\(sina+cosa=\sqrt{2}\Leftrightarrow\left(sina+cosa\right)^2=2\\ \)

\(\Leftrightarrow\sin^2a+2\sin a.cosa+cos^2a=2\)

\(\Leftrightarrow1+2.sina.cosa=2\)

\(\Leftrightarrow2.sina.cosa=2-1=1\)

\(\Leftrightarrow\sin a.cosa=\frac{1}{2}\)

Vậy  P=sina.cosa=\(\frac{1}{2}\)

\(Q=\sin^4a+cos^4a\)

\(\Leftrightarrow\left(sin^2a\right)^2+\left(cos^2a\right)^2\)

\(\Leftrightarrow\left(sin^2a+cos^2a\right)^2-2.sin^2a.cos^2a\)

\(\Leftrightarrow1^2-2.sin^2a.cos^2a\) tách tiếp rồi thế vào là được .tương tự phàn P ý
còn R thì tách sin^3a=sin^2a+sina tương tự cos mũ 3 a cụng vậy
theo tớ là như thế còn có sai thì đừng có ném đá ném gạch na

Giả sử các biểu thức đều xác định

a/ \(\frac{1-sina}{cosa}=\frac{cosa\left(1-sina\right)}{cos^2a}=\frac{cosa\left(1-sina\right)}{1-sin^2a}=\frac{cosa\left(1-sina\right)}{\left(1-sina\right)\left(1+sina\right)}=\frac{cosa}{1+sina}\)

b/ \(=\frac{sin^2a+\left(1+cosa\right)^2}{sina\left(1+cosa\right)}=\frac{sin^2a+cos^2a+2cosa+1}{sina\left(1+cosa\right)}=\frac{2\left(cosa+1\right)}{sina\left(1+cosa\right)}=\frac{2}{sina}\)

c/ \(=\frac{cosa\left(1-sina\right)+cosa\left(1+sina\right)}{\left(1-sina\right)\left(1+sina\right)}=\frac{2cosa}{1-sin^2a}=\frac{2cosa}{cos^2a}=\frac{2}{cosa}\)

Chứng minh các hằng đẳng thức trên

Lời giải:

1.

\(\cos ^2x+\cos ^2x\tan ^2x=\cos ^2x+\cos ^2x.(\frac{\sin x}{\cos x})^2\)

\(=\cos ^2x+\sin ^2x=1\)

2.

\(\frac{2\cos ^2a-1}{\sin a+\cos a}=\frac{2\cos ^2a-(\sin ^2a+\cos ^2a)}{\sin a+\cos a}=\frac{\cos ^2a-\sin ^2a}{\sin a+\cos a}=\frac{(\cos a-\sin a)(\cos a+\sin a)}{\sin a+\cos a}\)

\(=\cos a-\sin a\)

3.

\(\frac{1-2\sin ^2a}{\sin a-\cos a}=\frac{\cos ^2a+\sin ^2a-2\sin ^2a}{\sin a-\cos a}=\frac{\cos ^2a-\sin ^2a}{\sin a-\cos a}\)

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

4.

\(\frac{1+\sin a}{1-\sin a}-\frac{1-\sin a}{1+\sin a}=\frac{(1+\sin a)^2-(1-\sin a)^2}{(1-\sin a)(1+\sin a)}\)

\(=\frac{1+\sin ^2a+2\sin a-(1+\sin ^2a-2\sin a)}{1-\sin ^2a}=\frac{4\sin a}{\cos ^2a}=\frac{4\tan a}{\cos a}\)

\(\dfrac{sina}{sina-cosa}-\dfrac{cosa}{cosa-sina}=\dfrac{sina+cosa}{sina-cosa}=\dfrac{1+cota}{1-cota}=\dfrac{\left(1+cota\right)^2}{1-cot^2a}\)

Đề bài ko đúng

Lời giải:

a) Áp dụng công thức \(\sin ^2a+\cos ^2a=1\) thì:

\(P=3\sin ^2a+4\cos ^2a=3(\sin ^2a+\cos ^2a)+\cos ^2a\)

\(=3.1+(\frac{1}{3})^2=\frac{28}{9}\)

b)

\(\tan a=\frac{3}{4}\Rightarrow \cot a=\frac{1}{\tan a}=\frac{4}{3}\)

\(\frac{3}{4}=\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\frac{3}{4}\cos a\)

\(\Rightarrow \sin ^2a=\frac{9}{16}\cos ^2a\)

\(\Rightarrow \sin ^2a+\cos ^2a=\frac{25}{16}\cos ^2a\Rightarrow \frac{25}{16}\cos ^2a=1\)

\(\Rightarrow \cos ^2a=\frac{16}{25}\Rightarrow \cos a=\pm \frac{4}{5}\)

Nếu \(\Rightarrow \sin a=\pm \frac{3}{5}\) (theo thứ tự)

c)

\(\frac{1}{2}=\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\frac{\cos a}{2}\). Vì a góc nhọn nên \(\cos a\neq 0\)

Do đó:

\(\frac{\cos a-\sin a}{\cos a+\sin a}=\frac{\cos a-\frac{\cos a}{2}}{\cos a+\frac{\cos a}{2}}=\frac{\cos a(1-\frac{1}{2})}{\cos a(1+\frac{1}{2})}=\frac{1-\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3}\)

Xét ΔBAC vuông tại B có a = ^A ta có :

a) \(\frac{\sin\alpha}{\cos\alpha}=\frac{\sin A}{\cos A}=\frac{\frac{BC}{AB}}{\frac{AB}{AC}}=\frac{BC}{AB}\cdot\frac{AC}{AB}=\frac{BC}{AB}=\tan A=\tan\alpha\left(đpcm\right)\)

b) \(\frac{\cos\alpha}{\sin\alpha}=\frac{\cos A}{\sin A}=\frac{\frac{AB}{AC}}{\frac{BC}{AC}}=\frac{AB}{AC}\cdot\frac{AC}{BC}=\frac{AB}{BC}=\cot A=\cot\alpha\left(đpcm\right)\)

c) \(\tan\alpha\cdot\cot\alpha=\tan A\cdot\cot A=\frac{BC}{AB}\cdot\frac{AB}{BC}=1\left(đpcm\right)\)

d) \(\sin^2\alpha+\cos^2\alpha=\sin^2A+\cos^2A=\frac{BC^2}{AC^2}+\frac{AB^2}{AC^2}=\frac{AB^2+BC^2}{AC^2}=1\left(đpcm\right)\)

e) \(\frac{1}{\cos^2\alpha}=\frac{1}{\cos^2A}=\frac{1}{\frac{AB^2}{AC^2}}=\frac{AC^2}{AB^2};1+\tan^2\alpha=1+\tan^2A=1+\frac{BC^2}{AB^2}=\frac{AB^2+BC^2}{AB^2}=\frac{AC^2}{AB^2}\)

\(\Rightarrow1+\tan^2\alpha=\frac{1}{\cos^2\alpha}\left(đpcm\right)\)

f) \(\frac{1}{\sin^2\alpha}=\frac{1}{\sin^2A}=\frac{1}{\frac{BC^2}{AC^2}}=\frac{AC^2}{BC^2};1+\cot^2\alpha=1+\cot^2A=1+\frac{AB^2}{BC^2}=\frac{BC^2+AB^2}{BC^2}=\frac{AC^2}{BC^2}\)

\(\Rightarrow1+\cot^2\alpha=\frac{1}{\sin^2\alpha}\left(đpcm\right)\)

Lời giải:

a)

\(A=\frac{4\sin ^2a}{1-\cos ^2\frac{a}{2}}=\frac{4\sin ^2a}{\sin ^2\frac{a}{2}}=\frac{4(2\sin \frac{a}{2}\cos \frac{a}{2})^2}{\sin ^2\frac{a}{2}}=16\cos ^2\frac{a}{2}\)

b)

Sử dụng công thức: \(1-\cos 2a=2\sin ^2a; 1+\cos 2a=2\cos ^2a\) và \(\sin 2a=2\sin a\cos a\) ta có:

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

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

\(=\frac{-\cos \frac{a}{2}}{\sin \frac{a}{2}}=-\cot \frac{a}{2}\)

c) \(45-\frac{\pi}{2}\)??? sao đơn vị nó không thống nhất vậy?