# Trigonometric equations?

• Mar 9th 2011, 03:19 PM
homeylova223
Trigonometric equations?
I have two problems that I am not entirely sure how to

The first one is

(1/cosx-sinx)= cosx+sinx

Now I am not entirely sure how to this one but would I turn the right side into a fraction by placing one under it then cancelling it out?

My second question is

sinxcosx= 1/2

Would I do
sinxcosx-1/2=0 then what next as I am not sure how to proceed(Crying)
• Mar 9th 2011, 03:44 PM
pickslides
Quote:

Originally Posted by homeylova223

My second question is

sinxcosx= 1/2

Would I do
sinxcosx-1/2=0 then what next as I am not sure how to proceed(Crying)

Multiply both sides by 2 giving

$\displaystyle \sin x\cos x= \frac{1}{2}$

$\displaystyle 2\sin x\cos x= 1$

$\displaystyle \sin 2x= 1$

You have it from here?
• Mar 9th 2011, 03:49 PM
e^(i*pi)
Quote:

Originally Posted by homeylova223
I have two problems that I am not entirely sure how to

The first one is

(1/cosx-sinx)= cosx+sinx

Now I am not entirely sure how to this one but would I turn the right side into a fraction by placing one under it then cancelling it out?

My second question is

sinxcosx= 1/2

Would I do
sinxcosx-1/2=0 then what next as I am not sure how to proceed(Crying)

$\dfrac{1-\cos(x)\sin(x)}{\cos(x)} = \cos(x)+ \sin(x)$

$1 - \cos(x)\sin(x) = \cos^2(x)+\sin(x)\cos(x)$

$1 - \cos^2(x) = 2\sin(x)\cos(x) \implies \sin^2(x) = \sin(2x)$

Which is a nasty little equation to solve.

Did you happen to mean $\dfrac{1}{\cos(x)-\sin(x)} = \cos(x) + \sin(x)$ which is much easier to solve - multiply through and use the difference of two squares to get a double angel identity
• Mar 9th 2011, 03:57 PM
homeylova223
Yes I meant the one that is easier to solve. The one you wrote on the bottom part.

The thing is I am not sure how to multiply through. Is it like cross multiplication? Like 1 x 1 and then cosx-sinx times cosx+sinx
• Mar 9th 2011, 04:29 PM
e^(i*pi)
Yes, it's the same as cross multiplication but in this instance it's easier to solve by multiplying both sides by $\cos(x) - \sin(x)$

$\dfrac{1}{\cos(x)-\sin(x)} \times (\cos(x) - \sin(x)) = (\cos(x)+ \sin(x)) \times (\cos(x)-\sin(x))$

$\implies 1 = (\cos(x)+\sin(x))(\cos(x)-\sin(x))$

The RHS is of the form $(a-b)(a+b)$ so it's time for the difference of two squares to expand
• Mar 10th 2011, 05:01 PM
homeylova223
I have one follow up question

I managed to reduce it to
1= cosx^2-sinx^2 Would I do this?
1-cosx^2=-sinx^2
sinx^2=-sinx^2
sinx^2+sinx^2= 0 factor sinx^2
sinx^2=0
sinx=0
x= pi (Thinking)
• Mar 11th 2011, 03:19 AM
e^(i*pi)
Quote:

Originally Posted by homeylova223
I have one follow up question

I managed to reduce it to
1= cosx^2-sinx^2 Would I do this?
1-cosx^2=-sinx^2
sinx^2=-sinx^2
sinx^2+sinx^2= 0 factor sinx^2 -e^(i*pi): You're not factoring so much as dividing by 2
sinx^2=0
sinx=0
x= pi (Thinking)

Yes, I was thinking of $\cos(2x) = \cos^2(x)-\sin^2(x)$ giving $\cos(2x) = 1$ but your method is fine although while $x = \pi$ is a solution but it's not the only one. Check $\sin(0)$ (Wink)