How Do I Prove This Trig Identity?

[mathdrama]

I have no idea how to go about proving this trig identiy. I mean, I've been taught that it's a safe bet to convert everything to sines and cosines, but other than that, I've no clue.

Am I even on the right path?

 

[Prove It]

I agree with you up to $\displaystyle \begin{align*} \frac{1}{\cos{(\theta)}} + \frac{\sin{(\theta)}}{\cos{(\theta)}} \cdot \frac{\sin{(\theta)}}{1} \end{align*}$. This is NOT the same as $\displaystyle \begin{align*} \frac{1}{\cos{(\theta)}} + \frac{\sin^2{(\theta)}}{\cos{(\theta)}} \cdot \frac{\sin^2{(\theta)}}{\cos{(\theta)}} \end{align*}$. It's just $\displaystyle \begin{align*} \frac{1}{\cos{(\theta)}} + \frac{\sin^2{(\theta)}}{\cos{(\theta)}} \end{align*}$. Now, you have a common denominator, so the two fractions can be added. You should find that the top simplifies with the Pythagorean Identity.

 

[Deveno]

Yes, but...

You've already made some errors in your "transformation".

If you start with:

$\sec\theta - \tan\theta\sin\theta = \cos\theta$

your next step should be to change this to:

$\dfrac{1}{\cos\theta} - \dfrac{\sin^2\theta}{\cos\theta} = \cos\theta$.

Try multiplying through by $\cos\theta$ next.

 

[MarkFL]

I was taught to begin with the left side of the given identity and then through algebraic means and through the use of standard identities, transform the left side into the right.

I would begin be factoring $\sec(\theta)$ from the left side:

\(\displaystyle \sec(\theta)\left(1-\sin^2(\theta)\right)\)

Now apply a Pythagorean identity and simplify and you will get the right side.

 

[mathdrama]

I was taught to begin with the left side of the given identity and then through algebraic means and through the use of standard identities, transform the left side into the right.

I would begin be factoring $\sec(\theta)$ from the left side:

\(\displaystyle \sec(\theta)\left(1-\sin^2(\theta)\right)\)

Now apply a Pythagorean identity and simplify and you will get the right side.

I don't know how to apply a Pythagorean identity, can you help me?

 

[MarkFL]

I don't know how to apply a Pythagorean identity, can you help me?

Perhaps the best known Pythagorean identity is:

\(\displaystyle \sin^2(\theta)+\cos^2(\theta)=1\)

Now, can you arrange this such that you can make a substitution for:

\(\displaystyle 1-\sin^2(\theta)\) ?

 

[mathdrama]

Perhaps the best known Pythagorean identity is:

\(\displaystyle \sin^2(\theta)+\cos^2(\theta)=1\)

Now, can you arrange this such that you can make a substitution for:

\(\displaystyle 1-\sin^2(\theta)\) ?

Is it something like 1 - sin^2 = 1 = sin^2?

 

[MarkFL]

Is it something like 1 - sin^2 = 1 = sin^2?

No, if we begin with:

\(\displaystyle \sin^2(\theta)+\cos^2(\theta)=1\)

And then subtract $\sin^2(\theta)$ from both sides, we get the alternate form of the identity:

\(\displaystyle \cos^2(\theta)=1-\sin^2(\theta)\)

And so now we have (in the original identity we are trying to prove):

\(\displaystyle \sec(\theta)\cos^2(\theta)\)

 

[mathdrama]

Okay, but I don’t how to simplify that any further or even turn it into cosθ.

 

[MarkFL]

Okay, but I don’t how to simplify that any further or even turn it into cosθ.

Well, by definition, we have:

\(\displaystyle \sec(\theta)\equiv\frac{1}{\cos(\theta)}\)

And so we have:

\(\displaystyle \sec(\theta)\cos^2(\theta)=\frac{\cos(\theta)}{\cos(\theta)}\cos(\theta)=\cos(\theta)\)

 

[mathdrama]

Oh, thank you. I finally understand now.