trig identity

**math:/** · December 18th 2008, 04:56 PM

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

your help is much appreciated!

**Prove It** · December 18th 2008, 06:21 PM

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

your help is much appreciated!

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

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

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

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

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

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

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

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

$= \tan{A}$ .

**Prove It** · December 18th 2008, 06:26 PM

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

your help is much appreciated!

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

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

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

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

$\cos{x} + 1 = 0$

$\cos{x} = -1$

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

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

$x = \pi$ .

**Chop Suey** · December 18th 2008, 06:29 PM

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

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

==========

Use 2 to multiply by 1 to get:

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

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

**Prove It** · December 18th 2008, 06:32 PM

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

your help is much appreciated!

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

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

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

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

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

$= 2$ .

**Prove It** · December 18th 2008, 06:34 PM

Originally Posted by Chop Suey

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

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

==========

Use 2 to multiply by 1 to get:

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

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

Very nice

Thread: trig identity

LinkBack

Thread Tools

Display

trig identity

Similar Math Help Forum Discussions

trig identity help...

Another trig identity

Trig Identity

Trig identity

Odd Trig Identity

Search Tags