Math Is Fun Forum

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Trig identities

Given that sinx cosy + 3 cosx siny = 2 sinx siny - 4 cosx cosy, express tany in terms of tanx.

Thanks!

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Trig identities

I'm guessing it's the last line that got you stuck. It took me a while to figure out the trick.
Up until then, it's all standard manipulation:

sinx cosy + 3 cosx siny = 2 sinx siny - 4 cosx cosy
3 cosx siny - 2 sinx siny = -sinx cosy - 4 cosx cosy
siny (3 cosx - 2 sinx) = cosy (-sinx - 4 cosx)
tany (3 cosx - 2 sinx) = (-sinx - 4 cosx)
tany = (sinx + 4cosx)/(2 sinx - 3 cosx)

Now, divide both sides of the right-hand fraction by cosx:

tany = (tanx + 4)/(2 tanx - 3).

Edit: bossk's way is probably better. At least we agree on answers though.
Edit2: And we agree that agreeing is good!

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It wanted to be normal.

bossk171

Member

Registered: 2007-07-16

Posts: 305

Re: Trig identities

sinx cosy + 3 cosx siny = 2 sinx siny - 4 cosx cosy

dividing through by cosx:

tanx cosy + 3siny = 2 tanx siny - 4cosy

dividing through by cosy:

tanx + 3 tany = 2 tanx tany - 4

tanx +4 = 2 tanx tany - 3 tany

tanx + 4 = (2 tanx - 3) tany

(tanx + 4)/(2 tan x - 3) = tany

EDIT: Ahhh, I was too slow. I am glad to see I was right though.

Last edited by bossk171 (2007-10-23 07:00:20)

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There are 10 types of people in the world, those who understand binary, those who don't, and those who can use induction.

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Trig identities

Thanks both of you, it's really helpful to see two different methods.

And you're right mathsy, it was the last line I was getting stuck on!

Thanks