23PI/6= 2PI + 11PI/6=2PI+5PI/6 + PI=4PI-PI/6 which is easily figured, it is -30 deg, right?
To solve problems of finding exact trigonometric function values, you would need two concepts, the coterminal angle and the reference angle.
Trignometric angles are measured starting from the positive axis, called the initial side. The other side where the angle ends is called the terminal side.
An angle is coterminal with another when the terminal side of two angles coincide. Which means that the difference between the two angles are multiples of 2π. For example, π/4 and π/9 are coterminal.
The trigonometric function values of coterminal angles are identical.
For the given problem, the first step is to find an angle between 0 and 2π that is coterminal with the given angle, namely 23π/6. Try subtracting 2π to give 11π/6, which is less than 2π and greater than zero.
To find the functional values of an angle between 0 and 2π, we need to find the reference angle, which is the positive angle between 0 and π/2 that the angle makes with the x-axis (positive or negative), i.e. 0 or π.
The reference angle for 11π/6 is π/6, because it is the difference from 2π.
The value of sec(23π/6) is ±sec(π/6) = ± 1/cos(π6) = ± 1/(√3/2) = ± 2√3/3
To determine the sign of the value, we know that the angle 5π/6 lies in the fourth quadrant where the value of cosine is positive. Therefore we conclude that
sec(23π/6) = 2√3/3
...For example, π/4 and 9π/4 are coterminal.
... The value of sec(23π/6) is ±sec(π/6) = ± 1/cos(π/6) = ± 1/(√3/2) = ± 2√3/3
To determine the sign of the value, we know that the angle 11π/6 lies in the fourth
Oh, so basically.. I would subtract 2pi, which gives me 11pi/6. The cos of 11pi/6 is sqrt(3)/2. Since the sec is 1/cos(23pi/6), it would also be 1/(sqrt(3)/2)? Simplified, that gives me a final answer of 2sqrt(2)/3, which is the answer I was looking for. Thank you for your responses!
Basically, yes. But double-check the sign of the value. In this case, cos(11π/6)=cos(π/6), but not all the time. Sometimes they have opposite signs.
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