Thread: Stupid trig identities

the problem is tan(x)+cot(x)=2csc(2x). I can get to the point where I have (2sin(x)cos(x))/(sin(x)cos(x)). I am not quite sure what to do beyond this point.

Originally Posted by dnlstffrd

the problem is tan(x)+cot(x)=2csc(2x). I can get to the point where I have (2sin(x)cos(x))/(sin(x)cos(x)). I am not quite sure what to do beyond this point.

tan(x)+cot(x)=sin(x)/cos(x)+cos(x)/sin(x)=[sin^2(x)+cos^2(x)]/[cos(x)sin(x)]
............... =1/[cos(x)sin(x)]

but cos(x)sin(x)=(1/2)sin(2x), so:

tan(x)+cot(x)=2csc(2x).

RonL

wait a sec. It should be 2/sin(2x), not 1/2sin(2x) because that would equal (1/2)csc(2x), not 2csc(2x).

Originally Posted by dnlstffrd

wait a sec. It should be 2/sin(2x), not 1/2sin(2x) because that would equal (1/2)csc(2x), not 2csc(2x).

1/[cos(x)sin(x)]=1/[(1/2)sin(2x)]=2/sin(2x)=2csc(2x).

RonL

oops, I missed that part

I actually have another that I also cannot figure out for some reason:

[sin(x)+cos(2x)-1]/[cos(x)-sin(2x)]=tan(x)

Originally Posted by dnlstffrd

oops, I missed that part

I actually have another that I also cannot figure out for some reason:

[sin(x)+cos(2x)-1]/[cos(x)-sin(2x)]=tan(x)

You should post this in a new thread.

sin(2x) = 2sin(x)cos(x)
cos(2x) = 1 - 2[sin(x)]^2 = 2[cos(x)]^2 - 1 = [cos(x)]^2 - [sin(x)]^2

so
sin(x)+cos(2x)-1 = sin(x) + 1 - 2[sin(x)]^2 - 1 = -2[sin(x)]^2 + sin(x) = sin(x) {-2sin(x) + 1}

cos(x)-sin(2x) = cos(x) - 2sin(x)cos(x) = cos(x) {1 - 2sin(x)}

Dividing the two cancels the common 1 - 2sin(x) leaving sin(x)/cos(x) = tan(x).

-Dan

tan(x) + cot(x) = 2 csc(2x)


LHS = sin(x)/cos(x) + cos(x)/sin(x)

= (sin^2(x) + cos^2(x)) / (sin(x) cos(x))

= 1 / (sin(x) cos(x))


RHS = 2 / sin(2x)

= 2 / (2 sin(x) cos(x))

= 1 / (sin(x) cos(x))

= LHS