# Thread: Trig identities

1. ## Trig identities

1.cot^2x+sec^2x=tan^2x+csc^2x

2.cos3x=4cos^3x-3cosx

Can someone help guide me towards the answer but not actually give me the full solution to the answer. Thanks

2. 1. use the Pythagorean identities $\displaystyle \cot^2{x} + 1 = \csc^2{x}$ and $\displaystyle sec^2{x} = 1 + tan^2{x}$ in that order on the left side.
They might need a slight rearrangement before you use them.

2. use the sum identity $\displaystyle \cos(a+b) = \cos{a}\cos{b} - \sin{a}\sin{b}$ plus the fact that $\displaystyle \cos(3x) = \cos(2x+x)$.
you'll also need the double angle identity for cosine $\displaystyle \cos(2x) = 2\cos^2{x} - 1$

3. K I'm not really sure how to show what I mean without solving it. This a long way but I thought I show you a different way to work the problem.

Another way to see if the RHS is equal to the LHS is

$\displaystyle LHS-RHS = 0~\text{or}~RHS-LHS = 0$

$\displaystyle \cot^2{x} + \sec^2{x} = \tan^2{x} + \csc^2{x}$

$\displaystyle \cot^2{x} + \sec^2{x} - \tan^2{x} - \csc^2{x} = 0$

$\displaystyle \frac{\cos^2{x}}{\sin^2{x}} + \frac{1}{\cos^2{x}} - \frac{\sin^2{x}}{\cos^2{x}} - \frac{1}{\sin^2{x}} = 0$

$\displaystyle \frac{\cos^4{x} - \cos^2{x} + \sin^2{x} - \sin^4{x}}{\cos^2{x}\sin^2{x}} = 0$

$\displaystyle \frac{\cos^2{x}(\cos^2{x} -1)}{\cos^2{x}\sin^2{x}} + \frac{\sin^2{x}(1-\sin^2{x})}{\cos^2{x}\sin^2{x}}= 0$

$\displaystyle \frac{\cos^2{x} - 1}{\sin^2{x}} + \frac{\sin^2{x}\cos^2{x}}{\sin^2{x}\cos^2{x}} = 0$

$\displaystyle \frac{\cos^2{x} - 1}{1-\cos^2{x}} +1 = 0$

$\displaystyle -\bigg(\frac{-\cos^2{x} + 1}{1 - \cos^2{x}}\bigg) + 1 = 0$

$\displaystyle -1 + 1 = 0$

$\displaystyle 0 = 0$

4. Hello, johntuan!

$\displaystyle 1)\;\;\cot^2\!x+\sec^2\!x\;=\;\tan^2\!x+\csc^2\!x$
$\displaystyle \text{The left side is: }\:\overbrace{(\csc^2\!x - 1)}^{\cot^2\!x} + \overbrace{(\tan^2\!x + 1)}^{\sec^2\!x} \;=\; \tan^2\!x + \csc^2\!x$

5. Another way to do it using skeeter method

$\displaystyle \sec^2{x} - \tan^2{x}~=~\csc^2{x} - \cot^2{x}$

which simplfies to

$\displaystyle 1~=~1$