Trig Identity Proof: Show that 4(sin^4x + cos^4x) is Equivalent to cos(4x) + 3

[phospho]

Show that:

$4(\sin^4x+\cos^4x) \equiv \cos4x + 3$.

Really stuck with this, no idea how to go ahead with it. The book gives a hint: $\sin ^4 x = (\sin ^2 x)^2$ and use $\cos 2x = 1 - 2\sin ^2 x$

But I don't even understand the hint, where did they get
$\cos 2x = 1 - 2\sin ^2 x$ from?

 

[Mark44]

Show that:

$4(\sin^4x+\cos^4x) \equiv \cos4x + 3$.

Really stuck with this, no idea how to go ahead with it. The book gives a hint: $\sin ^4 x = (\sin ^2 x)^2$ and use $\cos 2x = 1 - 2\sin ^2 x$

But I don't even understand the hint, where did they get
$\cos 2x = 1 - 2\sin ^2 x$ from?

This is one of three double-angle identities for cos(2x).

The other two are
cos(2x) = cos²(x) - sin²(x)
cos(2x) = 2cos²(x) - 1

The 2nd one above and the one you're asking about can be obtained from the first one I showed by using the identity sin²(x) + cos²(x) = 1.

 

[CAF123]

But I don't even understand the hint, where did they get
cos2x=1−2sin2x from?

$$cos (x + x) = cos^2x - sin^2x = (1 - sin^2x) -sin^2x = 1 - 2sin^2x,$$ using the identity $sin^2x +cos^2x =1$

 

[CAF123]

To start you off, write the given equation as, $$4((\sin^2x)^2 + (\cos^2x)^2),$$ and use the double angle identities to get this solely in terms of $\cos.$

 

[Saitama]

You can start by using this:
a²+b²=(a+b)²-2ab
This would mean that:
$$4(\sin^4x+\cos^4x)=4((sin^2x+cos^2x)^2-2sin^2xcos^2x)$$
Now rest should be easy to solve.

 

[phospho]

To start you off, write the given equation as, $$4((\sin^2x)^2 + (\cos^2x)^2),$$ and use the double angle identities to get this solely in terms of $\cos.$

working :

$4((sin^2x)^2 + (cos^4x)$
$4((1-cos^2x)^2 + cos^4x))$
$4(1-2cos^2x + cos^4x + cos^4x)$
$4(1-2cos^2x + 2cos^4x)$

Pretty much stuck here.

You can start by using this:
a²+b²=(a+b)²-2ab
This would mean that:
$$4(\sin^4x+\cos^4x)=4((sin^2x+cos^2x)^2-2sin^2xcos^2x)$$
Now rest should be easy to solve.

continuing from your working: $4(1 - 2sin^2xcos^2x)$, but I don't really know any identities to go from here...

 

[Bacle2]

How about doing the angle sum on the RHS: Cos4x= Cos(2x+2x) , and expand?

 

[Clever-Name]

How about doing the angle sum on the RHS: Cos4x= Cos(2x+2x) , and expand?

He doesn't have $cos(4x)$, he has $cos^{4}(x)$, or $(cos(x))^{4}$

 

[ehild]

working :

$4((sin^2x)^2 + (cos^2x)^2)$

Use the following double-angle identities:
sin²(x)=0.5(1-cos(2x))
cos²(x)=0.5(1+cos(2x))

ehild

 

[Mark44]

How about doing the angle sum on the RHS: Cos4x= Cos(2x+2x) , and expand?

He doesn't have $cos(4x)$, he has $cos^{4}(x)$, or $(cos(x))^{4}$

Bacle2 explicitly was referring to the right side of the original identity, which has a cos(4x) term in it.

See below.

$4(\sin^4x+\cos^4x) \equiv \cos4x + 3$.

 

[Bacle2]

He doesn't have $cos(4x)$, he has $cos^{4}(x)$, or $(cos(x))^{4}$

Look at the RH side of the original post, it has a Cos(4x)+3.


Edit: Sorry, I did not see Mark44's post, and now I cannot see the 'delete' option.

 

[Saitama]

continuing from your working: $4(1 - 2sin^2xcos^2x)$, but I don't really know any identities to go from here...

Try using these identities:
i)sin(2x)=2sin(x)cos(x)
ii)cos(2x)=1-2sin²x

If you square both the sides of i identity, you may see a way through.

 

[CAF123]

Use what ehild suggested if attempting the way I started (which is just one way to tackle the problem - there are many others).
We have, as suggested in my last post,
$4((\sin^2x)^2 + (\cos^2x)^2)$

Now, use the double angle identities that ehild suggested, specifically $\sin^2x = -\frac{1}{2}(\cos2x - 1)$ and $\cos^2x = \frac{1}{2}(cos2x +1)$

Substitute these into the eqn, expand the brackets and simplify...