BN. This is another correct name for the segment.
The answer to your question is:
<span>there are holes at x= 3/2 and x=-1/3</span>
Answer:
cos(π/3)cos(π/5) + sin(π/3)sin(π/5) = cos(2π/15)
Step-by-step explanation:
We will make use of trig identities to solve this. Here are some common trig identities.
Cos (A + B) = cosAcosB – sinAsinB
Cos (A – B) = cosAcosB + sinAsinB
Sin (A + B) = sinAcosB + sinBcosA
Sin (A – B) = sinAcosB – sinBcosA
Given cos(π/3)cos(π/5) + sin(π/3)sin(π/5) if we let A = π/3 and B = π/5, it reduces to
cosAcosB + sinAsinB and we know that
cosAcosB + sinAsinB = cos(A – B). Therefore,
cos(π/3)cos(π/5) + sin(π/3)sin(π/5) = cos(π/3 – π/5) = cos(2π/15)
