a) √2 cos(x - π/4)

= √2.(cosx.cos π/4 + sinx.sin π/4)

= √2.(√2/2.cosx + √2/2.sinx)

= √2.√2/2.cosx + √2.√2/2.sinx

= cosx + sinx (đpcm)

b) √2.sin(x - π/4)

= √2.(sinx.cos π/4 - sin π/4.cosx )

= √2.(√2/2.sinx - √2/2.cosx )

= √2.√2/2.sinx - √2.√2/2.cosx

= sinx – cosx (đpcm).

\(\dfrac{\left(cosa-sina\right)^2-\left(cosa+sina\right)^2}{cosa\cdot sina}\)

\(=\dfrac{\left(cosa-sina-cosa-sina\right)\left(cosa-sina+cosa+sina\right)}{cosa\cdot sina}\)

\(=\dfrac{-2\cdot sina\cdot2\cdot cosa}{cosa\cdot sina}=-4\)

cos(\(\dfrac{3a}{2}\))*cos(\(\dfrac{a}{2}\))=\(\dfrac{1}{2}\left(cos\left(\dfrac{3a}{2}+\dfrac{a}{2}\right)+cos\left(\dfrac{3a}{2}-\dfrac{a}{2}\right)\right)\)=\(\dfrac{1}{2}\left(cos\left(2a\right)+cos\left(a\right)\right)\)=\(\dfrac{1}{2}\left(2cos^2a-1+cosa\right)\)=\(\dfrac{1}{2}\left(2\cdot\left(\dfrac{3}{4}\right)^2-1+\dfrac{3}{4}\right)=\dfrac{7}{16}\)

tana = 3/4.
=>cota=1/ tana =1:3/4=4/3
sina /cosa =tana
=> sina =tana .cosa =3/4. cosa
lại có sin^2(a)+cos^2(a)=1
<=>9/16cos^2(a)+cos^2=1
<=>25/16cos^2(a)=1
<=>cos^2(a)=16/25
=>[cosa =4/5=>sina =3/5
    [cosa =-4/5=> sina =-2/5

4

\(=\frac{\left(\sin a+\cos a-\sin a+\cos a\right)\left(\sin a+\cos a+\sin a-\cos a\right)}{\sin a.\cos a}=\frac{2.\cos a.2.\sin a}{\sin a.\cos a}=4\)

ĂN CHO CÒN NÓNG:NGON.

\(sina+cosa=\frac{5}{4}\Rightarrow\left(sina+cosa\right)^2=\frac{25}{16}\)

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

\(sina.cosa=\frac{\frac{25}{16}-1}{2}=\frac{9}{32}\)

b/ \(\left(sina-cosa\right)^2=sin^2a+cos^2a-2sinacosa\)

\(\left(sina-cosa\right)^2=1-2.\frac{9}{32}=\frac{7}{16}\)

\(\Rightarrow sina-cosa=\pm\frac{\sqrt{7}}{4}\)

c/ \(sin^3a-cos^3a=\left(sina-cosa\right)\left(sin^2a+cos^2a+sina.cosa\right)\)

\(=\left(sina-cosa\right)\left(1+\frac{9}{32}\right)=\pm\frac{41\sqrt{7}}{128}\)

Chọn A.

Ta có 

Thay  vào P ta được P = 4.