Trig: solve (cosx)^2 - 2sinxcosx - (sinx)^2 = 0 on [0,2pi]

[carly]

Hi, I'm working on some trig questions and can't get the answer for this last question; I have a hunch that it might be one of those obvious ones where you think too much :? ....

Either way, help would be much appreciated!

solve (cosx)^2 - 2sinxcosx - (sinx)^2 = 0, 0 ? x ? 2? giving the answer in exact form

 

[pka]

Re: Trig question

Use double angles:
$\displaystyle \begin{array}{rcl} \cos ^2 (x) - 2\sin (x)\cos (x) - \sin ^2 (x) &=& \left[ {\cos ^2 (x) - \sin ^2 (x)} \right] - 2\sin (x)\cos (x) \\ & =& \cos (2x) - \sin (2x) = 0 \\ \end{array}$.

Now can solve $\displaystyle \cos (2x) = \sin (2x)$?

 

[carly]

Re: Trig question

are both cos2x and sin2x zero?

 

[pka]

Re: Trig question

are both cos2x and sin2x zero?

Absolutely NOT.
I will give you one solution: $\displaystyle x=\frac{\pi}{8}$.
There are others.

 

[o_O]

Re: Trig question

From: $\displaystyle cos(2x) = sin(2x)$

What happens if you divide both sides by cos(2x)?

 

[carly]

Re: Trig question

oh! I think I've got it:

divide both sides by cos2x to get tan2x=1
work out four answers to be ?/8, 5?/8, 9?/8, and 13?/8

Is this correct?

 

[pka]

Re: Trig question

No, not quite. The tangent is negative at two of those.

 

[carly]

Re: Trig question

... is it? Isn't tan positive in the first and third quadrants?

 

[pka]

Re: Trig question

... is it? Isn't tan positive in the first and third quadrants?

Yes that is correct.
But $\displaystyle \frac {5\pi}{8}$ is in quad II.