The commutative property of addition states that a + b = b + a, in other words, the numbers in an addition problem can be flipped, but the result will be the same, the answer is C.
Answer:
C=y=sin1/2x
Step-by-step explanation:
As given in the graph:
Amplitude= 1
period=2π
Finding function of sin that have period of 4π and amplitude 1
A: y=1/2sinx
Using the formula asin(bx-c)+d to find the amplitude and period
a=1/2
b=1
c=0
d=0
Amplitude=|a|
=1/2
Period= 2π/b
=2π
B: y=sin2x
Using the formula asin(bx-c)+d to find the amplitude and period
a=1
b=2
c=0
d=0
Amplitude=|a|
=1
Period= 2π/2
=π
C: y=sin1/2x
Using the formula asin(bx-c)+d to find the amplitude and period
a=1
b=1/2
c=0
d=0
Amplitude=|a|
=1
Period= 2π/1/2
=4π
D: y=sin1/4x
Using the formula asin(bx-c)+d to find the amplitude and period
a=1
b=1/4
c=0
d=0
Amplitude=|a|
=1
Period= 2π/1/4
=8π
Hence only c: y=sin1/2x has period of 2π and amplitude 1
We have proven that the trigonometric identity [(tan θ)/(1 - cot θ)] + [(cot θ)/(1 - tan θ)] equals 1 + (secθ * cosec θ)
<h3>How to solve Trigonometric Identities?</h3>
We want to prove the trigonometric identity;
[(tan θ)/(1 - cot θ)] + [(cot θ)/(1 - tan θ)] = 1 + sec θ
The left hand side can be expressed as;
[(tan θ)/(1 - (1/tan θ)] + [(1/tan θ)/(1 - tan θ)]
⇒ [tan²θ/(tanθ - 1)] - [1/(tan θ(tanθ - 1)]
Taking the LCM and multiplying gives;
(tan³θ - 1)/(tanθ(tanθ - 1))
This can also be expressed as;
(tan³θ - 1³)/(tanθ(tanθ - 1))
By expansion of algebra this gives;
[(tanθ - 1)(tan²θ + tanθ.1 + 1²)]/[tanθ(tanθ(tanθ - 1))]
Solving Further gives;
(sec²θ + tanθ)/tanθ
⇒ sec²θ * cotθ + 1
⇒ (1/cos²θ * cos θ/sin θ) + 1
⇒ (1/cos θ * 1/sin θ) + 1
⇒ 1 + (secθ * cosec θ)
Read more about Trigonometric Identities at; brainly.com/question/7331447
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