a.

\(tana=\dfrac{sina}{cosa}=\dfrac{1}{15}\Rightarrow sina=\dfrac{cosa}{15}\)

\(\Rightarrow sin2a=2sina.cosa=\dfrac{2cosa}{15}.cosa=\dfrac{2}{15}cos^2a=\dfrac{2}{15}.\dfrac{1}{1+tan^2a}=\dfrac{2}{15}.\dfrac{1}{1+\dfrac{1}{15^2}}=\dfrac{15}{113}\)

b.

\(5^2=\left(3sina+4cosa\right)^2\le\left(3^2+4^2\right)\left(sin^2+cos^2a\right)=25\)

Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}\dfrac{sina}{3}=\dfrac{cosa}{4}\\3sina+4cosa=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}sina=\dfrac{3}{5}\\cosa=\dfrac{4}{5}\end{matrix}\right.\)

c.

\(\dfrac{1}{tan^2a}+\dfrac{1}{cot^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)

\(\Leftrightarrow\dfrac{cos^2a}{sin^2a}+\dfrac{sin^2a}{cos^2a}+\dfrac{1}{sin^2a}+\dfrac{1}{cos^2a}=7\)

\(\)\(\Leftrightarrow\dfrac{sin^4a+cos^4a}{sin^2a.cos^2a}+\dfrac{sin^2a+cos^2a}{sin^2a.cos^2a}=7\)

\(\Leftrightarrow\dfrac{\left(sin^2a+cos^2a\right)^2-2sin^2a.cos^2a}{sin^2a.cos^2a}+\dfrac{1}{sin^2a.cos^2a}=7\)

\(\Leftrightarrow\dfrac{2}{sin^2a.cos^2a}=9\)

\(\Leftrightarrow\dfrac{8}{\left(2sina.cosa\right)^2}=9\)

\(\Leftrightarrow\dfrac{8}{sin^22a}=9\)

\(\Leftrightarrow sin^22a=\dfrac{8}{9}\)

\(1+tan^2a=\frac{1}{cos^2a}\)       

\(1+3^2=\frac{1}{cos^2a}\) 

\(10=\frac{1}{cos^2a}\)  

\(cos^2a=\frac{1}{10}\)          

\(cosa=\pm\sqrt{\frac{1}{10}}\) 

\(sin^2a+cos^2a=1\)   

\(sin^2a+\frac{1}{10}=1\)   

\(sin^2a=\frac{9}{10}\)   

\(sina=+\sqrt{\frac{9}{10}}\) 

Vì tan dương nên có hai trường hợp : 

TH1 : cả sin và cos cùng dương : 

\(A=\frac{sina\cdot cosa}{sin^2a-cos^2a}\) 

\(=\frac{\sqrt{\frac{9}{10}}\cdot\sqrt{\frac{1}{10}}}{\frac{9}{10}-\frac{1}{10}}\) 

\(=\frac{\frac{3}{10}}{\frac{8}{10}}\)    

\(=\frac{3}{8}\)   

TH2 : cả sin và cos cùng âm 

\(A=\frac{sina\cdot cosa}{sin^2a-cos^2a}\)                   

\(=\frac{-\sqrt{\frac{9}{10}}\cdot-\sqrt{\frac{1}{10}}}{\frac{9}{10}-\frac{1}{10}}\)                 

\(=\frac{\frac{3}{10}}{\frac{8}{10}}\)      

\(=\frac{3}{8}\)            

Chọn A.

Áp dụng công thức biến đổi tổng thành tích và công thức nhân đôi; ta có

ta có sin a^2=4/9 =>cos a^2=1-4/9=5/9

C=5*5/9+2*4/9+11/3

Ta có: \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}\)

Lại có: \(\dfrac{1}{cot\alpha}=tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{sin^2\alpha}{cos\alpha.sin\alpha}=\dfrac{1}{\sqrt{5}}\)

\(\Rightarrow A=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}+\dfrac{sin^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}+\dfrac{1}{\sqrt{5}}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)

Ta có : cot α = \(\sqrt{5}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\sqrt{5}\Rightarrow cos\alpha=\sqrt{5}.sin\alpha\)

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

\(A=\dfrac{sin^2\alpha+\left(\sqrt{5}sin\alpha\right)^2}{sin\alpha.\sqrt{5}sin\alpha}=\dfrac{sin^2\alpha+5sin^2\alpha}{\sqrt{5}sin^2\alpha}\)

\(A=\dfrac{6sin^2\alpha}{\sqrt{5}sin^2\alpha}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)

Câu a)

Từ \(\tan a=3\Leftrightarrow \frac{\sin a}{\cos a}=3\Rightarrow \sin a=3\cos a\)

Do đó:

\(\frac{\sin a\cos a+\cos ^2a}{2\sin ^2a-\cos ^2a}=\frac{3\cos a\cos a+\cos ^2a}{2(3\cos a)^2-\cos ^2a}\)

\(=\frac{\cos ^2a(3+1)}{\cos ^2a(18-1)}=\frac{4}{17}\)

Câu b)

Có: \(\cot \left(\frac{\pi}{2}-x\right)=\tan x=\frac{\sin x}{\cos x}\)

\(\cos\left(\frac{\pi}{2}+x\right)=-\sin x\)

\(\Rightarrow \cot \left(\frac{\pi}{2}-x\right)\cos \left(\frac{\pi}{2}+x\right)=\frac{-\sin ^2x}{\cos x}\)

Và:

\(\frac{\sin (\pi-x)\cot x}{1-\sin ^2x}=\frac{\sin x\cot x}{\cos^2x}=\frac{\sin x.\frac{\cos x}{\sin x}}{\cos^2x}=\frac{1}{\cos x}\)

Do đó:

\(\Rightarrow \cot \left(\frac{\pi}{2}-x\right)\cos \left(\frac{\pi}{2}+x\right)+\frac{\sin (\pi-x)\cot x}{1-\sin ^2x}=\frac{1-\sin ^2x}{\cos x}=\frac{\cos ^2x}{\cos x}=\cos x\)

Ta có đpcm.

MN K BT?

Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)

mình làm r nha

https://hoc24.vn/cau-hoi/biet-cotadfrac12-gia-tri-bieu-thuc-adfrac4sinalpha5cosalpha2sinalpha-3cosalpha-bang-bao-nhieughi-ro-tung-loi-giai-nha.5724337531039