Question

Answer 1

a: \(cos^2\left(\frac{\pi}{2}-x\right)=\sin^2\left(3x+\frac{\pi}{4}\right)\)

=>\(\frac{1+cos\left\lbrack2\cdot\left(\frac{\pi}{2}-x\right)\right\rbrack}{2}=\frac{1-cos\left\lbrack2\cdot\left(3x+\frac{\pi}{4}\right)\right\rbrack}{2}\)

=>\(\frac{1+cos\left(\pi-2x\right)}{2}=\frac{1-cos\left(6x+\frac{\pi}{2}\right)}{2}\)

=>Đúng

b: \(cos\left(\pi-2x\right)=cos\left(\pi\right)\cdot cos2x+\sin\left(\pi\right)\cdot\sin2x=-cos2x\)

=>Sai

c: \(cos\left(6x+\frac{\pi}{2}\right)=cos6x\cdot cos\left(\frac{\pi}{2}\right)-\sin6x\cdot\sin\left(\frac{\pi}{2}\right)=-\sin6x\)

\(\frac{1+cos\left(\pi-2x\right)}{2}=\frac{1-cos\left(6x+\frac{\pi}{2}\right)}{2}\)

=>\(1+cos\left(\pi-2x\right)=1-cos\left(6x+\frac{\pi}{2}\right)\)

=>\(cos\left(\pi-2x\right)=-cos\left(6x+\frac{\pi}{2}\right)\)

=>\(-cos2x=-cos\left(6x+\frac{\pi}{2}\right)\)

=>\(cos2x=cos\left(6x+\frac{\pi}{2}\right)=-\sin6x<>cos6x\)

=>Sai

d: \(cos\left(6x+\frac{\pi}{2}\right)=cos2x\)

=>\(\left[\begin{array}{l}6x+\frac{\pi}{2}=2x+k2\pi\\ 6x+\frac{\pi}{2}=-2x+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}4x=-\frac{\pi}{2}+k2\pi\\ 8x=-\frac{\pi}{2}+k2\pi\end{array}\right.\)

=>\(x=-\frac{\pi}{16}+\frac{k\pi}{4}\)

=>Sai

Answer 2

a: c o s 2 ( π 2 − x ) = sin 2 ( 3 x + π 4 ) => 1 + c o s [ 2 ⋅ ( π 2 − x ) ] 2 = 1 − c o s [ 2 ⋅ ( 3 x + π 4 ) ] 2 => 1 + c o s ( π − 2 x ) 2 = 1 − c o s ( 6 x + π 2 ) 2 =>Đúng b: c o s ( π − 2 x ) = c o s ( π ) ⋅ c o s 2 x + sin ( π ) ⋅ sin 2 x = − c o s 2 x =>Sai c: c o s ( 6 x + π 2 ) = c o s 6 x ⋅ c o s ( π 2 ) − sin 6 x ⋅ sin ( π 2 ) = − sin 6 x 1 + c o s ( π − 2 x ) 2 = 1 − c o s ( 6 x + π 2 ) 2 => 1 + c o s ( π − 2 x ) = 1 − c o s ( 6 x + π 2 ) => c o s ( π − 2 x ) = − c o s ( 6 x + π 2 ) => − c o s 2 x = − c o s ( 6 x + π 2 ) => c o s 2 x = c o s ( 6 x + π 2 ) = − sin 6 x <> c o s 6 x =>Sai d: c o s ( 6 x + π 2 ) = c o s 2 x => [ 6 x + π 2 = 2 x + k 2 π 6 x + π 2 = − 2 x + k 2 π ⇒ [ 4 x = − π 2 + k 2 π 8 x = − π 2 + k 2 π => x = − π 16 + k π 4 =>Sai