ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: dundundundun on March 16, 2015, 06:59:27 pm
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Would really appreciate it if someone could demonstrate how to work these two questions out. Thanks in advance
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Wow! What a tricky little set of questions.
Alright, the first one is:
Let theta = arccos(2x)
Drawing a triangle with theta, the adjacent is 2x, the hypotenuse is 1 (arccos = adjacent/hypotenuse), and the opposite is therefore sqrt(1 - 4x^2)
And since we let theta = arccos(2x), we are left with sin(2 theta), which is equal to 2sin(theta)cos(theta). Working from the triangle above, we get 2 x 2x x sqrt(1 - 4x^2) = 4x sqrt(1 - 4x^2), so a=4, b=2, C.
Second one is:
Substitute cos^2 = 1 - sin^2 (I'm omitting the theta here for simplicity's sake), so after expanding, you would get
a - a sin^2 + b sin^2 = c ---> sin^2 = (c-a)/(b-a)
Substitute sin^2 = 1 - cos^2, so after expanding, you would get
a cos^2 + b - b cos^2 = c ---> cos^2 = (c-b)/(a-b)
tan^2 = sin^2/cos^2 so
tan^2 = (c-a)/(b-a) x (a-b)/(c-b)
=(c-a)/(b-a) x -(b-a)/(c-b)
=-(c-a)/(c-b)
Distributing the negative on top leads to (a-c)/(c-b) which is E.
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Wow! What a tricky little set of questions.
Alright, the first one is:
Let theta = arccos(2x)
Drawing a triangle with theta, the adjacent is 2x, the hypotenuse is 1 (arccos = adjacent/hypotenuse), and the opposite is therefore sqrt(1 - 4x^2)
And since we let theta = arccos(2x), we are left with sin(2 theta), which is equal to 2sin(theta)cos(theta). Working from the triangle above, we get 2 x 2x x sqrt(1 - 4x^2) = 4x sqrt(1 - 4x^2), so a=4, b=2, C.
Second one is:
Substitute cos^2 = 1 - sin^2 (I'm omitting the theta here for simplicity's sake), so after expanding, you would get
a - a sin^2 + b sin^2 = c ---> sin^2 = (c-a)/(b-a)
Substitute sin^2 = 1 - cos^2, so after expanding, you would get
a cos^2 + b - b cos^2 = c ---> cos^2 = (c-b)/(a-b)
tan^2 = sin^2/cos^2 so
tan^2 = (c-a)/(b-a) x (a-b)/(c-b)
=(c-a)/(b-a) x -(b-a)/(c-b)
=-(c-a)/(c-b)
Distributing the negative on top leads to (a-c)/(c-b) which is E.
Thanks so much, that was really helpful I would never have figured them out!
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I got the same answers as wobblywobbly, although I used a slightly different method for the second question:
acos^2(θ) + bsin^2(θ) = c
=> (a - b)cos^2(θ) + bcos^2(θ) + bsin^2(θ) = c
=> (a - b)cos^2(θ) + b = c
=> cos^2(θ) = (c - b) / (a - b)
Since 1 + tan^2(θ) = sec^2(θ) then tan^2(θ) = 1/cos^2(θ) - 1
Thus tan^2(θ) = (a - b) / (c - b) - 1
= (a - b) / (c - b) + (b - c) / (c - b)
= (a - c) / (c - b)
=> Option E
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sorry to bother, do you mind telling me where do you get those 2 questions from? :) :) thanks