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secondary 3 | A Maths
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Need help for (b). Thanks
Date Posted: 3 years ago
Views: 308
cos pi is -1 so...
Tan (x - 0.1) = 1
Basic angle
= Tan inverse 1
= pi/4
Since 1 is a positive number,
x - 0.1 = 1st or 3rd quad (and 3rd quad from the previous cycle, since the range given extends to -pi)
x - 0.1 = pi/4 or pi + pi/4 (and pi + pi/4 - 2pi)
x = pi/4 + 0.1 or 5pi/4 + 0.1 or -3pi/4 + 0.1
Tan (x - 0.1) = 1
Basic angle
= Tan inverse 1
= pi/4
Since 1 is a positive number,
x - 0.1 = 1st or 3rd quad (and 3rd quad from the previous cycle, since the range given extends to -pi)
x - 0.1 = pi/4 or pi + pi/4 (and pi + pi/4 - 2pi)
x = pi/4 + 0.1 or 5pi/4 + 0.1 or -3pi/4 + 0.1
Thanks Mr. Eric
Mr. Eric, can help with this qn?
a)Solve 1+sinx =root3 cosx for 0
My ans is 90 and 30 for a).Answer key says 30 only. Why 90 is not the ans for part a)
If I sketch graph, there is only 1 solution. Which must be 30 degree. Am I correct?
So for part b, what explanation we give?
a)Solve 1+sinx =root3 cosx for 0
My ans is 90 and 30 for a).Answer key says 30 only. Why 90 is not the ans for part a)
If I sketch graph, there is only 1 solution. Which must be 30 degree. Am I correct?
So for part b, what explanation we give?
Hang on, later at a very late time if I am not tired I will look at your question
Ok sure. If u have new maths 360, qns 12,13,14 (Exercise 11.2) page 280
That book I don’t actually have with me.
I will return close to 12:30 am, but I am not sure if I can spare the time.
I will return close to 12:30 am, but I am not sure if I can spare the time.
See 2 Answers
done
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So first of all you do as per normal of finding what the basic angle is. However look at the angle range for this tan(x-0.1). So basically it is like shifting the curve to the right in the +ve direction of the x-axis. Therefore you need to consider the change in principle angle too which is shown as the one that i corrected. Then while you solve the anti clockwise motion which you'll find 2 roots to the solution, you'll have to consider the clockwise motion which is the negative side of the angle. So there will be a total of 3 solutions to this question. (P.S. dont forget to change it to x form !! )
Date Posted:
3 years ago
Thank u
done
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Interestingly I did not obtain 90 at all this time.
Date Posted:
3 years ago
Thanks Mr Eric. I will send my working. Let me know where I am wrong. I send my working as a question (photo)
A and B are acute angles such that cos ( A-B) =4/5 and SinASinB =1/5.
Mr Eric, pls explain why the condition that A and B are acute angles has to be specified in the question?
Mr Eric, pls explain why the condition that A and B are acute angles has to be specified in the question?
Now, it could well be the case where A is, for example, a third quadrant angle of 200 and B is a second quadrant angle of 163, so that the angle difference is 37 (and cos 37 is roughly 4/5).
Supposing this is the case, then sin A would be negative while sin B would be positive. Their product is negative in this instance, unlike the given 1/5.
This is why the question must state that both A and B are acute, to avoid such ambiguities.
Supposing this is the case, then sin A would be negative while sin B would be positive. Their product is negative in this instance, unlike the given 1/5.
This is why the question must state that both A and B are acute, to avoid such ambiguities.
Oh ok
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