trig question

ryan750

2

express sin2x/(1-2cos2x) in terms of tanx

thanks

Reply 1

dvs

10

sin2x/(1-2cos2x)
= 2sinxcosx/[1-2(2cos²x-1)]
= 2sinxcosx/(3-4cos²x) --- now multiply be sec²x/sec²x (or divide the nom and denom by cos²x)
= 2tanx/(3sec²x-4)
= 2tanx/[3(1+tan²x)-4]
= 2tanx/(3tan²x-1)

Reply 2

ryan750

OP

2

dvs

sin2x/(1-2cos2x)
= 2sinxcosx/[1-2(2cos²x-1)]
= 2sinxcosx/(3-4cos²x) --- now multiply be sec²x/sec²x (or divide the nom and denom by cos²x)
= 2tanx/(3sec²x-4)
= 2tanx/[3(1+tan²x)-4]
= 2tanx/(3tan²x-1)

could u write the last part out in the way of dividin numerator and denominator by cos²x pls. it would be really helpful. thanks

Reply 3

dvs

10

It's the same thing since sec²x=1/cos²x, i.e. multiplying by sec²x is the same as multiplying by 1/cos²x which is basically dividing by cos²x.

Reply 4

ryan750

OP

2

dvs

It's the same thing since sec²x=1/cos²x, i.e. multiplying by sec²x is the same as multiplying by 1/cos²x which is basically dividing by cos²x.

ok - but i cant see how to do this. i cant work it out. pls show me. i wil rep u for it.

Reply 5

dvs

10

2sinxcosx/(3-4cos²x)
= 2sinxcosx.sec²x/[(3-4cos²x).sec²x]

2sinxcosx.sec²x
= 2sinxcosx/cos²x
= 2sinx/cosx
= 2tanx

(3-4cos²x).sec²x
= 3sec²x - 4cos²xsec²x
= 3sec²x - 4cos²x/cos²x
= 3sec²x - 4
= 3(1+tan²x) - 4
= 3 + 3tan²x - 4
= 3tan²x - 1

Reply 6

ryan750

OP

2

dvs

2sinxcosx/(3-4cos²x)
= 2sinxcosx.sec²x/[(3-4cos²x).sec²x]

2sinxcosx.sec²x
= 2sinxcosx/cos²x
= 2sinx/cosx
= 2tanx

(3-4cos²x).sec²x
= 3sec²x - 4cos²xsec²x
= 3sec²x - 4cos²x/cos²x
= 3sec²x - 4
= 3(1+tan²x) - 4
= 3 + 3tan²x - 4
= 3tan²x - 1

i do see how u r doing it - but in the book i havent done sec²x so how would i know that 1/cos²x is equal to 1+tan²x. Thats where i dont understand it. i promise rep if u can clarify this dude.

Reply 7

dvs

10

sin²x + cos²x = 1, now divide by cos²x to get:
sin²x/cos²x + cos²x/cos²x = 1/cos²x
tan²x + 1 = 1/cos²x = sec²x

secx = 1/cosx
cosecx = 1/sinx
cotx = 1/tanx

Reply 8