# Trig Identities Help? Thanks!

• Feb 11th 2010, 02:15 PM
Justinram22
Trig Identities Help? Thanks!
Ok, I have a couple of questions on my homework that i just can't figure out how they work....

"Verify that each equation is an identity"

Cos^4(s)-sin^4(s) = cos2(s)

I'm a little confused as the how this could equal each other? I get the double angle identity (cos^2(x)-sin^2(x)=cos2x) But how would powers to the 4th simplify down?

And this one...
(sin4t/4) = cos^3(t)sin(t)-sin^3(t)cos(t)

I don't really even know where to begin on this one...We are working with double-angle identities and half-angle identities....

Thanks to anyone who can help!!
• Feb 11th 2010, 02:22 PM
Prove It
Quote:

Originally Posted by Justinram22
Ok, I have a couple of questions on my homework that i just can't figure out how they work....

"Verify that each equation is an identity"

Cos^4(s)-sin^4(s) = cos2(s)

I'm a little confused as the how this could equal each other? I get the double angle identity (cos^2(x)-sin^2(x)=cos2x) But how would powers to the 4th simplify down?

And this one...
(sin4t/4) = cos^3(t)sin(t)-sin^3(t)cos(t)

I don't really even know where to begin on this one...We are working with double-angle identities and half-angle identities....

Thanks to anyone who can help!!

$\displaystyle \cos^4{s} - \sin^4{s} = (\cos^2{s} + \sin^2{s})(\cos^2{s} - \sin^2{s})$

$\displaystyle = 1(\cos^2{s} - \sin^2{s})$

$\displaystyle = \cos^2{s} - \sin^2{s}$

$\displaystyle = \cos{2s}$.
• Feb 11th 2010, 02:26 PM
Prove It
Quote:

Originally Posted by Justinram22
Ok, I have a couple of questions on my homework that i just can't figure out how they work....

"Verify that each equation is an identity"

Cos^4(s)-sin^4(s) = cos2(s)

I'm a little confused as the how this could equal each other? I get the double angle identity (cos^2(x)-sin^2(x)=cos2x) But how would powers to the 4th simplify down?

And this one...
(sin4t/4) = cos^3(t)sin(t)-sin^3(t)cos(t)

I don't really even know where to begin on this one...We are working with double-angle identities and half-angle identities....

Thanks to anyone who can help!!

I assume for the second one, the left hand side is $\displaystyle \frac{\sin{4t}}{4}$...

$\displaystyle \cos^3{t}\sin{t} - \sin^3{t}\cos{t} = \sin{t}\cos{t}(\cos^2{t} - \sin^2{t})$

$\displaystyle = \frac{1}{2}\cdot 2\sin{t}\cos{t}\cos{2t}$

$\displaystyle = \frac{1}{2}\sin{2t}\cos{2t}$

$\displaystyle = \frac{1}{4}\cdot 2\sin{2t}\cos{2t}$

$\displaystyle = \frac{1}{4}\sin{4t}$

$\displaystyle = \frac{\sin{4t}}{4}$.
• Feb 11th 2010, 02:55 PM
Justinram22
Thanks you!!!

Could i ask one more? Sorry, i thought i had this one but can't seem to get it...

cos2(y) = ((1-tan^2(y))/(1+tan^2(y))

Thanks again!

(Am i correct in thinking that you have to turn 1+tan^2(y) into sec^2(y))?
• Feb 11th 2010, 06:39 PM
Prove It
Quote:

Originally Posted by Justinram22
Thanks you!!!

Could i ask one more? Sorry, i thought i had this one but can't seem to get it...

cos2(y) = ((1-tan^2(y))/(1+tan^2(y))

Thanks again!

(Am i correct in thinking that you have to turn 1+tan^2(y) into sec^2(y))?

Use the identity $\displaystyle \tan^2{x} + 1 = \sec^2{x}$.

$\displaystyle \frac{1 - \tan^2{y}}{1 + \tan^2{y}} = \frac{1 - (\sec^2{y} - 1)}{\sec^2{y}}$

$\displaystyle = \frac{2 - \sec^2{y}}{\sec^2{y}}$

$\displaystyle = \frac{2}{\sec^2{y}} - 1$

$\displaystyle = 2\cos^2{y} - 1$

$\displaystyle = \cos^2{y} + \cos^2{y} - 1$

$\displaystyle = \cos^2{y} - (1 - \cos^2{y})$

$\displaystyle = \cos^2{y} - \sin^2{y}$

$\displaystyle = \cos{2y}$.