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Addmaths paper 1

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It could be done without substitution, you just had to deal with (-sin Q - cos Q) as one whole. I did not find it appropriate to follow working in this way so I substituted x.

I dont think i vil b getting marks on that one ... I just solved it sumhow, n wrote 'Proved' in the end :D
 
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Would it get two or two would be cut...?
I wrote the gradient as 2 instead of 1/2 and then multiplying by 6 coz i was in a hurry lol...
And how much in The Binomial ques?

You'd get 2 I think.
In binomial, for second part. 1 mark.
 
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For the k, gradient and binomial, you'll only get partial credit for k and binomial. 1-2 marks for the gradient one.
Ok so, I let x = -sin Q - cos Q.
Then,
=(1+x)^2 - 2(1-cos Q)(1-sin Q)
=1 + 2x + x^2 - 2(1-sin Q - cos Q + sin Q cos Q).
=1 + 2(-sin Q - cos Q) + (-sin Q - cos Q)^2 - 2 + 2 sin Q + 2 cos Q - 2 sin Q cos Q
=1 - 2 sin Q -cos Q + sin^2 Q + 2 sin Q cos Q + cos^2 Q - 2 + 2 sin Q + 2 cos Q - 2 sin Q cos Q
Rearranging all the terms correctly,
=1-2 -2 sin Q + 2 sin Q - 2 cos Q + 2 cos Q + 2 sin Q cos Q - 2 sin Q cos Q + (sin^2 Q + cos^2 Q)
=-1 + (1)
=0

I shifted one half of equation to rhs to form identity, then expanded LHS then factorized to form RHS!!!
 
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