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trig

Given that
8 tan x - 3 cos x = 0
show that
3 sin^x + 8 sinx - 3 = 0
Reply 1
Avatar for Notnek
Notnek
21
Original post by Katiekat2000
Given that
8 tan x - 3 cos x = 0
show that
3 sin^x + 8 sinx - 3 = 0

Start by changing tan(x) to sin(x)/cos(x). Then multiply the equation by cos(x). Please post your working if you get stuck.
Original post by notnek
Start by changing tan(x) to sin(x)/cos(x). Then multiply the equation by cos(x). Please post your working if you get stuck.

thanks so much
i did what you said and came up with sinx - 3cos^2x = 0
i then subbed 1 - sin^2x into where the 3cos^2x is
i was wondering if this was right and where to get the -3 from
Reply 3
Avatar for Notnek
Notnek
21
Original post by Katiekat2000
thanks so much
i did what you said and came up with sinx - 3cos^2x = 0
i then subbed 1 - sin^2x into where the 3cos^2x is
i was wondering if this was right and where to get the -3 from

Your method looks good but It should be

8sin⁡x−3cos⁡2x8 \sin x - 3\cos^2 x

and that becomes

8sin⁡x−3(1−sin⁡2x)8\sin x - 3\left(1-\sin^2 x\right)

If you expand the bracket you should see where the −3-3 comes from.
Original post by notnek
Your method looks good but It should be

8sin⁡x−3cos⁡2x8 \sin x - 3\cos^2 x

and that becomes

8sin⁡x−3(1−sin⁡2x)8\sin x - 3\left(1-\sin^2 x\right)

If you expand the bracket you should see where the −3-3 comes from.

that makes so much more sense now i didn't realize you carried the intergers aswell. thanks :smile:

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