We are asked to use the half-angle formulas to find the exact values of sine, cosine, and tangent of the angle [tex]\(\theta/2\)[/tex], given that [tex]\(\sin(\theta) = \frac{1}{2}\) and \(\cos(\theta) = \frac{1}{2}\)[/tex].
The half-angle formulas allow us to express trigonometric functions of an angle [tex]\(\theta/2\[/tex]) in terms of the trigonometric functions of[tex]\(\theta\)[/tex]. The formulas are as follows:
[tex]\(\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}}\)\(\cos(\frac{\theta}{2}) = \pm \sqrt{\frac{1 + \cos(\theta)}{2}}\)\(\tan(\frac{\theta}{2}) = \frac{\sin(\theta)}{1 + \cos(\theta)}\)[/tex]
Given that [tex]\(\sin(\theta) = \frac{1}{2}\) and \(\cos(\theta) = \frac{1}{2}\)[/tex], we can substitute these values into the half-angle formulas.
For [tex]\(\sin(\frac{\theta}{2})\)[/tex]:
[tex]\(\sin(\frac{\theta}{2}) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}} = \pm \sqrt{\frac{1 - \frac{1}{2}}{2}} = \pm \frac{1}{2}\)[/tex]
For [tex]\(\cos(\frac{\theta}{2})\):\(\cos(\frac{\theta}{2}) = \pm \sqrt{\frac{1 + \cos(\theta)}{2}} = \pm \sqrt{\frac{1 + \frac{1}{2}}{2}} = \pm \frac{\sqrt{3}}{2}\)[/tex]
For[tex]\(\tan(\frac{\theta}{2})\):\(\tan(\frac{\theta}{2}) = \frac{\sin(\theta)}{1 + \cos(\theta)} = \frac{\frac{1}{2}}{1 + \frac{1}{2}} = \frac{1}{3}\)[/tex]
Therefore, using the half-angle formulas, we find that \[tex](\sin(\frac{\theta}{2}) = \pm \frac{1}{2}\), \(\cos(\frac{\theta}{2}) = \pm \frac{\sqrt{3}}{2}\), and \(\tan(\frac{\theta}{2}) = \frac{1}{3}\).[/tex]
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1.Find the period of the following functions. a) f(t) = (7 cos t)² b) f(t) = cos (2φt²/m)
Period of the functions: The period of the function f(t) = (7 cos t)² is given by 2π/b where b is the period of cos t.The period of the function f(t) = cos (2φt²/m) is given by T = √(4πm/φ). The period of the function f(t) = (7 cos t)² is given by 2π/b where b is the period of cos t.
We know that cos (t) is periodic and has a period of 2π.∴ b = 2π∴ The period of the function f(t) =
(7 cos t)² = 2π/b = 2π/2π = 1.
The period of the function f(t) = cos (2φt²/m) is given by T = √(4πm/φ) Hence, the period of the function f(t) =
cos (2φt²/m) is √(4πm/φ).
The function f(t) = (7 cos t)² is a trigonometric function and it is periodic. The period of the function is given by 2π/b where b is the period of cos t. As cos (t) is periodic and has a period of 2π, the period of the function f(t) = (7 cos t)² is 2π/2π = 1. Hence, the period of the function f(t) = (7 cos t)² is 1.The function f(t) = cos (2φt²/m) is also a trigonometric function and is periodic. The period of this function is given by T = √(4πm/φ). Therefore, the period of the function f(t) = cos (2φt²/m) is √(4πm/φ).
The period of the function f(t) = (7 cos t)² is 1, and the period of the function f(t) = cos (2φt²/m) is √(4πm/φ).
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Jeff has 32,400 pairs of sunglasses. He wants to distribute them evenly among X people, where X is
a positive integer between 10 and 180, inclusive. For how many X is this possible?
Answer:
To distribute 32,400 pairs of sunglasses evenly among X people, we need to find the positive integer values of X that divide 32,400 without any remainder.
To determine the values of X for which this is possible, we can iterate through the positive integers from 10 to 180 and check if 32,400 is divisible by each integer.
Let's calculate:
Number of possible values for X = 0
For each value of X from 10 to 180, we check if 32,400 is divisible by X using the modulo operator (%):
for X = 10:
32,400 % 10 = 0 (divisible)
for X = 11:
32,400 % 11 = 9 (not divisible)
for X = 12:
32,400 % 12 = 0 (divisible)
...
for X = 180:
32,400 % 180 = 0 (divisible)
We continue this process for all values of X from 10 to 180. If the remainder is 0, it means that 32,400 is divisible by X.
In this case, the number of possible values for X is the count of the integers from 10 to 180 where 32,400 is divisible without a remainder.
After performing the calculations, we find that 32,400 is divisible by the following values of X: 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, 40, 45, 48, 50, 54, 60, 64, 72, 75, 80, 90, 96, 100, 108, 120, 128, 135, 144, 150, 160, 180.
Therefore, there are 33 possible values for X between 10 and 180 (inclusive) for which it is possible to distribute 32,400 pairs of sunglasses evenly.
Hope it helps!