Simplifying trig functions

[garytse86]

I am currently learning how to simplify trig functions, but is there a way to know which formulae to use?

In my text book there are three formulae:

cos^2ttheta + sin^2theta = 1
1 + tan^2theta = sec^2theta
cot^2theta + 1 = cosec^2theta

I am also stuck in this question:

prove this identity:

cosA / (1 - tanA) + sinA / (1 - cotA) = Sin A + cosA

Since this requires a proof I need to work on the LHS and RHS separately. Can someone help me start this question? I really appreciate your help.

[KLscilevothma]

First off, work on the complicated side, which is the LHS.

Usually if we see tan A and cot A in an expression together with sin A and cos A, we change tan A to sin A/cos A and cot A to cos A/ sin A before further simplifying the expression. In this question, you don't need to use the 3 formulae listed.

[garytse86]

well, the best I have done is this stage:

(cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA)

[KLscilevothma]

Please show your working as I think the expression you got isn't correct. You can try to substitute A=30 degrees to (cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) and (Sin A + cosA), you'll get 2 different values.

[garytse86]

sorry the LHS was this:

(cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) + sinA + cosA

[KLscilevothma]

Originally posted by garytse86
sorry the LHS was this:

(cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) + sinA + cosA

If (cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) + sinA + cosA = sin A + cos A

then,
(cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) = 0
which is also not correct

Your orginal question is to prove
cosA / (1 - tanA) + sinA / (1 - cotA) = Sin A + cosA

Let's focus on cosA / (1 - tanA) first. After changing tan A to sin A/cos A, what do you get ? (Hint: you need to simply the denominator first)

[garytse86]

you get:

(cosA / 1- (sinA/cosA)) + (sinA / 1- (cosA/sinA))

= cosA ( 1- (cosA/sinA)) + sinA (1- (sinA/cosA))

= cosA - (cos^2A/sinA) + sinA - (sin^2A/cosA)

[KLscilevothma]

Originally posted by garytse86
(cosA / 1- (sinA/cosA)) + (sinA / 1- (cosA/sinA))

= cosA ( 1- (cosA/sinA)) + sinA (1- (sinA/cosA)) [/B]

This step is wrong.

The correct steps should be (I'll only do [cosA/(1 - tanA)] )
1-tan A
= 1- (sin A / cos A)
= (cos A - sin A) / cos A

therefore,
cos A / (1- tan A)
= cos A / [(cos A - sin A) / cos A]
= cos2A / (cos A - sin A)

Can you do the second part yourself and simplify the LHS ?

[garytse86]

the second part would be:

sin^2A / (sinA - cosA)

[KLscilevothma]

Bingo! So can you prove the identity now ? And do you know what you have done wrong?

[garytse86]

yeah, I know what I have done wrong, but how does this prove that identity = sin A + cos A?

[KLscilevothma]

cos2A / (cos A - sin A) + sin2A / (sinA - cosA)
= cos2A / (cos A - sin A) - sin2A / (cos A - sin A)
= (cos2A - sin2 A)/(cos A - sin A)
= (cos A - sin A)(cos A + sin A)/ (cos A - sin A)
= cos A + sin A

[garytse86]

thanks very much for your help, kl.[;)]