This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ) + A cos(ωx) sin(φ) + B Once again, we may take ω = 2 and B = 0 so that
cos(2x) − √ 3 sin(2x) = A sin(2x) cos(φ) + A cos(2x) sin(φ) √
We equate9 the coeﬃcients of cos(2x) on either side and get A sin(φ) = 1 and A cos(φ) = − 3.
Using A2 cos2 (φ) + A2 sin2 (φ) = A2 as before, we get A = ±2, and again we choose A = 2.
This means 2 sin(φ) = 1, or sin(φ) = 1 , and 2 cos(φ) = − 3, which means cos(φ) = − 23 .
One such angle which meets these criteria is φ = 56 . Hence, we have f (x) = 2 sin 2x + 56 .
Checking our work analytically, we have
f (x) = 2 sin 2x + 5π
6 = 2 sin(2x) cos 5π
2 √ + cos(2x) sin + cos(2x)
= 2 sin(2x) −
= cos(2x) − 3 sin(2x) 5π
2 Graphing the three formulas for f (x) result in the identical curve, verifying our analytic work. It is important to note that in order for the technique presented in Example 10.5.3 to ﬁt a function
into one of the forms in Theorem √
10.23, the arguments of the cosine and sine√
function much match.
That is, while f (x) = cos(2x)...
View Full Document