Trig: how to factor this : cos^4 x - sin^4 x please help i am so lost on this problem and it seems easy D;

and you can help verify this identity too if you want:

sin(A+B) / cosAcosB = tanA + tanB

cos⁴x - sin⁴x
= (cos²x - sin²x)(cos²x + sin²x)
= (cosx - sinx)(cosx + sinx)(1)
= (cosx - sinx)(cosx + sinx)

Note: cos²x + sin²x = 1

But unless you wanna substitute cos²x - sin²x = cos2x...then:
cos⁴x - sin⁴x
= (cos²x - sin²x)(cos²x + sin²x)
= (cos2x)(1)
= cos2x

Hope this helps :D




sin(A + B) / cosAcosB
= (sinAcosB + sinBcosA) / cosAcosB
= (sinAcosB)/(cosAcosB) + (sinBcosA)/(cosAcosB)

The cosB in the first fraction cancel out, the cosA in the second fraction cancel out.
= (cosB/cosB)(sinA/cosA) + (cosA/cosA)(sinB/cosB)
= (1)(sinA/cosA) + (1)(sinB/cosB)
= sinA/cosA + sinB/cosB
= tanA + tanB

:D

For the first question:

This is simply a difference of squares:

cos(x)^4-sin(x)^4 = (cos(x)^2-sin(x)^2)(cos(x)^2+sin(x)^2)

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

= cos(x)^2-sin(x)^2 = (cos(x)-sin(x))(cosIx)+sin(x)) = cos(2x)

For the second problem:

sin(A+B) = sin(A)cos(B)+sin(B)cos(A)

Therefore, sin(A+B)/(cos(A)cos(B)) = (sin(A)cos(B)+sin(B)cos(A))/(cos(A)cos(B…

Split the fraction around the "+" sign, and the cosines will divide out leaving:

sin(A)/cos(A) + sin(B)/cos(B) = tan(A)+tan(B)