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secondary 3 | A Maths
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please help with trigo
Date Posted: 1 year ago
Views: 138
We are basically looking for one angle for principal value questions. Let me explain part (a).
The value of sin 5π/6 is 1/2, if you remember your special angle list. So, the question simplifies into finding the value of sin inverse of (-1/2).
We are finding a specific principal angle, θ, such that θ = sin inverse (-1/2). The principal range of sin inverse of a value is -π/2 ≤ θ ≤ π/2.
In other words, we are essentially solving the equation sin θ = -1/2, where -π/2 ≤ θ ≤ π/2. So, the angle we wish to look for is either in the fourth quadrant or the first quadrant. Since the right side value is negative, our required angle must be in the fourth quadrant.
The basic/reference angle to this is π/6, which is pretty much expected since we are taking the sine inverse of a sine.
The angle we wish to look for is a fourth , quadrant angle with basic angle π/6, or in short, -π/6.
The value of sin 5π/6 is 1/2, if you remember your special angle list. So, the question simplifies into finding the value of sin inverse of (-1/2).
We are finding a specific principal angle, θ, such that θ = sin inverse (-1/2). The principal range of sin inverse of a value is -π/2 ≤ θ ≤ π/2.
In other words, we are essentially solving the equation sin θ = -1/2, where -π/2 ≤ θ ≤ π/2. So, the angle we wish to look for is either in the fourth quadrant or the first quadrant. Since the right side value is negative, our required angle must be in the fourth quadrant.
The basic/reference angle to this is π/6, which is pretty much expected since we are taking the sine inverse of a sine.
The angle we wish to look for is a fourth , quadrant angle with basic angle π/6, or in short, -π/6.
The cos version is similar, but the principal value range is 0 ≤ θ ≤ π. Your final answer must either be in the first quadrant or second quadrant, but cos 5π/3 has a positive value, so the negative makes -cos 5π/3 negative.
You know what to do now.
You know what to do now.
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done
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I haven’t done principle values in a while, I hope this is correct and helps you :)
Date Posted:
1 year ago
Do note that for part (a), the principal angles for arcsin (inverse sin) are in the range -90° ≤ θ ≤ 90° in the case of degrees or -π/2 ≤ θ ≤ π/2 in the case of radians.
Correct for part (a) is -π/6, which is basically a fourth quadrant version of the basic reference angle π/6.
Do note that for part (b), the principal angles for arccos (inverse cos) are in the range 0° ≤ θ ≤ 180° in the case of degrees or 0 ≤ θ ≤ π in the case of radians.
Correct for part (b) is 2π/3, which is basically a second quadrant version of the basic reference angle π/3.
Correct for part (a) is -π/6, which is basically a fourth quadrant version of the basic reference angle π/6.
Do note that for part (b), the principal angles for arccos (inverse cos) are in the range 0° ≤ θ ≤ 180° in the case of degrees or 0 ≤ θ ≤ π in the case of radians.
Correct for part (b) is 2π/3, which is basically a second quadrant version of the basic reference angle π/3.
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