C3 Trig questions.

Hello, I have a couple questions on trig,

(1) If tan2A = 12/5 , find the possible values of tan A.

(2) Prove the identity tan2x -tanx = tanxsec2x

(3) Prove that (cosA + sinA)(cos2A + sin2A) = cosA + sin3A

Thanks a lot
From john

Reply 1

westhamfan

Neo1

Hello, I have a couple questions on trig,

(1) If tan2A = 12/5 , find the possible values of tan A.

(2) Prove the identity tan2x -tanx = tanxsec2x

(3) Prove that (cosA + sinA)(cos2A + sin2A) = cosA + sin3A

Thanks a lot
From john

(1) tan2a=(2tanA)/(1-tan^2.A)

Easy from there (quadratic in tanA)

dno about the other two, ive only done c2 lol

Reply 2

Neo1

OP

4

yea thats right i actually converted it into quadratic form before and thought it wasnt possible to solve it but yeah ive managed to do it.

The other 2 i dont know where to start any help please.?

Thanks a lot

Reply 3

silent ninja

19

Neo1

Hello, I have a couple questions on trig,

(1) If tan2A = 12/5 , find the possible values of tan A.

(2) Prove the identity tan2x -tanx = tanxsec2x

(3) Prove that (cosA + sinA)(cos2A + sin2A) = cosA + sin3A

Thanks a lot
From john

(1) No identities required. Inverse tan(12/5) to get the 'base' value and go from there. I dont have a calculator to hand, but 2A= whatever value you get, then figure out the others using a sketch graph or the circle thingy.

(2) tan2x- tanx = sin2x/cos2x - tanx

sub in sin2x= 2sinxcosx and subtract tanx (turn into one fraction)
= (2sinxcos²x - sinxcos2x)/cos2xcosx
= (2sinxcos²x - 2sinxcos²x + sinx)/cos2xcosx (using cos2x= 2cos²x-1 for the top part)
= sinx/cos2xcosx
= tanx/cos2x
= tanxsec2x

Reply 4

silent ninja

19

Neo1

(3) Prove that (cosA + sinA)(cos2A + sin2A) = cosA + sin3A

expanding
= cos2AcosA + sin2AcosA + cos2AsinA + sin2AsinA
= cos2AcosA + sin2AsinA + sin2AcosA + cos2AsinA

but sin2AcosA + cos2AsinA= sin(2A+A)= sin3A
and cos2AcosA + sin2AsinA= cos(2A-A)=cosA

= cosA + sin3A

C3 Trig questions.

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