# Trig identies

• Apr 12th 2008, 08:21 PM
Trig identies
If acos^2x + bsin^2x = c, then tan^2x in terms of a,b and c is given by?

Okay, this obviously involves an identity but the only one i can see to use is cos^2x + sin^2x = 1, but that assumes that a and b are of the same value. can someone guide me on the right track?
• Apr 12th 2008, 08:41 PM
o_O
You can divide both sides by $\cos^{2} x$ to get your $\tan^{2} x$ in your solution. However, is your answer supposed to be ONLY in terms of a,b, and c? If so, I don't see anything at the moment.
• Apr 12th 2008, 09:21 PM
angel.white
Quote:

If acos^2x + bsin^2x = c, then tan^2x in terms of a,b and c is given by?

Okay, this obviously involves an identity but the only one i can see to use is cos^2x + sin^2x = 1, but that assumes that a and b are of the same value. can someone guide me on the right track?

$a * \frac 12(1 + cos(2x)) + b * \frac 12(1-cos(2x)) = c$

$a + a*cos(2x) + b -b*cos(2x) = 2c$

$a + b + cos(2x)(a -b) = 2c$

$cos(2x)(a -b) = 2c-a-b$

$cos(2x) = \frac {2c-a-b}{a -b}$

$x = \frac 12arccos(\frac {2c-a-b}{a -b})$

$tan ~x = tan \left( \frac 12 ~arccos(\frac {2c-a-b}{a -b})\right)$

From here, I'll let you take over, b/c my sister is bugging me. (Might want to check my work, she's been bugging me for a while now)

Use this to simplify it to trig function of arccos(...) Tangent half-angle formula - Wikipedia, the free encyclopedia You should be able to then simplify it further, probably removing all trig functions from the RHS. Then square both sides.
• Apr 12th 2008, 10:12 PM
Soroban

Quote:

If $a\cos^2\!x + b\sin^2\!x \:=\: c$, express $\tan^2\!x$ in terms of $a,b,c$

Divide by $\cos^2\!x\!:\;\;\;\frac{a\cos^2\!x}{\cos^2\!x} + \frac{b\sin^2\!x}{\cos^2\!x} \;=\;\frac{c}{\cos^2\!x} \quad\Rightarrow\quad a + b\tan^2\!x \:=\:c\sec^2\!x$

. . $a + b\tan^2\!x \;=\;c(\tan^2\!x + 1) \quad\Rightarrow\quad a + b\tan^2\!x \;=\;c\tan^2\!x + c
$

. . $b\tan^2\!x - c\tan^2\!x \;=\;c - a \quad\Rightarrow\quad (b - c)\tan^2\!x \;=\;c-a$

Therefore: . $\boxed{\tan^2\!x \;=\;\frac{c-a}{b-c}}$

• Apr 13th 2008, 02:53 AM
Yer that looks right, but apparently the answer is a-c/c-b. Im sure that the answer given is an error though. One question though, what made u decide to divide both sides by cos^2x? It makes sense what u have done yet i never thought to even try that. Any tips you can offer me when faced with a question like this?
• Apr 13th 2008, 03:32 AM
Moo
Hello,

Quote:

Yer that looks right, but apparently the answer is a-c/c-b. Im sure that the answer given is an error though.

$\frac{a-c}{c-b}=\frac{-(c-a)}{-(b-c)}=\frac{c-a}{b-c}$

There is no mistake ^^

Quote:

One question though, what made u decide to divide both sides by cos^2x? It makes sense what u have done yet i never thought to even try that. Any tips you can offer me when faced with a question like this?
Well, you've got equations with cos and sin. You want to get to an expression including tan.
Knowing that tan=cos/sin, you can either make appear the fraction cos/sin or sin/cos, which you obtain here by dividing the thing by sinē or cosē
• Apr 13th 2008, 04:15 AM
$\frac{\sin 3x}{\sin x}-\frac{\cos 3x}{\cos x}=2$