I need help with Geometry. If you can help comment and I will message you.
1 answer:
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Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π and cos A = cos B · cos C
scratchwork:
A + B + C = π
A = π - (B + C)
cos A = cos [π - (B + C)] Apply cos
= - cos (B + C) Simplify
= -(cos B · cos C - sin B · sin C) Sum Identity
= sin B · sin C - cos B · cos C Simplify
cos B · cos C = sin B · sin C - cos B · cos C Substitution
2cos B · cos C = sin B · sin C Addition
Division
2 = tan B · tan C
<u>Proof LHS → RHS</u>
Given: A + B + C = π
Subtraction: A = π - (B + C)
Apply tan: tan A = tan(π - (B + C))
Simplify: = - tan (B + C)
Substitution: = -(tan B + tan C)/(1 - 2)
Simplify: = -(tan B + tan C)/-1
= tan B + tan C
LHS = RHS: tan B + tan C = tan B + tan C
For the given geometric sequence, we have:
So the correct option is B.
<h3>Which is the recursive formula?</h3>Here we have the geometric sequence:
To get the initial value of the geometric sequence, we just need to replace n by 1, so we get:
Now, notice that the common ratio is (-1/3), this means that each term of the sequence is (-1/3) times the previous term.
Then the correct option is B.
If you want to learn more about geometric sequences:
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