Can I ask you eutas1? Drawing is not required in the problem. Why do you draw?Wait am I explaining what I did in attachment T? Or the way I did it originally?
Can I ask you eutas1? Drawing is not required in the problem. Why do you draw?Wait am I explaining what I did in attachment T? Or the way I did it originally?
Assuming that by (c) you mean part (c) as posted in reply #1: 18(c) \(\cos\left(\dfrac{\pi}{2}-x\right)\) given \(x\in \left(\dfrac{\pi}{2},\pi\right)\).I don't understand your solution to (c)... Could you please explain step-by-step?
We have learnt to draw for simple questions, and it makes it easier to visually see what is going on.Can I ask you eutas1? Drawing is not required in the problem. Why do you draw?
I have not learnt angle sum and difference identities...Assuming that by (c) you mean part (c) as posted in reply #1: 18(c) \(\cos\left(\dfrac{\pi}{2}-x\right)\) given \(x\in \left(\dfrac{\pi}{2},\pi\right)\).
In my opinion the given is totally extraneous. \(\cos(a-b)=sin(a) sin(b) + cos(a) cos(b)\)
We should know that \(\cos\left(\dfrac{\pi}{2}\right)=0~\&~\sin\left(\dfrac{\pi}{2}\right)=1\).
Taken together we get simply \(\cos\left(\dfrac{\pi}{2}-x\right)=\sin(x)\). SEE HERE
I want to know what you are being taught (and others want to know, too!). You said that what you showed in attachment T was what the teacher showed you for (c), so I want to know what was said about it. I don't care what you did earlier that was wrong; I want to understand what you were told is right.Wait am I explaining what I did in attachment T? Or the way I did it originally?
My understanding is that, in context, the problem expects this particular approach.Can I ask you eutas1? Drawing is not required in the problem. Why do you draw?
Please refer to my attachment:I want to know what you are being taught (and others want to know, too!). You said that what you showed in attachment T was what the teacher showed you for (c), so I want to know what was said about it. I don't care what you did earlier that was wrong; I want to understand what you were told is right.
My understanding is that you are early in the process of learning about trig functions in terms of the unit circle, which is why a visual approach is appropriate; there is a lot you haven't learned yet. I think this problem goes too far with that, but I have certainly taught such methods, using reference angles and a triangle, to explain basic ideas such as symmetries in the functions.
Oh yeah... does that mean it is wrong?The trouble is, this "x" is not in the right domain; I can only guess that it is not really meant to be the x of the problem. It appears to be the reference angle all the way through.
Isn't it clear that the angle marked as x is acute?? The angle that is in the right domain is the unmarked angle from the positive x axis to the final yellow line. (A line is not an angle.)Oh yeah... does that mean it is wrong?
And why is x not in the right domain? Isn't the domain (pi/2, pi)?
Ahhhhhhhhh I am not sure what to do............Isn't it clear that the angle marked as x is acute?? The angle that is in the right domain is the unmarked angle from the positive x axis to the final yellow line. (A line is not an angle.)
My sense is that your teacher is not writing what they mean. What they mean, and perhaps even what they say to you in explaining it, may be right, but a math teacher really should teach how to write what you mean. The picture does not communicate truth.
I don't get the part after 'Because' ....This problem is EASY if you use basic identities. Why are you even discussing reference angles?
\(\displaystyle sin( \theta ) = cos \left ( \dfrac{\pi}{2} - \theta \right ).\)
Why is that relevant? Because
\(\displaystyle sin \left ( \dfrac{3 \pi}{2} - x \right ) = cos \left \{ \dfrac{\pi}{2} - \left ( \dfrac{3 \pi}{2} - x \right ) \right \} = cos(x - \pi).\)
Do you agree thatI don't get the part after 'Because' ....
Ohh, I see now.Do you agree that
\(\displaystyle sin ( \theta ) = \cos \left ( \dfrac{\pi}{2} - \theta \right ).\)
\(\displaystyle \text {Let } \theta = \dfrac{3 \pi}{2} - x \implies\)
\(\displaystyle \dfrac{\pi}{2} - \theta = \dfrac{\pi}{2} - \left ( \dfrac{3 \pi}{2} - x \right ) = \dfrac{\pi(1 - 3)}{2} + x = x - \pi.\)
Basic algebra.
\(\displaystyle \therefore sin \left ( \dfrac{3 \pi}{2} - x \right ) = cos(x - \pi).\)
What is there not to understand?