1. ## trig identity

here are the problems I am having problem with... no pun intended...

(1 + tan^2X)(1 + cos2x) = 2 (prove identity)

(1 + sinA - cos2A) / (cosA + sin2A) = tanA

and Solve the equation for 0 < or = x < or = 2pi

cosx + cos2x + 2sin^2x = 0

2. Originally Posted by math:/
here are the problems I am having problem with... no pun intended...

(1 + tan^2X)(1 + cos2x) = 2 (prove identity)

(1 + sinA - cos2A) / (cosA + sin2A) = tanA

and Solve the equation for 0 < or = x < or = 2pi

cosx + cos2x + 2sin^2x = 0

Hint, whenever you see $\displaystyle \sin{2x}$ or $\displaystyle \cos{2x}$ you usually have to use double-angle identities.

2. Use the identities $\displaystyle \cos{2A} = \cos^2{A} - \sin^2{A}$, $\displaystyle \sin{2A} = 2\sin{A}\cos{A}$ and $\displaystyle \sin^2{A} + \cos^2{A} = 1$.

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

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

$\displaystyle = \frac{\sin^2{A} + \sin^2{A} + \sin{A}}{\cos{A}(2\sin{A} + 1)}$

$\displaystyle = \frac{2\sin^2{A} + \sin{A}}{\cos{A}(2\sin{A} + 1)}$

$\displaystyle = \frac{\sin{A}(2\sin{A} + 1)}{\cos{A}(2\sin{A} + 1)}$

$\displaystyle = \frac{\sin{A}}{\cos{A}}$

$\displaystyle = \tan{A}$.

3. Originally Posted by math:/
here are the problems I am having problem with... no pun intended...

(1 + tan^2X)(1 + cos2x) = 2 (prove identity)

(1 + sinA - cos2A) / (cosA + sin2A) = tanA

and Solve the equation for 0 < or = x < or = 2pi

cosx + cos2x + 2sin^2x = 0

3. Use the identities $\displaystyle \cos{2x} = \cos^2{x} - \sin^2{x}$ and $\displaystyle \sin^2{x} + \cos^2{x} = 1$.

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

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

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

$\displaystyle \cos{x} + 1 = 0$

$\displaystyle \cos{x} = -1$

$\displaystyle x = \pi + 2n\pi, n \in \mathbf{Z}$.

In the domain $\displaystyle 0 \leq x \leq 2\pi$, the only solution is

$\displaystyle x = \pi$.

4. 1. use the identities:
$\displaystyle \frac{1+\cos{2x}}{2} = \cos^2{x}$

$\displaystyle 1+\tan^2{x} = \sec^2{x}$

==========

Use 2 to multiply by 1 to get:

$\displaystyle 2(1+tan^2{x})\left(\frac{1+\cos{2x}}{2}\right)$

This simplifies to $\displaystyle 2(\sec^2{x})(\cos^2{x}$) and the result follows.

5. Originally Posted by math:/
here are the problems I am having problem with... no pun intended...

(1 + tan^2X)(1 + cos2x) = 2 (prove identity)

(1 + sinA - cos2A) / (cosA + sin2A) = tanA

and Solve the equation for 0 < or = x < or = 2pi

cosx + cos2x + 2sin^2x = 0

1. Use the identities $\displaystyle \frac{\sin{x}}{\cos{x}} = \tan{x}$ and $\displaystyle \cos{2x} = \cos^2{x} - \sin^2{x}$.

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

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

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

$\displaystyle = \frac{1}{\cos^2{x}}(2\cos^2{x})$

$\displaystyle = 2$.

6. Originally Posted by Chop Suey
1. use the identities:
$\displaystyle \frac{1+\cos{2x}}{2} = \cos^2{x}$

$\displaystyle 1+\tan^2{x} = \sec^2{x}$

==========

Use 2 to multiply by 1 to get:

$\displaystyle 2(1+tan^2{x})\left(\frac{1+\cos{2x}}{2}\right)$

This simplifies to $\displaystyle 2(\sec^2{x})(\cos^2{x}$) and the result follows.
Very nice