# help with trig identities!

• Dec 16th 2010, 05:23 PM
Joker
help with trig identities!
1. solve : (sinX)(cosX)-1.2cosx = 0

2. find identity or counter example : csc(3pi/4 + A) = -secA

3. find identity or counter example : cosX = (cscXtanX)/(1+tan^2x)

4. find identity or counter example : (sinX)cosx = tanx

5. find identity or counter example : cos^2x = (sin^2x+cos^2x)/tan^2x + 1
• Dec 16th 2010, 06:42 PM
TheCoffeeMachine
Quote:

Originally Posted by Joker
1. solve : (sinX)(cosX)-1.2cosx = 0

$\displaystyle \sin{x}\cos{x}-1.2\cos{x} = 0 \Rightarrow \cos{x}\left(\sin{x}-1.2\right) = 0 \Rightarrow \cos{x} = 0$, or $\displaystyle \sin{x} = 1.2$.

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2. find identity or counter example : csc(3pi/4 + A) = -secA
Take $\displaystyle A = 0$.

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3. find identity or counter example : cosX = (cscXtanX)/(1+tan^2x)
$\displaystyle \displaystyle \frac{\csc{x}\tan{x}}{1+\tan^2{x}} = \frac{1}{\cos{x}\left(1+\frac{\sin^2{x}}{\cos^2{x} }\right)}$ $\displaystyle \displaystyle = \frac{1}{\cos{x}\left(1+\frac{1-\cos^2{x}}{\cos^2{x}}\right)} = \frac{1}{\cos{x}\left(\frac{\cos^2{x}+1-\cos^2{x}}{\cos^2{x}}\right)} = \frac{1}{\cos{x}\left(\frac{1}{\cos^2{x}\right)}} = \cos{x}$

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4. find identity or counter example : (sinx)cosx = tanx
Oh, come on! Think about it for a bit!

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5. find identity or counter example : cos^2x = (sin^2x+cos^2x)/tan^2x + 1
$\displaystyle \displaystyle \frac{\sin^2{x}+\cos^2{x}}{\tan^2{x}+1} = \frac{1}{\tan^2{x}+1} = \frac{1}{\frac{\sin^2{x}}{\cos^2{x}}+1} = \frac{1}{\frac{1-\cos^2{x}}{\cos^2{x}}+1} = \cdots$
• Dec 16th 2010, 06:52 PM
Joker
thanks for the two posts. this should help a lot! the first one was suppose to be 1/2cosx lol but i think i can handle it. Thanks Again
• Dec 16th 2010, 08:09 PM
Prove It
Quote:

Originally Posted by Joker
thanks for the two posts. this should help a lot! the first one was suppose to be 1/2cosx lol but i think i can handle it. Thanks Again

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

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

$\displaystyle \displaystyle \cos{x} = 0$ or $\displaystyle \displaystyle \sin{x} + \frac{1}{2} = 0$.

Case 1: $\displaystyle \displaystyle \cos{x} = 0$

$\displaystyle \displaystyle x = \left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\} + 2\pi n$ where $\displaystyle \displaystyle n \in \mathbf{Z}$.

Case 2: $\displaystyle \displaystyle \sin{x} + \frac{1}{2} = 0$

$\displaystyle \displaystyle \sin{x} = -\frac{1}{2}$

$\displaystyle \displaystyle x = \left\{\pi + \frac{\pi}{6}, 2\pi - \frac{\pi}{6}\right\} + 2\pi n$ where $\displaystyle \displaystyle n \in \mathbf{Z}$

$\displaystyle \displaystyle x = \left\{\frac{7\pi}{6}, \frac{11\pi}{6}\right\} + 2\pi n$.

So the solution is $\displaystyle \displaystyle x = \left\{\frac{\pi}{2}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}\right\} + 2\pi n$.