proving trig identity problem (2 Viewers)

[Aznmichael92]

hey i was going through the fitzpatrick textbook and didnt know how to do one of its question. Can someone please help me?

The question is:

Prove (sin2A + 1)/cos2A = (cosA + sinA)/(cosA - sinA)

thanks

 

[Just.Snaz]

(sinA)^2 + (cosA)^2 = 1
so sub the 1 into the LHS
LHS = (sin2A + sinA^2 + cosA^2)/cos2A
= [2sinAcosA + sinA^2 + cosA^2]/cos2A
= [cosA + sinA]^2 / (cosA)^2 - (sinA)^2
= [cosA + sinA]^2 / [CosA + SinA][CosA - sinA]
= (cosA + sin A)/(cosA - sin A)
= RHS

 

[ronnknee]

Okay what I would first is compare both sides in my head just to see what approach to take

You notice how if you change the denominator to (cos² A - sin² A) ie. (cos A - sin A)(cos A + sin A) <- difference of two squares
And you compare it to the right, (cos A + sin A) is missing from the left?
So somehow we need to make the numerator (cos A + sin A) to get rid of it

So:

LHS
= (sin 2A + 1) / cos 2A
= 2sin A cos A + 1 / (cos A - sin A)(cos A + sin A)

Now I want cos A + sin A at the top so I change the 1 to (sin² A + cos² A)

= 2 sin A cos A + sin² A + cos² A / (cos A - sin A)(cos A + sin A)

Now if we arrange the numerator like this:

= cos² A + 2 sin A cos A + sin² A / (cos A - sin A)(cos A + sin A)

Can you see that it's a perfect square?

= (cos A + sin A)² / (cos A - sin A)(cos A + sin A)
= cos A + sin A / cos A - sin A
= RHS


Alternatively you can start from RHS and prove it is equal to LHS
I think that would be easier as there's no trick to it (the perfect square)
Just start with RHS
Multiple top and bottom by (cos A + sin A) and then work backwards in my solution

 

[bored of sc]

Prove (sin2A + 1)/cos2A = (cosA + sinA)/(cosA - sinA)

LHS = (2sinAcosA + sin²A + cos²A)/(cos²A-sin²A) ------ rewrite double angle formulae and recognise 1 as the fundamental identity
= (sinA+cosA)²/[(cosA-sinA)(cosA+sinA)] ------- recognise numerator as perfect square and denominator as difference of two squares
= (cosA + sinA)/(cosA - sinA) ------ cancel
= RHS

P.S That's a lovely question. I enjoyed it.

 

[leoyh]

just multiply top and bottom of the RHS by (CosA + SinA), bottom becomes (CosA)^2 - (SinA)^2 which is cos 2A and top you should see becomes Sin2A + 1

 

[Aznmichael92]

thanks all for the help. I got the working out correct but didnt pick up the (sin+cos)^2 bit in my working out. lol. I guess I need to be more observant. Thanks once again!