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secondary 3 | A Maths
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Hi, could you kindly advise me the working solution for this question please? Thanks :)
Date Posted: 3 years ago
Views: 309
(cos³ A - sin³ A)/(cos A - sin A)
= (cos A - sin A)(cos² A + cos A sin A + sin² A) / (cos A - sin A)
(Using formula a³ - b³ = (a - b)(a² + ab + b²) )
= 1 + cos A sin A
(Using cos² A + sin² A = 1)
= 1 + ½ sin 2A
(Using property sin 2A = 2 sin A cos A, so ½ sin 2A = sin A cos A)
= (2 + sin 2A)/2
(Common denominator)
(Proved)
= (cos A - sin A)(cos² A + cos A sin A + sin² A) / (cos A - sin A)
(Using formula a³ - b³ = (a - b)(a² + ab + b²) )
= 1 + cos A sin A
(Using cos² A + sin² A = 1)
= 1 + ½ sin 2A
(Using property sin 2A = 2 sin A cos A, so ½ sin 2A = sin A cos A)
= (2 + sin 2A)/2
(Common denominator)
(Proved)
Good morning and thank you Mr J :)
I think there is a minor typo error and I know where it is.
Thank you very much for responding to my question.
I think there is a minor typo error and I know where it is.
Thank you very much for responding to my question.
Ah yes. Last two lines.
Edited
Thanks Mr J :)
Welcome.
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done
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Good evening Candice! Here are my workings for this question.
Date Posted:
3 years ago
Good morning Mr Eric :)
Thank you so much for working solution.
I got it now :D
Thank you so much for working solution.
I got it now :D
If you want to use the substitution , you'll have to first state ' sub a = cos A, b = sin A'
Presenting like this will result in loss of marks.
Candice do take note.
Presenting like this will result in loss of marks.
Candice do take note.
Noted with thanks, Mr J
Usually for the small details I skip the formalities. The students should more or less be prepared that I have used a substitution.
It's always better to lead by example and show full proper working. You never know if a student copies verbatim and then gets marked down
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