1. ## 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?

2. 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.

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.

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

5. 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?

6. Hello,

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 ^^

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

7. Ahhh k, that makes sense. Dont suppose you guys would possible no any questions that are similar to this (as in having to use trig identities)? Just trig is my weakest topic and i would really like to get better at it.

8. Well, you can start by looking here : http://www.mathhelpforum.com/math-help/trigonometry/ selecting topics that seem interesting by their title ^^

$\frac{\sin 3x}{\sin x}-\frac{\cos 3x}{\cos x}=2$

The solution is here : http://www.mathhelpforum.com/math-he...-identity.html

There are many...
If you search for the questions asked by Thelema ( http://www.mathhelpforum.com/math-he...&starteronly=1 ) and kelsey3 ( http://www.mathhelpforum.com/math-he...&starteronly=1 ) and looi76 ( http://www.mathhelpforum.com/math-he...&starteronly=1 ), it can give you further exercises :-)