some more trig identities

**jayshizwiz** · September 4th 2010, 02:57 PM

hi. I think i'm getting the hang of these problems but I still need some assistance:

1) $cos^2x+sinx=1$
$cos^2x+sinx=cos^2x+sin^2x \rightarrow sinx - sin^2x=0 \rightarrow sinx(1-sinx)=0$
$sinx_1 = 0 \rightarrow x_1=180n$
$1-sinx_2=0 \rightarrow sinx_2=1 \rightarrow x_2=90+360n$

2) $4sin^3x-sinx=0$
$sinx(4sin^2x-1)=0$
$sinx_1 = 180n$
$4sin^2x_2-1=0 \rightarrow sin^2x=1/4 \rightarrow sinx=1/2 \rightarrow x_2 = +-30+180n$

3) $2cos^2x-5sinx-4=0$
$2-2sin^2x -5sinx-4=0$ ???

are 1 and 2 correct? number 2 I wasn't sure about (sinx)^2=1/4 - - - sinx=1/2. can i just do a square root like that? and how do I solve 3?

Thanks

**Archie Meade** · September 4th 2010, 04:50 PM

Originally Posted by jayshizwiz

hi. I think i'm getting the hang of these problems but I still need some assistance:

1) $cos^2x+sinx=1$
$cos^2x+sinx=cos^2x+sin^2x \rightarrow sinx - sin^2x=0 \rightarrow sinx(1-sinx)=0$
$sinx_1 = 0 \rightarrow x_1=180n$
$1-sinx_2=0 \rightarrow sinx_2=1 \rightarrow x_2=90+360n$

2) $4sin^3x-sinx=0$
$sinx(4sin^2x-1)=0$
$sinx_1 = 180n$
$4sin^2x_2-1=0 \rightarrow sin^2x=1/4 \rightarrow sinx=1/2 \rightarrow x_2 = +-30+180n$

3) $2cos^2x-5sinx-4=0$
$2-2sin^2x -5sinx-4=0$ ???

are 1 and 2 correct? number 2 I wasn't sure about (sinx)^2=1/4 - - - sinx=1/2. can i just do a square root like that? and how do I solve 3?

Thanks

(1) is good...

For (2), you neglected to write the negative square root of 1/4, which is -1/2, but you found the angles nonetheless.

(3) can written as a quadratic equation, using

$-2Sin^2x-5Sinx-2=0$

$2Sin^2x+5Sinx+2=0$

$(2Sinx+1)(Sinx+2)=0$

and work from there

**jayshizwiz** · September 4th 2010, 11:55 PM

thanks. i can't believe i didn't realize that. here's what I did...

$sinx(2sinx+5)=-2$ and then i was stuck...lol

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