Answer:
Here you go let me know if you need anything else
Step-by-step explanation:
Answer:
We have to prove
sin(α+β)-sin(α-β)=2 cos α sin β
We will take the left hand side to prove it equal to right hand side
So,
=sin(α+β)-sin(α-β) Eqn 1
We will use the following identities:
sin(α+β)=sin α cos β+cos α sin β
and
sin(α-β)=sin α cos β-cos α sin β
Putting the identities in eqn 1
=sin(α+β)-sin(α-β)
=[ sin α cos β+cos α sin β ]-[sin α cos β-cos α sin β ]
=sin α cos β+cos α sin β- sinα cos β+cos α sin β
sinα cosβ will be cancelled.
=cos α sin β+ cos α sin β
=2 cos α sin β
Hence,
sin(α+β)-sin(α-β)=2 cos α sin β
The answer to #1 is C. The answer to #2 is 9 checks. Hope that helps! :)
M<y = 30 is the correct answer
Answer:
The nth term of the geometric sequence 7, 14, 28, ... is:

Step-by-step explanation:
Given the geometric sequence
7, 14, 28, ...
We know that a geometric sequence has a constant ratio 'r' and is defined by

where a₁ is the first term and r is the common ratio
Computing the ratios of all the adjacent terms

The ratio of all the adjacent terms is the same and equal to

now substituting r = 2 and a₁ = 7 in the nth term


Therefore, the nth term of the geometric sequence 7, 14, 28, ... is:
