[SOLVED] proving sin, and confirmation needed.

**MathBlaster47** · January 29th 2010, 01:46 PM

I have a weird proof(?) that I can't figure out how to work.

It goes as follows:
Show that $\sin \theta=\cos\theta\tan\theta$

I also have a few true/false questions I wanted to make sure I have right:

1: $\sin\theta=\cos(90-\theta)$ my answer: False
2: $\tan\theta=-\cot(90-\theta)$ my answer: False
3: $\sin\theta=\frac{1}{\csc\theta}$ my answer: True

Many thanks for all of your help so far!

**e^(i*pi)** · January 29th 2010, 02:03 PM

Originally Posted by MathBlaster47

I have a weird proof(?) that I can't figure out how to work.

It goes as follows:
Show that $\sin \theta=\cos\theta\tan\theta$

I also have a few true/false questions I wanted to make sure I have right:

1: $\sin\theta=\cos(90-\theta)$ my answer: False
2: $\tan\theta=-\cot(90-\theta)$ my answer: False
3: $\sin\theta=\frac{1}{\csc\theta}$ my answer: True

Many thanks for all of your help so far!

I'd use the basic definitions of the trig ratios:

$\frac{\text {opp}}{\text {hyp}} = \frac{\text {adj}{\text {hyp}} \times \frac{\text {opp}}{\text {adj}}$

The RHS cancels to give the LHS proving the identity.

$cos(90-\theta) = cos90cos\theta + sin90sin\theta$

This identity is true

**Archie Meade** · January 29th 2010, 02:10 PM

Originally Posted by MathBlaster47

I have a weird proof(?) that I can't figure out how to work.

It goes as follows:
Show that $\sin \theta=\cos\theta\tan\theta$

I also have a few true/false questions I wanted to make sure I have right:

1: $\sin\theta=\cos(90-\theta)$ my answer: False
2: $\tan\theta=-\cot(90-\theta)$ my answer: False
3: $\sin\theta=\frac{1}{\csc\theta}$ my answer: True

Many thanks for all of your help so far!

Hi Mathblaster,

Write $tan\theta$ using $Sin\theta$ and $Cos\theta$

$Cos\theta\ tan\theta=Cos\theta\ \frac{Sin\theta}{Cos\theta}$

For the next one, draw a right-angled triangle with $\theta$ in one corner.
The acute angle in the other corner is $90-\theta$

Now label the sides "a", "b", "c" and discover whether your statement is true or false.

For the next, write $tan\theta$ and $cot(90-\theta)$ using $Sin\theta$ and $Cos\theta$

and $Sin(90-\theta)$ and $Cos(90-\theta)$

You were correct that time but were you guessing?

Your final answer is also correct.

**MathBlaster47** · January 29th 2010, 03:08 PM

Originally Posted by Archie Meade

Originally Posted by MathBlaster47

I have a weird proof(?) that I can't figure out how to work.

It goes as follows:
Show that $\sin \theta=\cos\theta\tan\theta$

I also have a few true/false questions I wanted to make sure I have right:

1: $\sin\theta=\cos(90-\theta)$ my answer: False
2: $\tan\theta=-\cot(90-\theta)$ my answer: False
3: $\sin\theta=\frac{1}{\csc\theta}$ my answer: True

Many thanks for all of your help so far!

Hi Mathblaster,

Write $tan\theta$ using $Sin\theta$ and $Cos\theta$

$Cos\theta\ tan\theta=Cos\theta\ \frac{Sin\theta}{Cos\theta}$

For the next one, draw a right-angled triangle with $\theta$ in one corner.
The acute angle in the other corner is $90-\theta$

Now label the sides "a", "b", "c" and discover whether your statement is true or false.

For the next, write $tan\theta$ and $cot(90-\theta)$ using $Sin\theta$ and $Cos\theta$

and $Sin(90-\theta)$ and $Cos(90-\theta)$

You were correct that time but were you guessing?

Your final answer is also correct.

Thanks Archie!
What I did for all of the true/false questions was:
I took an arbitrary value for $\theta$ (10) evaluated for the measure of given function, then I worked the other side of the equation in terms of the given function.

Let me make sure I have what you said straight:All of the true/false questions are correct, and all I have to do for the identity is rewrite the equation as: $Cos\theta\ tan\theta=Cos\theta\ \frac{Sin\theta}{Cos\theta}$ ?

**Archie Meade** · January 29th 2010, 03:18 PM

Yes,
because tan is sine divided by cosine.

You could also do what $e^{i\pi}$ recommended,
which is to use the identity for Cos(A-B) and also Sin(A-B) for the tan and cotan, working from there.

In using an arbitrary value, you are proving the equation for that value only,
not in general.

You should go the extra mile and either draw the right-angled triangle
or use identities as $e^{i\pi}$ recommended.

Your original answers to numbers 2. and 3. were correct,
however I'd recommend you work on them a bit more.

I have to go to sleep but I will continue with this tomorrow,
in the meantime try.

**MathBlaster47** · January 29th 2010, 03:47 PM

Originally Posted by Archie Meade

Yes,
because tan is sine divided by cosine.

You could also do what $e^{i\pi}$ recommended,
which is to use the identity for Cos(A-B) and also Sin(A-B) for the tan and cotan, working from there.

In using an arbitrary value, you are proving the equation for that value only,
not in general.

You should go the extra mile and either draw the right-angled triangle
or use identities as $e^{i\pi}$ recommended.

Your original answers to numbers 2. and 3. were correct,
however I'd recommend you work on them a bit more.

I have to go to sleep but I will continue with this tomorrow,
in the meantime try.

Thanks so much Archie!I'll be sure to do that!

**Archie Meade** · January 30th 2010, 04:24 AM

Hi MathBlaster47,

attached is a sketch to help work through those examples...
We use a right-angled triangle for these (one corner is 90 degrees)

Since

$Sin\theta=\frac{b}{c},\ Cos\theta=\frac{a}{c}\ and\ Tan\theta=\frac{b}{a}$

then

$Cos\theta\ Tan\theta=\frac{a}{c}\ \frac{b}{a}=\frac{a}{a}\ \frac{b}{c}=\frac{b}{c}=Sin\theta$

or...

$Tan\theta=\frac{Sin\theta}{Cos\theta}$

$Cos\theta\ Tan\theta=Cos\theta\frac{Sin\theta}{Cos\theta}=\fr ac{Cos\theta\ Sin\theta}{Cos\theta}=Sin\theta$

True/false 1.

On the attachment,

$Sin\theta=\frac{b}{c}$

the side marked "b" is opposite the top corner angle and the side marked "a" is adjacent to it, while "c" is still the hypotenuse...

$Cos(90-\theta)=\frac{b}{c},\ so\ Sin\theta=Cos(90-\theta)$

Alternatively, you can a the trigonometric identity for this...

$Cos(X-Y)=CosXCosY+SinXSinY$

$X=90,\ Y=\theta$

$Cos90^o=0\ and\ Sin90^o=1$

$Cos(90^o)Cos\theta+Sin(90^o)Sin\theta=(0)Cos\theta +(1)Sin\theta=Sin\theta$

True\false 2.

$Tan\theta=\frac{a}{b},\ Tan(90^o-\theta)=\frac{b}{a}$

$Cot(90^o-\theta)=\frac{1}{Tan(90^o)-\theta}=\frac{a}{b}$

$Cot(90^0-\theta)=Tan\theta$

hence

$Tan\theta\ \ne\ -Cot(90^o-\theta)$

True/false 3.

$Cosec\theta=\frac{1}{Sin\theta}$

so

$Sin\theta=\frac{1}{Cosec\theta}$

**MathBlaster47** · February 1st 2010, 02:35 PM

AHHH!
Thank you!!
I made the mistake of not remembering that those fancy words (Sine, Cosine, Tangent, and so on) stand for the relationship between the sides of a triangle.
So correct me if I'm wrong but....had I remembered that, all I would have had to do to answer the question is multiply the fractional form of the function(s) and multiply or divide them as prescribed in the question?

**Archie Meade** · February 1st 2010, 02:59 PM

Originally Posted by MathBlaster47

AHHH!
Thank you!!
I made the mistake of not remembering that those fancy words (Sine, Cosine, Tangent, and so on) stand for the relationship between the sides of a triangle.
More specifically, the ratios of the sides of a right-angled triangle, to put it simply, though the idea is extended to cover all angles.
You will see Sine and Cosine being used for non-right-angled triangles, for example in the Sine Rule and Cosine Rule, but they are not defined by the ratio of the sides of those triangles.

So correct me if I'm wrong but....had I remembered that, all I would have had to do to answer the question is multiply the fractional form of the function(s) and multiply or divide them as prescribed in the question? Yes

Or, use the "trigonometric identities", best to know both ways.

**MathBlaster47** · February 2nd 2010, 08:45 AM

Originally Posted by MathBlaster47

AHHH!
Thank you!!
I made the mistake of not remembering that those fancy words (Sine, Cosine, Tangent, and so on) stand for the relationship between the sides of a triangle.
More specifically, the ratios of the sides of a right-angled triangle, to put it simply, though the idea is extended to cover all angles.
You will see Sine and Cosine being used for non-right-angled triangles, for example in the Sine Rule and Cosine Rule, but they are not defined by the ratio of the sides of those triangles.

So correct me if I'm wrong but....had I remembered that, all I would have had to do to answer the question is multiply the fractional form of the function(s) and multiply or divide them as prescribed in the question? Yes

This brings me to another question: What would I have done if the question had said "assume that the triangle is non-right angled" or something to that effect?

**Archie Meade** · February 2nd 2010, 11:08 AM

All the questions you were given initially allow you to use the definitions of Sine, Cosine, Tangent, Cosec, Sec and Cotan as per the right-angled triangle.

If you were given a diagram of a non-right-angled triangle or asked to model a real life situation in terms of angles and lengths and there was no way to find a right-angle,

then you would need to examine the clues for alternative ways,
such as "The Law of Sines" or "The Law of Cosines", and other ways.
Even though Sine and Cosine is still used in these, the ratio of the sides of non-right-angled triangles are not SinA or CosA or TanA etc...

You just need to practice and become familiar with the differences.
You need visual comprehension of all that.

**MathBlaster47** · February 2nd 2010, 11:20 AM

Ah....ok so practice, practice, practice is the best way to do it......
Makes sense, it depends on the question.
Thank you very, very much for your help Archie!

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