How Do You Prove Trigonometric Identities with Minimal Equations?

[alexandria]

Homework Statement

prove the following identity [/B]

Homework Equations

no equations required

The Attempt at a Solution

I've been trying to prove this identity, but no matter what I do, I can't seem to make both sides the same
here is my answer to this qts so far: can someone please tell me what I have to do next?
Im pretty sure I'm supposed to be getting tan(x) - 1 on both sides. So, should I leave the left side the same (keep it as tan(x) -1) and work only with the right side, until both sides are equal?
any help would be appreciated. Thanks.

Sorry for accidently posting the same thread twice :p

 

[ehild]

Homework Statement

prove the following identity [/B]
View attachment 104676

Homework Equations

no equations required

The Attempt at a Solution

I've been trying to prove this identity, but no matter what I do, I can't seem to make both sides the same
here is my answer to this qts so far: can someone please tell me what I have to do next?
Im pretty sure I'm supposed to be getting tan(x) - 1 on both sides. So, should I leave the left side the same (keep it as tan(x) -1) and work only with the right side, until both sides are equal?
any help would be appreciated. Thanks.
View attachment 104677Sorry for accidently posting the same thread twice :p

When you divide a sum or difference, you have to divide all terms. $\frac{\sin(θ)-\cos(θ)}{\cos(θ)}$ is not sin(θ)-1. And you made the same error in the right side.
Leave the left side as it is, and work with the right side. Divide both numerator and denominator with cos²(θ), for example.

 

[alexandria]

Thanks for the quick reply!
Ok so i understand that i am supposed to leave the left side the same, but what do i do with the right side. You mentioned dividing both numerator and denominator with cos^2(θ), how do i do that?
so is it:
sin^2(x) - cos^2(x) /cos^2(x)
-------------------------------------
sin(x) cos(x) + cos^2(x) / cos^2(x)

i was originally using trigonometric identities:
for ex: sin^2(x) + cos^2(x) = 1 so i was assuming that sin^2(x) - cos^2(x) = -1
and so i replaced sin^2(x) - cos^2(x) with -1
is this correct?

 

[ehild]

Thanks for the quick reply!
Ok so i understand that i am supposed to leave the left side the same, but what do i do with the right side. You mentioned dividing both numerator and denominator with cos^2(θ), how do i do that?
so is it:
((sin^2(x) - cos^2(x)) /cos^2(x)
-------------------------------------
(sin(x) cos(x) + cos^2(x)) / cos^2(x)

No. You have to divide the whole expressions, not only the last terms. Do not omit the parentheses.

i was originally using trigonometric identities:
for ex: sin^2(x) + cos^2(x) = 1 so i was assuming that sin^2(x) - cos^2(x) = -1
and so i replaced sin^2(x) - cos^2(x) with -1
is this correct?

No. It is wrong. sin^2(x) - cos^2(x) is not -1. Try with numbers. 2/3 + 1/3 = 1, is it true, that 2/3 - 1/3 = -1?

 

[member 587159]

Thanks for the quick reply!
Ok so i understand that i am supposed to leave the left side the same, but what do i do with the right side. You mentioned dividing both numerator and denominator with cos^2(θ), how do i do that?
so is it:
sin^2(x) - cos^2(x) /cos^2(x)
-------------------------------------
sin(x) cos(x) + cos^2(x) / cos^2(x)

i was originally using trigonometric identities:
for ex: sin^2(x) + cos^2(x) = 1 so i was assuming that sin^2(x) - cos^2(x) = -1
and so i replaced sin^2(x) - cos^2(x) with -1
is this correct?

sin^2 x + cos^2 x = 1
=> -sin^2 x - cos^2 x = -1 and not what you wrote.

Leave the left side as it is. In the right side, we notice that we can use the formula: a^2 - b^2 = (a-b)(a+b) in the numerator. Try to find the factors of the denumerator and the problem becomes quite trivial then.

 

[alexandria]

ok so for the right side,if I use the formula: a^2 - b^2 = (a-b)(a+b)
then it becomes:

(sin(x) - cos(x)) (sin(x) + cos(x))
-------------------------------------
sin(x) cos(x) + cos^2(x)

how do i solve this??

 

[dextercioby]

You're not getting the right push so far. You need to show that a = b/c. Thus ac = b. Cross-multiply (tan theta - 1) by the denominator (sin theta cos theta + cos^2 theta) theta and simplify.

 

[alexandria]

srry I am very confused, what do you mean by cross multiply and why would I need to cross multiply tan(x) - 1 by the denominator of the right side?
Can you provide an example!

 

[alexandria]

ok so i came up with an answer, can someone tell me if I am on the right track?

((sin^2(x) - cos^2(x)) /cos^2(x)
-------------------------------------
(sin(x) cos(x) + cos^2(x)) / cos^2(x)

= tan^2(x)
----------
tan(x) + 1

= tan(x) tan(x)
-------------
tan(x) + 1

= tan(x) + 1

im not sure what to do from here.

 

[ehild]

ok so i came up with an answer, can someone tell me if I am on the right track?

((sin^2(x) - cos^2(x)) /cos^2(x)
-------------------------------------
(sin(x) cos(x) + cos^2(x)) / cos^2(x)

= tan^2(x)
----------
tan(x) + 1

The last equation is wrong.

 

[alexandria]

can you please elaborate?

 

[SammyS]

ok so i came up with an answer, can someone tell me if I am on the right track?

((sin^2(x) - cos^2(x)) /cos^2(x)
-------------------------------------
(sin(x) cos(x) + cos^2(x)) / cos^2(x)

Look at that numerator again !

What is $\displaystyle \frac{\sin^2(x) - \cos^2(x)}{\cos^2(x)} \$ ?

It's not

= tan^2(x)

You may need to review: adding and subtracting fractions !

 

[alexandria]

for the numerator:

sin^2(x) - cos^2(x) / cos^2 (x)
inorder to find the answer, I have to divide cos^2(x) by each term in the expression:
so sin^2(x) / cos^2(x) = tan^2(x) (since sin(x)/cos(x) = tan(x))
and -cos^2(x) /cos^2(x) = -1

so is the answer tan^2(x) - 1
?

 

[member 587159]

for the numerator:

sin^2(x) - cos^2(x) / cos^2 (x)
inorder to find the answer, I have to divide cos^2(x) by each term in the expression:
so sin^2(x) / cos^2(x) = tan^2(x) (since sin(x)/cos(x) = tan(x))
and -cos^2(x) /cos^2(x) = -1

so is the answer tan^2(x) - 1
?

ok so for the right side,if I use the formula: a^2 - b^2 = (a-b)(a+b)
then it becomes:

(sin(x) - cos(x)) (sin(x) + cos(x))
-------------------------------------
sin(x) cos(x) + cos^2(x)

how do i solve this??

Factor out a cosx in the denumerator:

[sin^2 (x) - cos^2(x)]/[sinxcosx + cos^2 x]
= [(sinx - cosx)(sinx + cosx)]/[cosx(sinx + cosx)]

Then use (a+b)/c = a/c + b/c

 

[SammyS]

for the numerator:

(sin^2(x) - cos^2(x)) / cos^2 (x)
inorder to find the answer, I have to divide cos^2(x) by each term in the expression:
so sin^2(x) / cos^2(x) = tan^2(x) (since sin(x)/cos(x) = tan(x))
and -cos^2(x) /cos^2(x) = -1

so is the answer tan^2(x) - 1
?

Yes.

So make that correction in the following.

ok so i came up with an answer, can someone tell me if I am on the right track?

((sin^2(x) - cos^2(x)) /cos^2(x)
-------------------------------------
(sin(x) cos(x) + cos^2(x)) / cos^2(x)

= tan^2(x)
----------
tan(x) + 1

...

That gives

$\displaystyle \frac{\tan^2(x) - 1}{\tan(x)+1} \$

 

[alexandria]

((sin^2(x) - cos^2(x)) /cos^2(x)
-------------------------------------
(sin(x) cos(x) + cos^2(x)) / cos^2(x)= tan^2(x) - 1
--------------
tan(x) + 1

= tan(x) - 1

is this correct?

 

[SammyS]

((sin^2(x) - cos^2(x)) /cos^2(x)
-------------------------------------
(sin(x) cos(x) + cos^2(x)) / cos^2(x)= tan^2(x) - 1
--------------
tan(x) + 1

= tan(x) - 1

is this correct?

That is correct.

Can you justify that last step ?

 

[alexandria]

ok so,

tan^2(x) - 1
---------------
tan(x) + 1

= tan(x) tan(x) - 1
------------------ ( I am pretty sure tan(x) crosses out at this part)
tan(x) + 1

= tan(x) - 1 / 1

= tan(x) - 1

is this process correct?

 

[SammyS]

ok so,

tan^2(x) - 1
---------------
tan(x) + 1

= tan(x) tan(x) - 1
------------------ ( I am pretty sure tan(x) crosses out at this part)
tan(x) + 1

= tan(x) - 1 / 1

= tan(x) - 1

is this process correct?

That's what I suspected.

That is not the correct way to obtain the result.

a² -1 factors to (a - 1)(a + 1)..

So, can you factor $\ \tan^2(x) - 1 \ ?$

 

[alexandria]

ok so,

(tan(x) - 1) (tan(x) + 1) (cross out the expression tan(x) + 1)
-----------------------------
tan(x) + 1

= tan(x) - 1

 

[SammyS]

ok so,

(tan(x) - 1) (tan(x) + 1) (cross out the expression tan(x) + 1)
-----------------------------
tan(x) + 1

= tan(x) - 1

That's more like it !

 

[alexandria]

thanks so much for all the help, I really appreciate it!

 

[epenguin]

Well take RHS.
cos θ is obviously a factor of the denominator. So factorise it.
Now look at the numerator. This is the difference of two squares. Know how you can factorise - OK how you can write difference of two squares? (a² - b²)?
Do that then you see one of the factors is the same as the factor of the denominator. Then it all falls out easily.

Why can I do that in my head, and you have spent an hour on it? Well partly it's because I don't have to do it. So I'm relaxed about it You might need to relax
Secondly I'm using a Polya 'How To Solve It' principle - asking self "have you seen anything like this before?".
Sure you see trig the formula and you start thinking of other trig formulae – you might not recognise immediately the resemblance with (a² - b²). That may be where being relaxed helps

 

[Evangeline101]

Homework Statement

Prove the identity:

Homework Equations

The Attempt at a Solution

Can someone tell me if this is correct? I am not sure if I should have worked on both the right and left side until they were the same, instead of just the left side?

 

[Student100]

Homework Statement

Prove the identity:
View attachment 104794

Homework Equations

View attachment 104795

The Attempt at a Solution

View attachment 104793

Can someone tell me if this is correct? I am not sure if I should have worked on both the right and left side until they were the same, instead of just the left side?

Looks okay,

The second step on the left hand side is a bit strange, but then you fixed it the next step, so probably just a typo.

 

[Evangeline101]

The second step on the left hand side is a bit strange, but then you fixed it the next step, so probably just a typo.

This step?

should I just take it out then?

 

[Student100]

This step?
View attachment 104796
should I just take it out then?

The next step, it's not equal to $\frac{\sin{x}+\sin{x}}{\cos{x}}$

 

[SammyS]

It seems to me that working the right hand side would be easier than working the left hand side.

 

[Evangeline101]

The next step, it's not equal to sinx+sinxcosx

Okay, I edited it:

Is this better?

 

[Student100]

Okay, I edited it:

View attachment 104797
View attachment 104798

Is this better?

That's fine. You just need to be careful, the = is saying something very precise.