right)+...+\sin^245^0\)

\(=\left[\left(\sin^21^0-\cos^289^0\right)+\left(\sin^289^0+\cos^289^0\right)\right]+\)

\(\left[\left(\sin^22^0-\cos^288^0\right)+\left(\sin^288^0+\cos^288^0\right)\right]+...+\sin^245^0\)

\(=\left(0+1\right)+\left(0+1\right)+...+\frac{\sqrt{2}}{2}=\frac{44+\sqrt{2}}{2}\)