Using sum and difference identities from trigonometric identities shows that; Asin(ωt)cos(φ) +Acos(ωt)sin(φ) = Asin(ωt + φ)
<h3>How to prove Trigonometric Identities?</h3>
We know from sum and difference identities that;
sin (α + β) = sin(α)cos(β) + cos(α)sin(β)
sin (α - β) = sin(α)cos(β) - cos(α)sin(β)
c₂ = Acos(φ)
c₁ = Asin(φ)
The Pythagorean identity can be invoked to simplify the sum of squares:
c₁² + c₂² =
(Asin(φ))² + (Acos(φ))²
= A²(sin(φ)² +cos(φ)²)
= A² * 1
= A²
Using common factor as shown in the trigonometric identity above for Asin(ωt)cos(φ) +Acos(ωt)sin(φ) gives us; Asin(ωt + φ)
Complete Question is;
y(t) = distance of weight from equilibrium position
ω = Angular Frequency (measured in radians per second)
A = Amplitude
φ = Phase shift
c₂ = Acos(φ)
c₁ = Asin(φ)
Use the information above and the trigonometric identities to prove that
Asin(ωt + φ) = Asin(ωt)cos(φ) +Acos(ωt)sin(φ)
Read more about Trigonometric Identities at; brainly.com/question/7331447
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Answer:
He wants Rohan to concentrate on the race.
Explanation:
Answer:
yea u can but im not going to
Explanation:
The sampling distribution of the difference in sample means x⁻₁ - x⁻₂ is; 3.5 - 3.5 = 0
<h3>Difference in sample means</h3>
We are told that;
- A fair six-sided die, with sides numbered 1 through 6, will be rolled a total of 15 times.
- x⁻₁ represents the average of the first ten rolls.
- x⁻₂ represents the average of the remaining five rolls.
Now, the average of the largest and lowest numbers of the six sided die is;
E(x) = (6 + 1)/2 = 3.5
Thus, the average mean of the first ten rows is expressed as;
E(x⁻₁) = (n * E(x))/n
E(x⁻₁) = (10 * 3.5)/10
E(x⁻₁) = 3.5
The average mean of the last five rolls will be;
E(x⁻₂) = (n * E(x))/n
E(x⁻₂) = (5 * 3.5)/5
E(x⁻₂) = 3.5
Thus;
x⁻₁ - x⁻₂ = 3.5 - 3.5 = 0
Read more about difference in sample means at; brainly.com/question/16428987
