Thread: Trig equations and proof of identity

1. Trig equations and proof of identity

Can anyone help with this question?

I have done all of the parts except the last bit, kinda went a bit over my head there :S

There is also this question that I have gotten really stumped with, I have to try and prove the identitiy, we have only covered the first trig identity so far but we have been set this question, so I am not quite sure what I can do :S

$\displaystyle 1+\tan^2\theta=\frac{1}{\cos^2\theta}$

2. What did you get for part d) ?

Look closely between the relationship in the equations in parts d) and e)

There is no difference given $\displaystyle x = 2\theta$

3. Originally Posted by pickslides
What did you get for part d) ?

Look closely between the relationship in the equations in parts d) and e)

There is no difference given $\displaystyle x = 2\theta$
Okay so with part d) I got solutions to = 0, pi and 2pi solutions which I got from: (x=cox^-1..{from cos^2x=1}

So basically my solutions to part d would be the same as $\displaystyle x = 2\theta$ ?

4. Originally Posted by rhysiboy
Can anyone help with this question?

I have done all of the parts except the last bit, kinda went a bit over my head there :S

There is also this question that I have gotten really stumped with, I have to try and prove the identitiy, we have only covered the first trig identity so far but we have been set this question, so I am not quite sure what I can do :S

$\displaystyle 1+\tan^2\theta=\frac{1}{\cos^2\theta}$
$\displaystyle 1+\tan^2\theta=\frac{1}{\cos^2\theta}$

$\displaystyle 1+\frac{sin^2\theta}{\cos^2\theta}=\frac{1}{\cos^2 \theta}$

Do you see the proof now?

5. Originally Posted by harish21
$\displaystyle 1+\tan^2\theta=\frac{1}{\cos^2\theta}$

$\displaystyle 1+\frac{sin^2\theta}{\cos^2\theta}=\frac{1}{\cos^2 \theta}$

Do you see the proof now?
okay so what I did was:

$\displaystyle 1+\frac{sin^2\theta}{\cos^2\theta}=\frac{1}{\cos^2 \theta}$

So I multiplied the $\displaystyle {\cos^2\theta}$ on the left side to the right side, leaving me with $\displaystyle 1+sin^2\theta=\frac{1*\cos^2\theta}{\cos^2\theta}$

simplified: $\displaystyle 1+sin^2\theta=1$ so effectively $\displaystyle sin^2\theta=0$... That proves the identitiy doesn't it?

6. Originally Posted by rhysiboy

simplified: $\displaystyle 1+sin^2\theta=1$ so effectively $\displaystyle sin^2\theta=0$... That proves the identitiy doesn't it?
No it doesn't

When working with a proof you must alter only one side of the equation to equate the other.

As harish21 has already shown and only changing the LHS

$\displaystyle 1+\tan^2\theta$

$\displaystyle =1+\frac{\sin^2\theta}{\cos^2\theta}$

$\displaystyle =\frac{\cos^2\theta}{\cos^2\theta}+\frac{\sin^2\th eta}{\cos^2\theta}$

$\displaystyle =\frac{\cos^2\theta}{\cos^2\theta}+\frac{1-\cos^2\theta}{\cos^2\theta}$

$\displaystyle =\frac{\cos^2\theta+1-\cos^2\theta}{\cos^2\theta}$

$\displaystyle =\frac{1}{\cos^2\theta}$

as required.