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secondary 4 | A Maths
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Can some one pls help, thank
Date Posted: 2 years ago
Views: 222
"Use the result in part (i)..."
We start from the left side.
Make the numerator in part ii become (sin A + cos A)².
Convert cos 2A into cos² A - sin² A before factorising it into (cos A - sin A) (cos A + sin A).
Then the LHS becomes (cos A + sin A) / (cos A - sin A).
Divide both the numerator and the denominator each by cos A.
You will obtain the right side expression.
We start from the left side.
Make the numerator in part ii become (sin A + cos A)².
Convert cos 2A into cos² A - sin² A before factorising it into (cos A - sin A) (cos A + sin A).
Then the LHS becomes (cos A + sin A) / (cos A - sin A).
Divide both the numerator and the denominator each by cos A.
You will obtain the right side expression.
Thank a lot
Can further help, still cannot solve
cos 2A has three forms.
The cosine-sine version is useful because of the factorisation of the difference of two squares algebraic identity.
The cosine-only version is useful in the situation of 1 + cos 2A where the “1” will vanish later on.
The sine-only version is useful in the situation of 1 - cos 2A where the “1” will vanish later on.
The cosine-sine version is useful because of the factorisation of the difference of two squares algebraic identity.
The cosine-only version is useful in the situation of 1 + cos 2A where the “1” will vanish later on.
The sine-only version is useful in the situation of 1 - cos 2A where the “1” will vanish later on.
Thank for sharing
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done
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The step in which I introduced “1/cos A” in both the numerator and denominator (equivalent to dividing by cos A on both the numerator and denominator) is vital in obtaining the final format.
Date Posted:
2 years ago
Thank a lot
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