Transforming Trigonometric expression

[Drain Brain]

transform the first member to the second member

$\sqrt{\frac{\sec^{2}(\phi)-1}{\sec^{2}(\phi)(1+\cot^{2}(\phi))}}+\frac{\csc^{2}(\phi)}{\csc(\phi)}\sqrt{\frac{\csc^{2}(\phi)-1}{\csc^{2}(\phi)}} = (1+\cot(\phi))(1-\sin(\phi)\cos(\phi))$this is what I tried so far,

$\sin^{2}(\phi)+\cot(\phi)\cos^{2}(\phi) = (1+\cot(\phi))(1-\sin(\phi)\cos(\phi))$

I maybe out of idea on how to transform the left side expression to right side expression please help me. thanks!

 

[Drain Brain]

transform the first member to the second member

$\sqrt{\frac{\sec^{2}(\phi)-1}{\sec^{2}(\phi)(1+\cot^{2}(\phi))}}+\frac{\csc^{2}(\phi)}{\csc(\phi)}\sqrt{\frac{\csc^{2}(\phi)-1}{\csc^{2}(\phi)}} = (1+\cot(\phi))(1-\sin(\phi)\cos(\phi))$

there's a typo in my first post it should be

$\sqrt{\frac{\sec^{2}(\phi)-1}{\sec^{2}(\phi)(1+\cot^{2}(\phi))}}+\frac{\cot^{2}(\phi)}{\csc(\phi)}\sqrt{\frac{\csc^{2}(\phi)-1}{\csc^{2}(\phi)}} = \left(1+\cot(\phi)\right)\left(1-\sin(\phi)\cos(\phi)\right)$

Help please!

 

[paulmdrdo]

Maybe it's just me,the first member cannot be transformed into the second member. But we can still prove that the given identity is true by some using algebraic manipulation.

 

[Drain Brain]

What? How come my trig book gives a problem like that. I think there's still a way around it.

 

[paulmdrdo]

What? How come my trig book gives a problem like that. I think there's still a way around it.

Again, I think it's just me. Just wait for the Math Helpers to tell you how to go about it.
Regards!

 

[MarkFL]

Begin by simplifying the radicals using various Pythagorean identities.

Then simplify the resulting expressions be expressing in terms of sine and cosine only.

Then combine terms and factor the numerator using the sum of cubes formula.

Further distribution/simplification will give the desired result.

 

[Drain Brain]

Begin by simplifying the radicals using various Pythagorean identities.

Then simplify the resulting expressions be expressing in terms of sine and cosine only.

Then combine terms and factor the numerator using the sum of cubes formula.

Further distribution/simplification will give the desired result.

By doing so I end up getting,

$\frac{\sin^{3}(\phi)+\cos^{3}(\phi)}{\sin(\phi)}=1-\sin(\phi)\cos(\phi)+\cot(\phi)-\cos^{2}(\phi)$I still could not get the desired form.

 

[MarkFL]

By doing so I end up getting,

$\frac{\sin^{3}(\phi)+\cos^{3}(\phi)}{\sin(\phi)}=1-\sin(\phi)\cos(\phi)+\cot(\phi)-\cos^{2}(\phi)$I still could not get the desired form.

Okay, good, I too wound up with:

\(\displaystyle \frac{\sin^{3}(\phi)+\cos^{3}(\phi)}{\sin(\phi)}\)

Now what do you get when factor the numerator as the sum of cubes?

 

[Drain Brain]

Okay, good, I too wound up with:

\(\displaystyle \frac{\sin^{3}(\phi)+\cos^{3}(\phi)}{\sin(\phi)}\)t

Now what do you get when factor the numerator as the sum of cubes?

$\frac{\sin^{3}(\phi)+\cos^{3}(\phi)}{\sin(\phi)}=\frac{(\sin(\phi)+\cos(\phi))(\sin^{2}(\phi)-\sin(\phi)\cos(\phi)+\cos^{2}(\phi))}{\sin(\phi)}=
1-\sin(\phi)\cos(\phi)+\cot(\phi)-\cos^{2}(\phi)$

Still not the form that I want.

 

[Euge]

$\frac{\sin^{3}(\phi)+\cos^{3}(\phi)}{\sin(\phi)}=\frac{(\sin(\phi)+\cos(\phi))(\sin^{2}(\phi)-\sin(\phi)\cos(\phi)+\cos^{2}(\phi))}{\sin(\phi)}=
1-\sin(\phi)\cos(\phi)+\cot(\phi)-\cos^{2}(\phi)$

Still not the form that I want.

Hi Drain Brain,

There's still more you can do. Note that

\(\displaystyle \cos^2(\phi) = \cot(\phi) \tan(\phi) \cos^2(\phi) = \cot(\phi) [\tan(\phi) \cos(\phi)] \cos(\phi) = \cot(\phi) \sin(\phi) \cos(\phi).\)

Using this, you can regroup the terms and factor to get the result you want.

 

[MarkFL]

$\frac{\sin^{3}(\phi)+\cos^{3}(\phi)}{\sin(\phi)}=\frac{(\sin(\phi)+\cos(\phi))(\sin^{2}(\phi)-\sin(\phi)\cos(\phi)+\cos^{2}(\phi))}{\sin(\phi)}=
1-\sin(\phi)\cos(\phi)+\cot(\phi)-\cos^{2}(\phi)$

Still not the form that I want.

Okay, look at:

\(\displaystyle \frac{(\sin(\phi)+\cos(\phi))(\sin^{2}(\phi)-\sin(\phi)\cos(\phi)+\cos^{2}(\phi))}{\sin(\phi)}\)

Write this as:

\(\displaystyle \frac{\sin(\phi)+\cos(\phi)}{\sin(\phi)}\left(\sin^{2}(\phi)-\sin(\phi)\cos(\phi)+\cos^{2}(\phi)\right)\)

Now...simplify the first factor by dividing each term in the numerator by $\sin(\phi)$, and apply a Pythagorean identity to the second factor...what do you get?

 

[Drain Brain]

Okay, look at:

\(\displaystyle \frac{(\sin(\phi)+\cos(\phi))(\sin^{2}(\phi)-\sin(\phi)\cos(\phi)+\cos^{2}(\phi))}{\sin(\phi)}\)

Write this as:

\(\displaystyle \frac{\sin(\phi)+\cos(\phi)}{\sin(\phi)}\left(\sin^{2}(\phi)-\sin(\phi)\cos(\phi)+\cos^{2}(\phi)\right)\)

Now...simplify the first factor by dividing each term in the numerator by $\sin(\phi)$, and apply a Pythagorean identity to the second factor...what do you get?

I would get

\(\displaystyle \frac{\sin(\phi)}{\sin(\phi)}+\frac{\cos(\phi)}{\sin(\phi)}\left(\sin^{2}(\phi)-\sin(\phi)\cos(\phi)+\cos^{2}(\phi)\right) = \left(1+\cot(\phi)\right) \left(1-\sin(\phi)\cos(\phi)\right)\)