Math Trig Identities?


by Stanc
Tags: identities, math, trig
Stanc
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#1
Dec6-12, 02:14 PM
P: 60
Hi im in need of some help for this question:

1+tanx/1-tanx = tan(x+(∏/4))

It is easy to solve with the tan trig identites on the right side however, my teacher had told me to do it with SIN and COS only. Im not sure if its possible and was looking for some insight

Left Side:
Cosx+Sinx/Cosx-Sinx


Right Side:
End up with
{(SinxCos∏/4) + (CosxSin∏/4)}/ (CosxCos∏/4) - (SinxSin∏/4)

Can anyone give me some insight?
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pasmith
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#2
Dec6-12, 02:30 PM
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A good first step is to replace cos([itex]\pi[/itex]/4) and sin([itex]\pi[/itex]/4) by their actual values.
Stanc
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#3
Dec6-12, 02:38 PM
P: 60
Quote Quote by pasmith View Post
A good first step is to replace cos([itex]\pi[/itex]/4) and sin([itex]\pi[/itex]/4) by their actual values.
Yes. I have tried that but then i get stuck at:

(Sinx(√2/2)) + (Cosx(√2/2)) / (Cosx(√2/2)) - (Sinx(√2/2))

IS there anything i can do?

AlephZero
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#4
Dec6-12, 02:52 PM
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Math Trig Identities?


Don't get confused by all the notation. This is just the same as simplfying something like
$$\frac{3a + 3b}{3c - 3d}$$
Stanc
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#5
Dec6-12, 03:22 PM
P: 60
Quote Quote by AlephZero View Post
Don't get confused by all the notation. This is just the same as simplfying something like
$$\frac{3a + 3b}{3c - 3d}$$
I dont think im catching it. I understand how to simplify the example you had but i cannot factor out a sin or a cos?
Stanc
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#6
Dec6-12, 03:24 PM
P: 60
Nevermind, I feel reallydumb... i got it, thanks guys
Mark44
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#7
Dec6-12, 03:43 PM
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Quote Quote by Stanc View Post
Hi im in need of some help for this question:

1+tanx/1-tanx = tan(x+(∏/4))
The left side needs parentheses!

What you wrote is 1 + (tanx)/1 - tanx, which is equal to 1.

What you meant was (1 + tanx)/(1 - tanx).


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