How can I prove the trigonometry identity \sin^4 - \cos^4 = 1 - 2\cos^2?

[pavadrin]

i need to prove the following identity

\sin^4 - \cos^4 = 1 - 2\cos^2

i have tried approaching this identity by solving the RHS (right hand side) of the equation but this lead no where. however i am unsure on how to maniplulate the LHS of this equation because of the power of 4. would somebody please be able to give me a slight hint in which direction to head to prove this?
many thanks,
Pavadrin

 

[pizzasky]

Do you know how to factorize a^2-b^2? How can you use this to help you factorize a^4-b^4?

Also, the identity should have a variable (i.e. instead of sin^4, it should be sin^4x).

All the best!

 

[pavadrin]

oh, my mistake i forgot the \theta sign after the trigonometric functions. i am not sure on how top factorise, sorry :(

 

[HallsofIvy]

Long, long, long, before you learn about trigonometry, you should have learned that a²- b²= (a- b)(a+ b). And if we replace a by x² and b by y², we get that
x⁴- y⁴= (x²- y²)(x²+ y²). And what is x²+ y² in this case?

 

[pavadrin]

can x^2 + y^2 be factorised? I've tried (x + y)(x + y), (x - y)(x - y), but neither have worked. am i completely missing the point here? its been a while since i have had to factorise. thanks

 

[BobG]

can x^2 + y^2 be factorised? I've tried (x + y)(x + y), (x - y)(x - y), but neither have worked. am i completely missing the point here? its been a while since i have had to factorise. thanks

In this case you don't have to, since x^2=sin^2\theta and y^2=cos^2\theta.

 

[sdekivit]

we don't need to factorize. Just use:

sin^{4} \theta = sin^{2} \theta \cdot sin^{2} \theta

sin^{2} \theta = 1 - cos^{2} \theta

(1-cos^{2} \theta) \cdot (1-cos^{2} \theta) = 1 - 2\cdot cos^{2} \theta + cos^{4} \theta

Now substract cos^{4} \theta and you got your proof.

 

[pavadrin]

okay thanks for the help people, sorry to be a little slow

 

[hmm?]

In addition to what these people said: you should try to remember identities that will serve as shortcuts when proving identities.

like: Sin^2 + Cos^2=1 <= this may be the most important one

Sin^2= 1 - Cos^2

Cos^2= 1 - Sin^2

 

[VietDao29]

like: Sin^2 + Cos^2=1 <= this may be the most important one

Sin^2= 1 - Cos^2

Cos^2= 1 - Sin^2

You're missing an angle here.
It should read sin²x, not just sin² :)

 

[hmm?]

You're missing an angle here.
It should read sin²x, not just sin² :)

You're absolutely right--all those -1 from omitted thetas and xs from test questions still haven't sunk in :/...haha.

 

[HallsofIvy]

Unfortunately, I have had students declare that
\frac{sin x}{sin y}= \frac{x}{y}!

 

[Office_Shredder]

Unfortunately, I have had students declare that
\frac{sin x}{sin y}= \frac{x}{y}!

There aren't any parentheses, so it could just be that s, i, and n are all variables

 

[VietDao29]

Unfortunately, I have had students declare that
\frac{sin x}{sin y}= \frac{x}{y}!

... *no comments* :)

 

[BobG]

Unfortunately, I have had students declare that
\frac{sin x}{sin y}= \frac{x}{y}!

Well, I used to think \frac{sin x}{x} = sin but then I found out it's worse than that. It's not just any sin, it's an X-rated video filmed in someone's kitchen sinc.