Question

sin2x+cos4x express as product​

Answer 1

Answer:

Step-by-step explanation:

sin(2x) + cos(4x) =>  

sin(3x - x) + cos(3x + x) =>  

sin(3x)cos(x) - sin(x)cos(3x) + cos(3x)cos(x) - sin(3x)sin(x) =>  

sin(3x) * (cos(x) - sin(x)) + cos(3x) * (cos(x) - sin(x)) =>  

(sin(3x) + cos(3x)) * (cos(x) - sin(x)) =>  

sqrt(2) * (sin(3x) * sqrt(2)/2 + cos(3x) * sqrt(2)/2) * sqrt(2) * (cos(x) * sqrt(2)/2 - sin(x) * sqrt(2)/2) =>  

2 * (sin(3x)cos(pi/4) + cos(3x)sin(pi/4)) * (cos(x)cos(pi/4) - sin(x)sin(pi/4)) =>  

2 * sin(3x + pi/4) * cos(x + pi/4)

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