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Trig Issue - Is this step even allowed?

Hi there,

here is the question:

The issue I have is this step:

I don't understand how they are able to multiply ONLY the denominator to make it the same when surely you have to multiply the numerator too??

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Reply 1

HapaxOromenon3

Original post by CrazyFool229

Hi there,

here is the question:

The issue I have is this step:

I don't understand how they are able to multiply ONLY the denominator to make it the same when surely you have to multiply the numerator too??

Multiplying the numerator and denominator of 1/1 by cos 2u gives cos 2u / cos 2u.
Thus we have cos 2u / cos 2u + 1 / cos 2u = (cos 2u + 1) / cos 2u, as required.

Reply 2

the bear

i would factorise out a cosx on the top & bottom then cancel... this leaves ( 1 - cosx ) / ( 1 + cosx )... then look at cosx as being cos( 2*x/2) and work with identities for cos2θ

Reply 3

username2488725

Original post by HapaxOromenon3

Multiplying the numerator and denominator of 1/1 by cos 2u gives cos 2u / cos 2u.
Thus we have cos 2u / cos 2u + 1 / cos 2u = (cos 2u + 1) / cos 2u, as required.

Dear god... Thanks a lot

I just realised when I was looking at my working I made a very basic mistake with my working out which made it very confusing to follow. Your explanation cleared it all up. Thanks again!