Express sin4xcos3x as a sum or differences of sines and cosines
use your product-to-sum formulas:
sinAcosB = 1/2 (sin(A+B) + sin(A-B))
sin4x cos3x = 1/2 (sin7x + sin(x))
To express sin(4x)cos(3x) as a sum or difference of sines and cosines, we will need to use some trigonometric identities.
We can use the identity:
sin(a)cos(b) = (1/2) [sin(a+b) + sin(a-b)]
Let's rewrite sin(4x)cos(3x) using this identity:
sin(4x)cos(3x) = (1/2) [sin(4x+3x) + sin(4x-3x)]
This simplifies to:
(1/2) [sin(7x) + sin(x)]
Therefore, sin(4x)cos(3x) can be expressed as the sum of sin(7x) and sin(x).
To express sin(4x)cos(3x) as a sum or difference of sines and cosines, we will use the angle addition formula for sine.
The angle addition formula for sine states that sin(A + B) = sin(A)cos(B) + cos(A)sin(B).
Let's rewrite sin(4x) as sin(3x + x) and apply the angle addition formula.
sin(3x + x) = sin(3x)cos(x) + cos(3x)sin(x)
Now, we need to express cos(3x) in terms of sine and cosine using the Pythagorean identity:
cos^2(A) + sin^2(A) = 1
cos^2(3x) + sin^2(3x) = 1
cos^2(3x) = 1 - sin^2(3x)
cos^2(3x) = cos^2(x) using the Pythagorean identity: sin^2(A) = 1 - cos^2(A)
Taking the square root of both sides:
cos(3x) = ± √(1 - sin^2(3x))
Now, let's substitute this into our expression:
sin(3x)cos(x) + cos(3x)sin(x) = sin(3x)cos(x) + (± √(1 - sin^2(3x)))sin(x)
Finally, we simplify:
sin(3x)cos(x) + (± √(1 - sin^2(3x)))sin(x) = sin(3x)cos(x) ± sin(x) √(1 - sin^2(3x))
Therefore, sin(4x)cos(3x) can be expressed as a sum or difference of sines and cosines as sin(3x)cos(x) ± sin(x) √(1 - sin^2(3x)).