Trigonometric proofs and exact values.

**MathBlaster47** · March 2nd 2010, 04:30 PM

I have a couple questions that I need a little direction on.

1:
Prove that $\sin^2 x(1+\cot^2 x)=1$

Method:
I figured out that I should convert $\sin^2 x(1+\cot^2 x)=1$ to $\sin^2 x (\csc^2 x)=1$ . But beyond that, I'm a little lost....I know there is more to be done though. I just don't quite understand how reciprocal and Pythagorean identities are related, I know that $\sin x \csc x=1$ but I'm not sure if the same holds true for their Pythagorean counterparts or if that would be a valid proof.

2:
Prove that $\tan x(\cot x +\tan x)=\sec^2 x$ .

Method:
I made the equation $\tan x(\cot x +\tan x)=\sec^2 x$ into $\tan x(\frac{1}{\tan x} + \tan x)= sec^2 x$ .
Does the equation become $1+\tan^2=\sec^2$ , or am I missing something?
I think I have the right answer, as I see that the Pythagorean identity for $\sec^2 x$ coinsides with my converted equation.

3:
Find the exact value of $\cos 105$ with out using tables or a calculator.
Be sure to show your work.

Method:
Since the question asks for the value of a unusual degree measure I have to use an addition formula, namely: $\cos(\alpha - \beta)=\cos\alpha \cos \beta+\sin \alpha \sin \beta$ .
For \alpha I used value 135 and for \beta I used 30 so $\cos(\alpha - \beta) \rightarrow \cos(135-30)$ , therefore the whole formula changes to become: $\cos(135 - 30)=\cos 135 \cos 30 + \sin 135 \sin 30$ .
So, $\cos 135 \cos 30 + \sin 135 \sin 30 \rightarrow -\frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \frac{1}{2} \rightarrow \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \rightarrow \frac{\sqrt{2} +\sqrt{6}}{4}$ , have I gotten the right answer?

4:
Find the direct value of $\tan 15$ without using tables or a calculator.
Be sure to to show your work.

Method:
The book said nothing about the addition formula for $\tan$ so I searched the internet and found this: $\tan(\alpha-\beta)= \frac{\tan \alpha - \tan \beta}{1+ \tan\alpha \tan\beta}$
I decided to use this formula since I could figure out no other to use, I entered $\tan 45$ for $\alpha$ and $\tan 30$ for $\beta$ , but I dont know where to go from here....

Thanks.

**sa-ri-ga-ma** · March 2nd 2010, 04:43 PM

You first three methods are correct.
Last one,
tan15 = (tan45 - tan30)/(1-tan45*tan30)
Substitute the values to get tan 15.

**MollyMillions** · March 2nd 2010, 04:49 PM

For the second one, it might help to write $tan(x)$ as $\frac {sin(x)}{cos(x)}$ and $cot(x)$ as $\frac {cos(x)}{sin(x)}$

Edit: Ah, never mind, I didn't read your question properly. If you can use those identities in your proof then you're right.

**MathBlaster47** · March 3rd 2010, 12:34 PM

Originally Posted by sa-ri-ga-ma

You first three methods are correct.
Last one,
tan15 = (tan45 - tan30)/(1-tan45*tan30)
Substitute the values to get tan 15.

Ok! Thank you!

Just curious, but is there a simpler way to do this question?
I know I can and will do it this way--don't get me wrong, but my text book said nothing about this way of solving the question, so I think that they may want a different method to be used. Like I said, they did not give one though...

**icemanfan** · March 3rd 2010, 12:51 PM

Originally Posted by MathBlaster47

Ok! Thank you!

Just curious, but is there a simpler way to do this question?
I know I can and will do it this way--don't get me wrong, but my text book said nothing about this way of solving the question, so I think that they may want a different method to be used. Like I said, they did not give one though...

You don't need the addition formula for tan. A slightly longer method is to use the subtraction formulas for sin and cos:

$\sin(a - b) = \sin a \cos b - \cos a \sin b$

$\cos(a - b) = \cos a \cos b + \sin a \sin b$

$\tan(a - b) = \frac{\sin(a - b)}{\cos(a - b)}$

**MathBlaster47** · March 7th 2010, 03:04 PM

Originally Posted by icemanfan

You don't need the addition formula for tan. A slightly longer method is to use the subtraction formulas for sin and cos:

$\sin(a - b) = \sin a \cos b - \cos a \sin b$

$\cos(a - b) = \cos a \cos b + \sin a \sin b$

$\tan(a - b) = \frac{\sin(a - b)}{\cos(a - b)}$

Thank you!
I was beginning to wonder if I was going crazy, or I if I just didn't read my book properly....

Originally Posted by MathBlaster47

1:
Prove that $\sin^2 x(1+\cot^2 x)=1$

Method:
I figured out that I should convert $\sin^2 x(1+\cot^2 x)=1$ to $\sin^2 x (\csc^2 x)=1$ . But beyond that, I'm a little lost....I know there is more to be done though. I just don't quite understand how reciprocal and Pythagorean identities are related, I know that $\sin x \csc x=1$ but I'm not sure if the same holds true for their Pythagorean counterparts or if that would be a valid proof.

OK! just so I'm clear, my answer to the question I have above is correct?
Am I right in inferring that $\sin^2 x (\csc^2 x)=1$ because $\sin x (\csc x)=1$ ?

**mtt0216** · March 7th 2010, 04:05 PM

$<br /> <br /> \sin^2 x (\csc^2 x)=1<br />$

$\frac{1}{cscx} \frac{1}{cscx} (\csc^2 x)$

or

$\frac{1}{csc^2x} (\csc^2 x)$

$\frac{1}{csc^2x} \frac{{csc^2x} }{1}$

$\frac{csc^2x}{csc^2x}$ =1

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