a) Đầu tiên bạn tự đi chứng minh hai công thức sau, do quá dài nên bạn có thể lên mạng tham khảo cách chứng minh:

\(\sin2a=2\sin a.\cos a\)

\(cos2a=cos^2a-sin^2a\)

Áp dụng hai công thức trên ta có:

\(sin30^o=2sin15^ocos15^o\Leftrightarrow sin15^ocos15^o=\frac{1}{4}\Leftrightarrow cos15^o=\frac{1}{4sin15^o}\)

\(cos30^o=cos^215^o-sin^215^o\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}=cos^215^o-sin^215^o\)

\(\Leftrightarrow\left(\frac{1}{4sin^215^o}\right)^2-sin^215^o=\frac{\sqrt{3}}{2}\)

\(\Leftrightarrow\frac{1}{16sin^415^o}-sin^215^o=\frac{\sqrt{3}}{2}\)

\(\Leftrightarrow-32sin^415^o-16sin^215^o\sqrt{3}+2=0\)

\(\Leftrightarrow sin^215^o=\frac{2-\sqrt{3}}{4}\left(sin^215^o\ge0\right)\)

\(\Leftrightarrow sin15^o=\sqrt{\frac{2-\sqrt{3}}{4}}=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{4\sqrt{2}}}=\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}\left(đpcm\right)\)

Vẽ \(\Delta ABC\) vuông tại A có \(\widehat{C}=30^0\), đường phân giác CD.

CD là tia phân giác của \(\widehat{ACB}\)

\(\Rightarrow\widehat{ACD}=\widehat{BCD}=\dfrac{\widehat{ACD}}{2}=15^0\)

\(\Delta ABC\) vuông tại A có \(\widehat{C}=30^0\)

\(\Rightarrow\Delta ABC\) là nửa tam giác đều

\(\Rightarrow\left\{{}\begin{matrix}AB=\dfrac{1}{2}BC\\AC=\dfrac{\sqrt{3}}{2}BC\end{matrix}\right.\)

\(\Delta ABC\) có CD là đường phân giác

\(\Rightarrow\tan15^0=\tan\widehat{ACD}=\dfrac{AD}{AC}=\dfrac{BD}{BC}=\dfrac{AD+BD}{AC+BC}=\dfrac{\dfrac{1}{2}BC}{\dfrac{\sqrt{3}}{2}BC+BC}=2-\sqrt{3}\) (áp dụng tính chất của dãy tỉ số bằng nhau_

\(1+\tan^215^0=\dfrac{1}{\cos^215^0}\)

\(\Rightarrow\cos15^0=\sqrt{\dfrac{1}{1+\tan^215^0}}=\sqrt{\dfrac{1}{1+\left(2-\sqrt{3}\right)^2}}\)

\(=\sqrt{\dfrac{1}{8-4\sqrt{3}}}=\sqrt{\dfrac{\left(8+4\sqrt{3}\right)}{64-48}}=\sqrt{\dfrac{\left(\sqrt{6}+\sqrt{2}\right)^2}{4^2}}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\left(\text{đ}pcm\right)\)

\(0< 15^0< 90^0\Rightarrow sin,cos,tan\) đều dương

\(cos15=\sqrt{1-sin^215}=\sqrt{1-\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)^2}=\frac{\sqrt{6}+\sqrt{2}}{4}\)

\(tan15=\frac{sin15}{cos15}=2+\sqrt{3}\)

\(cot15=\frac{1}{tan15}=2-\sqrt{3}\)

a) AM ứng với cạnh huyền BC nên AM = \(\frac{1}{2}\) x BC = \(\frac{4}{2}\) = 2 cm

AH = tan\(\widehat{ACH}\)x HM = tan 15⁰ x 2 = \(4-2\sqrt{3}\)cm

Sin \(\widehat{AMH}\)= \(\frac{AH}{AM}\)= \(\frac{4-2\sqrt{3}}{2}\)  = \(2-\sqrt{3}\)    cm

Định lí Pitago : AM² = AH² + HM²

HC = tan \(\widehat{ACH}\)x AH

Bài 1 :

a) \(Cos30^o=Cos\left(2.15^o\right)=2cos^215^o-1\)

\(\Rightarrow cos^215^o=\dfrac{cos30^o+1}{2}\)

\(\Rightarrow cos^215^o=\dfrac{\dfrac{\sqrt[]{3}}{2}+1}{2}\)

\(\Rightarrow cos^215^o=\dfrac{\sqrt[]{3}+2}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{\sqrt[]{3}+2}}{2}\)

\(\Rightarrow cos15^o=\dfrac{2\sqrt[]{\sqrt[]{3}+2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{4\sqrt[]{3}+8}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{6+2.2\sqrt[]{2}\sqrt[]{6}+2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{\left(\sqrt[]{6}+\sqrt[]{2}\right)^2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{6}+\sqrt[]{2}^{ }}{4}\left(dpcm\right)\)

a)

 Dựng tam giác ABC vuông tại A với \(\widehat{C}=15^o\). Trên đoạn thẳng AC lấy điểm D sao cho \(\widehat{CBD}=15^o\). Không mất tính tổng quát, ta chuẩn hóa \(AB=1\). \(\Rightarrow\left\{{}\begin{matrix}BD=\dfrac{AB}{cos60^o}=2\\AD=AB.tan60^o=\sqrt{3}\end{matrix}\right.\)

 Dễ thấy tam giác DBC cân tại D \(\Rightarrow BD=CD=2\) \(\Rightarrow AC=AD+DC=2+\sqrt{3}\)

  \(\Rightarrow tanC=\dfrac{AB}{AC}=\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}\) 

\(\Rightarrow\dfrac{sinC}{cosC}=2-\sqrt{3}\)

\(\Rightarrow sinC=\left(2-\sqrt{3}\right)cosC\)

Mà \(sin^2C+cos^2C=1\)

\(\Rightarrow\left(7-4\sqrt{3}\right)cos^2C+cos^2C=1\)

\(\Leftrightarrow\left(8-4\sqrt{3}\right)cos^2C=1\)

\(\Leftrightarrow cos^2C=\dfrac{1}{8-4\sqrt{3}}=\dfrac{2+\sqrt{3}}{4}\)

\(\Leftrightarrow cosC=\sqrt{\dfrac{2+\sqrt{3}}{4}}\) \(=\dfrac{\sqrt{2+\sqrt{3}}}{2}=\dfrac{\sqrt{8+4\sqrt{3}}}{4}\) \(=\dfrac{\sqrt{6}+\sqrt{2}}{4}\) 

\(\Rightarrow cos15^o=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)

b) \(A=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{32cos^415^o-10-8\sqrt[]{3}}}\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{32.\dfrac{1}{4^4}\left(\sqrt[]{6}+\sqrt[]{2}\right)^4-10-8\sqrt[]{3}}}\) \(\left(Cos15^o=\dfrac{\sqrt[]{6}-\sqrt[]{2}}{4}\right)\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{\dfrac{1}{8}\left(8+2\sqrt[]{12}\right)^2-10-8\sqrt[]{3}}}\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{\dfrac{1}{8}\left[64+32\sqrt[]{12}+48-80-64\sqrt[]{3}\right]}}\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\dfrac{1}{2}\sqrt[3]{32}}\)

\(\Rightarrow A=\dfrac{1}{2+\sqrt[3]{2}+\dfrac{1}{2}.2\sqrt[3]{4}}=\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{4}}\)

Trước hết , ta cần chứng minh \(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}\)(*) (Bạn tự chứng minh)

Đặt \(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}\)

\(\Rightarrow2A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{5}+\sqrt{6}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{79}+\sqrt{80}}\)

\(>\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+...+\frac{1}{\sqrt{79}+\sqrt{80}}+\frac{1}{\sqrt{80}+\sqrt{81}}\)

Áp dụng (*) :\(\Rightarrow2A>\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+\left(\sqrt{4}-\sqrt{3}\right)+\left(\sqrt{5}-\sqrt{4}\right)+...+\left(\sqrt{80}-\sqrt{79}\right)+\left(\sqrt{81}-\sqrt{80}\right)\)

\(\Rightarrow2A>\sqrt{81}-1=8\Rightarrow A>4\)(đpcm)