write an equation expressing 19cos(theta-50degrees) as a sum of sine and cosine.
Using the identity cos(a-b) = cos a cos b + sin a sin b
cos(θ-50) = cosθcos50 + sinθsin50
= .64cosθ + .77sinθ
so,
19cos(θ-50) = 19(.64cosθ + .77sinθ)
= 12.16cosθ + 14.63sinθ
To express 19cos(theta-50 degrees) as a sum of sine and cosine, we can use the trigonometric identity:
cos(A - B) = cos(A)cos(B) + sin(A)sin(B)
In this case, A = theta and B = 50 degrees. Thus, the equation becomes:
19cos(theta-50 degrees) = 19(cos(theta)cos(50 degrees) + sin(theta)sin(50 degrees))
To express 19cos(theta-50°) as a sum of sine and cosine, we can use the trigonometric identity known as the difference of angles formula. This formula states that cos(theta - phi) = cos(theta)cos(phi) + sin(theta)sin(phi).
In this case, theta represents the angle in question, and phi represents the angle we want to subtract (in this case, 50°). Let's substitute these values into the formula:
cos(theta - 50°) = cos(theta)cos(50°) + sin(theta)sin(50°).
Now, since we want to express the given expression in terms of sine and cosine, let's substitute cos(50°) and sin(50°) using their trigonometric identities. The cosine identity is cos^2(phi) + sin^2(phi) = 1.
cos(50°) can be rewritten as √[(1 + cos(100°))/2], and sin(50°) can be rewritten as √[(1 - cos(100°))/2].
So, substituting these values into the previous equation, we have:
19cos(theta - 50°) = 19cos(theta) * √[(1 + cos(100°))/2] + 19sin(theta) * √[(1 - cos(100°))/2].
Therefore, the equation expressing 19cos(theta - 50°) as a sum of sine and cosine is:
19cos(theta - 50°) = 19cos(theta) * √[(1 + cos(100°))/2] + 19sin(theta) * √[(1 - cos(100°))/2].