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secondary 4 | A Maths
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Sonia
Sonia

secondary 4 chevron_right A Maths chevron_right Singapore

Hello! pleas help me with these ^^ Thank u sooo much:))!!!

Date Posted: 4 years ago
Views: 185

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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
Good evening Sonia!!! Here are my workings for the first question.

The “3x” in the angle “3x - 75” (applies for cos and tan mainly, sin is trickier than this at times) suggests that there are 6 solutions altogether (each x will give 2 solutions in the range 0 to 360 degrees).

This is because x is limited to one cycle 0 to 360, so 3x is limited to three cycles 0 to 1080, with two solutions per cycle since two Quadrants per cycle are positive and two Quadrants are negative.

Let me know if you need more explanation and I will do my best to explain them again!
Eric Nicholas K
Eric Nicholas K
4 years ago
The three cycles include the current cycle (0 to 360), one cycle forward (360 to 720) and two cycles forward (720 to 1080).

There are 6 solutions altogether no matter what.

I standby one cycle backward (-360 to 0) for a replacement value in case one value from the “two cycles forward” get rejected.

We should also standby three cycles forward (1080 to 1440) for a replacement value in case one value from the current cycle (0 to 360) gets rejected.
Sonia
Sonia
4 years ago
Hi Eric! Thanks so much for your fast reply, i’m not very sure about how you got the negative 60 as one of your answers ?
Eric Nicholas K
Eric Nicholas K
4 years ago
Hi Sonia!!!

For the angle 3x - 75, its value must lie between -75 and 1005, so that x will lie between 0 and 360.

The fact that there is the region from -75 to 0 in the range -75 to 1005 means that I may need to extract angles “from one cycle before”, so that I would account for all angles. As it turns out, there does exist one angle from -75 to 0.

Now, cos (3x - 75) = 0.5, and 0.5 is positive, so 3x - 75 must lie in the first quadrant or the fourth quadrant, which is characterised by alpha (for the first quadrant) and 360 - alpha (for the fourth quadrant).

Here, alpha is cos-1 (0.5) = 60 degrees so 360 - alpha is 300.

Because we are going back one cycle, we need to include two angles from one cycle back, which we have to subtract 360 from the two angles.

The two angles, one cycle back, are 60 - 360 = -300 and 300 - 360 = -60.

-60 lies in the range -75 to 1005, so we have to accept this angle.
Sonia
Sonia
4 years ago
Thank you so much Eric!!! Your explanations are so clear and i understand them! :) Have a great remaining friday and a blessed weekend !!
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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
Hi Sonia! Here are my workings for the second question.

Suppose we are solving for x in the equation

sin x = -1/2.

Regardless of whether the right side value is positive or negative, we make the basic angle as sin inverse of the positive value).

So the basic angle is sin-1 (1/2).

The Quadrants will take care of the actual positive or negative sign on the right side.

First quadrant is characterised by x. Second quadrant is characterised by 180 - x. Third quadrant is characterised by 180 + x. Fourth quadrant is characterised by 360 - x.

Since the right side is negative, x lies in the third or fourth quadrant.

Had the right side been positive, x would lie in the first or second quadrant.

Let me know if you need more explanation and I will do my best to explain them again!
Eric Nicholas K
Eric Nicholas K
4 years ago
Hi Sonia! Here is a minor correction to my text explanation.

First quadrant is characterised by x = alpha, where alpha is the basic acute angle.

Second quadrant is characterised by x = 180 - alpha.

Third quadrant is characterised by x = 180 + alpha.

Fourth quadrant is characterised by x = 360 - alpha.

As a side note, for each round that you need to go back (to collect some angles), you need to subtract 360 degrees from the angles that you have found in the range 0 to 360. Going one round back means subtracting 360. Two rounds backward is subtracting 720. And so on.

For each round that you need to go forward (to collect some angles), you need to add 360 degrees from the angles that you have found in the range 0 to 360. Going one round forward means adding 360. Two rounds forward is adding 720. And so on.

Usually they are not so evil to make you go beyond three rounds forward.