hello there strangers...slightly stuck and would totes love it if someone could help me.

"if cos 2x = 7/25 and 0<x<2pie find exact values of cosx and sinx"

I have this so far.

cos 2x= 7/25

since cos2x = 2cos^2x-1 then

2cos^2 x-1 = 7/25

2cos^2 x = 7/25 + 1 = 32/25

cos^2 x = 32/50

its going from there to the last step,how do I get rid of the square on the left hand side the SQa marking says cos x = 4/5 and sin x = 3/5 but I can't get it to come out right

thanks peepsies <3

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# Past paper 2005 noncalc Q9

Started by Sammie, May 03 2008 01:03 PM

1 reply to this topic

### #1

Posted 03 May 2008 - 01:03 PM

"Crowds are won and lost and won again, but our hearts beat for the diehards"

### #2

Posted 18 May 2008 - 10:02 AM

you've done good to this point, now its just rearranging the equation, just watch your last step up there ^

2cos^2 x-1 = 7/25

2cos^2 x = 7/25 + 1 = 32/25

cos^2 x = 16/25 ( as 2cos^2 x = 32/25 , bring the 2 to the other side so you will divide here by 2 )

cos x = 4/5 ( square root your answer here )

now if you use ur trig triangles filling in that cos x = 4/5, you'll realise its a 3,4,5 triangle, therefore sin x = 3/5

2cos^2 x-1 = 7/25

2cos^2 x = 7/25 + 1 = 32/25

cos^2 x = 16/25 ( as 2cos^2 x = 32/25 , bring the 2 to the other side so you will divide here by 2 )

cos x = 4/5 ( square root your answer here )

now if you use ur trig triangles filling in that cos x = 4/5, you'll realise its a 3,4,5 triangle, therefore sin x = 3/5

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