Trigonometric Identities, I don't know where to start

[James Ibarra]

Hi, I'm an 11th grade student having problems on pre calculus
Our instructorr gave us this exercise to atleast have an idea of what trigonometric identities look like because we were about to discuss this topic but due covid19 we were not able to. Can someone guide me into understanding this? Thanks!

 

[MarkFL]

Hello, and welcome to FMH!

Let's look at the first expression:

[MATH]\sec(\theta)-\cos(\theta)[/MATH]
Let's write this as:

[MATH]\frac{1}{\cos(\theta)}-\cos(\theta)[/MATH]
Now, what do you get when you combine terms using a common denominator?

 

[James Ibarra]

I literally have no idea, we haven't discussed one bit of this yet. I just wanted to get a head start before classes resume

 

[MarkFL]

I literally have no idea, we haven't discussed one bit of this yet. I just wanted to get a head start before classes resume

Let's talk algebra for a moment here...what if I asked you to combine:

[MATH]\frac{1}{x}-x[/MATH]
What would you get?

 

[James Ibarra]

is it 1 - cos(theta)/cos(theta)?

 

[MarkFL]

is it 1 - cos(theta)/cos(theta)?

Let's check:

[MATH]\frac{1-\cos(\theta)}{\cos(\theta)}=\frac{1}{\cos(\theta)}-\frac{\cos(\theta)}{\cos(\theta)}=\sec(\theta)-1[/MATH]
That's not what we started out with. We need to get a common denominator:

[MATH]\frac{1}{\cos(\theta)}-\cos(\theta)=\frac{1}{\cos(\theta)}-\cos(\theta)\cdot\frac{\cos(\theta)}{\cos(\theta)}=\frac{1-\cos^2(\theta)}{\cos(\theta)}[/MATH]
Does that make sense so far?

 

[James Ibarra]

Ah yes, sorry I forgot to add the square on the cos(theta) on the numerator.

 

[MarkFL]

Okay, the numerator we have now should be familiar to you...consider the Pythagorean identity:

[MATH]\sin^2(\theta)+\cos^2(\theta)=1[/MATH]
What if you subtract \(\cos^2(\theta)\) from both sides...

 

[James Ibarra]

it becomes, sin^2(theta) = 1 - cos^2(theta)

 

[MarkFL]

Yes, so now we have:

[MATH]\frac{\sin^2(\theta)}{\cos(\theta)}[/MATH]
A quick check reveals this is not one of the choices, so we still have some work to do. What about:

[MATH]\frac{\sin^2(\theta)}{\cos(\theta)}=\sin(\theta)\cdot\frac{\sin(\theta)}{\cos(\theta)}[/MATH]
Can you simplify that?

 

[James Ibarra]

whoaaaaaaaa, so i substitute the value of sin^2(theta) = 1 - cos^2(theta) to the equation i'm working on then it becomes, sin^2(theta)/
cos(theta).

 

[James Ibarra]

so from the equation sin^2(θ)/cos(θ) i take away the square from the sign then it becomes
sin(θ)⋅sin(θ)/cos(θ) is that right? then if I simplify that equation it then becomes sin^2(θ)/sin(θ)cos(θ)

 

[MarkFL]

so from the equation sin^2(θ)/cos(θ) i take away the square from the sign then it becomes
sin(θ)⋅sin(θ)/cos(θ) is that right?

Yes, that's right.

then if I simplify that equation it then becomes sin^2(θ)/sin(θ)cos(θ)

That's not right. Let's go back to:

[MATH]\frac{\sin(\theta)}{\cos(\theta)}\cdot\sin(\theta)[/MATH]
Now, let's look specifically at:

[MATH]\frac{\sin(\theta)}{\cos(\theta)}[/MATH]
You should recognize that this can be written as a single trig. function.

 

[James Ibarra]

Oh, is this where it becomes another trig function? it becomes a tangent right? our instructor also mention something about a hexagon, he said that it should help us memorize all the different trig functions.

 

[MarkFL]

Oh, is this where it becomes another trig function? it becomes a tangent right? our instructor also mention something about a hexagon, he said that it should help us memorize all the different trig functions.

Yes, we have:

[MATH]\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}[/MATH]
So, now you can see which choice matches this expression. That is, we may write:

[MATH]\sec(\theta)-\cos(\theta)=\tan(\theta)\sin(\theta)[/MATH]

 

[James Ibarra]

Yes, we have:

[MATH]\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}[/MATH]
So, now you can see which choice matches this expression. That is, we may write:

[MATH]\sec(\theta)-\cos(\theta)=\tan(\theta)\sin(\theta)[/MATH]

So i basically make one equation to exactly look like the other?

 

[MarkFL]

So i basically make one equation to exactly look like the other?

What you're trying to do is see if you can make the expressions on the left match the expressions on the right by rewriting them into equivalent expressions using identities.

 

[James Ibarra]

Thanks for you help mark! I will try to answer the other expressions and hopefully get better at it. If i have troubles should i reply to this thread or should i make another one? thanks again! have a nice day and stay safe during the pandemic

 

[Deleted member 4993]

Thanks for you help mark! I will try to answer the other expressions and hopefully get better at it. If i have troubles should i reply to this thread or should i make another one? thanks again! have a nice day and stay safe during the pandemic

Please make a new thread for new problem.

 

[James Ibarra]

can anybody help me with some of these, I got up to number 3 then got stuck on number 4.