Question

Proof:

Answer 1

\[\frac{ \cos³ \alpha +\sin³ \alpha }{ \cos \alpha + \sin \alpha } = 1 - \frac{ 1 }{ 2 } \sin2 \alpha\]

Answer 2

I was thinking of starting off with the formula of 1/2sin2a = sin a * cos a for the RHS

Answer 3

but not sure how to get rid of the ³'s in LHS

Answer 4

\[\frac{ cosx+sinx(\cos^2x-sinxcosx+\sin^x) }{ cosx+sinx }\]
\[\cos^2-sinxcosx+\sin^2x\]
We know that sin^2x+cos^x = 1
Hence\[1-cosxsinx\]
Therefore,
\[1-\frac{ 1 }{ 2 }sinxcosx\]

Answer 5

yu gave complete solution @Yttrium ...not the way of approach only

Answer 6

No problem, I'm not just copying it. Been working on exercises like these for 2 hours now, got a test this week and I just read it through to understand it.

Answer 7

I believe @Mieeeel can understand it.

Answer 8

I mean, easily he can understand. Because he is in there, so far.

Answer 9

Yea I understand it, sometimes I just lack insight in exercises like these, hence I try making them

Answer 10

So how's it? Just be honest if you don't understand something.

Answer 11

I just didn't know how to get started, I was looking for ways to scratch the ³ by the division, which is not allowed because there are no multiplications in the fracture.

Answer 12

Just review your algebra. :))

Answer 13

ohh..i just thought it is square term...just noticed that its cubic ...then yu can use the A^3+B^3 identity...applying that and solving progessively ,yu will get RHS

Answer 14

Wait, I don't get the last step.

You get 1 - sinx cos x from the baseformula, I get that

But then you go on to 1 - 1/2 (sinx cos x). Where does the 1/2 come from?

Answer 15

Ohh. It's an error. It should be 1-1/2 (sin2x)
It is becase sin2x = 2sinxcosx
What we have is just sinxcosx, hence it is 1/2 (sin2x)