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rsin(theta-a)

The answer in the book is different to what I'm getting. Not sure if it's a typo or if I have made a silly mistake. Any help is appreciated.IMG_20200904_162315-compressed.jpg.jpeg
Original post by Cpt Avocado
The answer in the book is different to what I'm getting. Not sure if it's a typo or if I have made a silly mistake. Any help is appreciated.IMG_20200904_162315-compressed.jpg.jpeg

Since sin⁡α<0\sin \alpha < 0 and cos⁡α<0\cos \alpha < 0 it means that π<α<3π2\pi < \alpha < \dfrac{3\pi}{2} (and adding multiples of 2pi also gives a valid α\alpha).

Your error is that you found α\alpha from rsin⁡α=−3r\sin\alpha = -\sqrt{3}, but you haven't checked if it satisfies rcos⁡α=−1r\cos\alpha = -1 ... It does not.

So you need to choose a different solution, precisely the one in the interval I say.

Also, the book's answer is incorrect.

https://www.desmos.com/calculator/0mz7x503pv (Red curve is what you want to replicate. Blue is the book's answer, and green is your answer. Neither one is replicating the red curve)
(edited 3 years ago)
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Cpt Avocado
OP
9
Original post by RDKGames
Since sin⁡α<0\sin \alpha < 0 and cos⁡α<0\cos \alpha < 0 it means that π<α<3π4\pi < \alpha < \dfrac{3\pi}{4} (and adding multiples of 2pi also gives a valid α\alpha).

Your error is that you found α\alpha from rsin⁡α=−3r\sin\alpha = -\sqrt{3}, but you haven't checked if it satisfies rcos⁡α=−1r\cos\alpha = -1 ... It does not.

So you need to choose a different solution, precisely the one in the interval I say.

Also, the book's answer is incorrect.

https://www.desmos.com/calculator/0mz7x503pv (Red curve is what you want to replicate. Blue is the book's answer, and green is your answer. Neither one is replicating the red curve)

In the question it says that π<α<pi2\pi < \alpha < \dfrac{pi}{2} . However there are no values of alpha that satisfy both this and sina = -root3/2 as far as I am aware?
Original post by Cpt Avocado
In the question it says that π<α<pi2\pi < \alpha < \dfrac{pi}{2} . However there are no values of alpha that satisfy both this and sina = -root3/2 as far as I am aware?

Sure there is.

Remember that solving a trig equation gives many solutions. But you just need the one in this interval. So if you remember how to obtain further solutions to trig equations, you just do that here and there's your alpha.
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Cpt Avocado
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Original post by RDKGames
Sure there is.

Remember that solving a trig equation gives many solutions. But you just need the one in this interval. So if you remember how to obtain further solutions to trig equations, you just do that here and there's your alpha.

to obtain further solutions to sin you do pi-ans and then +or- 2pi to first and second answers. starting with -pi/3 i get 5pi/3 and 4pi/3. 4pi/3 satisfies both the sin and cos equation and gives the correct graph on the website you linked. however it is not less than pi/2.
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davros
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Original post by Cpt Avocado
In the question it says that π<α<pi2\pi < \alpha < \dfrac{pi}{2} . However there are no values of alpha that satisfy both this and sina = -root3/2 as far as I am aware?

That interval doesn't seem to make sense! Can you upload a picture of the question from the book?
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Cpt Avocado
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image-6d0e2c0c-5fb3-48f9-a23a-89bdd500e6a52847695440639510214-compressed.jpg.jpeg question 5
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davros
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Original post by Cpt Avocado
image-6d0e2c0c-5fb3-48f9-a23a-89bdd500e6a52847695440639510214-compressed.jpg.jpeg question 5

OK as @RDKGames pointed out above your book seems to be seriously screwed here!

You need sin⁡α<0\sin \alpha < 0 and cos⁡α<0\cos \alpha < 0 which puts alpha in the 3rd quadrant if you look at a CAST diagram i.e. it doesn't belong in the interval quoted in the question.

The answer that you've quoted from the book solves a different problem:

2sin⁡(θ−π6)=3sin⁡θ−cos⁡θ2\sin(\theta - \dfrac{\pi}{6}) = \sqrt{3}\sin \theta - \cos \theta

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