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junior college 1 | H2 Maths
2 Answers Below
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Hi, kindly advise. Thanks alot!
Date Posted: 1 year ago
Views: 350
hint:
cos(π/2-x) = sin(x)
so,
[cos((n-1)π/2n)]^2
=[cos(π/2-π/2n)]^2
=[sin(π/2n)]^2
consider the expression inside the braces,
apply [sin(x)]^2+[cos(x)]^2 = 1 identity to ...
1st & last term,
2nd & 2nd last term,
etc.
cos(π/2-x) = sin(x)
so,
[cos((n-1)π/2n)]^2
=[cos(π/2-π/2n)]^2
=[sin(π/2n)]^2
consider the expression inside the braces,
apply [sin(x)]^2+[cos(x)]^2 = 1 identity to ...
1st & last term,
2nd & 2nd last term,
etc.
Thank you, Sir!
See 2 Answers
done
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Good morning Candice! Here is part one of my workings.
Date Posted:
1 year ago
Good morning, Mr Eric!
Thank you very much for coming to my help :)
Really appreciate so much on your detailed working solutions and explanations!
I would like to take this opportunity to wish you a Merry Christmas too...
Thank you very much for coming to my help :)
Really appreciate so much on your detailed working solutions and explanations!
I would like to take this opportunity to wish you a Merry Christmas too...
done
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Good morning Candice! Here is the second part of my workings.
Date Posted:
1 year ago
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