Use the phase shift formula to rewrite

Question

6cos2x+ 8sin2x

Answer 1

6cos2x + 8sin2x = ?

 

Multiply and divide through by √(6² + 8²) = √(36 + 64) = 10

 

6cos2x + 8sin2x = 10 [(6/10) cos2x + (8/10) sin2x]

 

Now consider a right triangle with hypotenuse of length 10 and perpendicular sides of lengths 6 and 10.  Then you can think of 6/10 as the cosine of some angle α and 8/10 as the sine of α.  Then

 

6cos2x + 8sin2x = 10(cosα cos2x + sinα sin2x) = 10cos(α - 2x) = 10cos(2x - α)

 

6cos2x + 8sin2x = 10cos(2x - α), where α = cos^-1(6/10) = sin^-1(8/10) = 53.13°.  The phase shift -53.13º.

 

You could also have said 6/10 = sinβ and 8/10 = cosβ.  Then

 

6cos2x + 8sin2x = 10(sinβ cos2x + cosβ sin2x) = 10sin(2x + β)

 

6cos2x + 8sin2x = 10sin(2x + β), where β = sin^-1(6/10) = cos^-1(8/10) = 36.87° is the phase shift

 

Note that 

 

10cos(2x - 53.13°) = 10cos(53.13° - 2x) = 10sin(2x + 36.87º)

 

because cosine is even and the property that cos(90° - θ) = sinθ.