Thursday

December 18, 2014

December 18, 2014

Posted by **Hailey** on Thursday, April 10, 2008 at 8:43am.

1.) 2tanx-(1+tanx)^2 = -secx

2.) cos(x-330degrees)=1/2(3cosx-sinx)

*the is only over the 3*

3.)

1+cos2x/sinx=cotx

any help is greatly appreciated!!

- Math (Trig) typo edit -
**Hailey**, Thursday, April 10, 2008 at 8:49amis supposed to represent square root.

so for number 2

it's 1/2 (square root 3 cosx-sinx)

- Math (Trig) a.s.a.p, thanks! -
**Reiny**, Thursday, April 10, 2008 at 11:57am1.

LS = 2sinx/cosx - (1+sinx/cosx)^2

= 2sinx/cosx - ((cosx + sinx)/cosx)

= 2sinx/cosx - (cos^2 x + 2sinxcosx + sin^2 x)/cos^2 x

= (2sinxcosx -(1 + 2sinxcosx))/cos^2 x

= -1/cos^2 x

= -sec^2 x

you had -secx on the right side

I think you made a typo

2. LS = cosxcos330 + sinxsin330

but sin 330 = -sin30 = -1/2

and cos 330 = cos30 = √3/2

so above

= (√3/2)cosx - 1/2 sinx

= 1/2(√3 cosx - sinx)

= RS

Why don't you try the third one.

Change everything to sines and cosines

let me know how you did.

- Math (Trig) a.s.a.p, thanks! -
**Reiny**, Thursday, April 10, 2008 at 12:07pmfor the third one you had

1+cos2x/sinx=cotx

Your lack of brackets made this ambiguous.

I tried (1+cos2x)/sinx=cotx and tested with 40, did not work

I tried 1+cos^2 x/sinx=cotx and tested with 40, did not work

I tried 1+(cos2x/sinx)=cotx and tested with 40, did not work

check your typing of the question.

- Math (Trig) a.s.a.p, thanks! -

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