Question

prove that
\[\cos^4θ−\sin^4θ=\cos^2θ−\sin^2θ\]

Answer 1

Think about the following two identities:

\[a^2 - b^2 = (a+b) (a-b)\]
and
\[\cos^2\theta + \sin^2\theta = 1\]

Answer 2

Difference of squares. So, you have:
(cos^2(theta) - sin^2(theta))(cos^2(theta) + sin^2(theta))

Since that latter factor is 1 by trig identity, you got the first factor as the result.

Answer 3

Um Ok, and how would you show that?

Answer 4

How would you show sin^ + cos^ = 1?

Answer 5

It's just a rule, isn;t it? I mean, how would you show the workings? Not sure I completely get this topic yet.

Answer 6

Yes. It is a rule. But, the proof has to do with the right triangle and pythagorean theorem.

Answer 7

Yes, that's true. But this question didn't want me to answer with the triangle shown... so In algebraic terms, w/o the diagram, how would the steps be shown?

Answer 8

It is as I showed:
(cos^2(theta) - sin^2(theta))(cos^2(theta) + sin^2(theta))

It is like using (a^2-b^2) = (a-b)(a+b) except that here a and b are "sin^2" and "cos^2".

Answer 9

What GT said is a good way to go, substitute Sin and Cos with 'a' and 'b' to show the working.

If you want to prove the rule, here's how

Take a right triangle. Let one of the acute angles be A.
Let the base (smaller side of the angle) be 'b'
Let the perpendicular (side opposite to the angle) be 'p' and
Let the hypotenuse (longest side of the triangle) be 'h'

Sin A = p/h
Cos A = b/h

\[\sin^2A = \frac{p^2}{h^2}\]
\[\cos ^2A = \frac{b^2}{h^2}\]

\[\sin^2A + \cos^2A= \frac{p^2}{h^2} + \frac{b^2}{h^2}\]
\[\sin^2A + \cos^2A= \frac{p^2+b^2}{h^2}\]

now p^2 + b^2 = h^2 according to pythagoras theorem

Answer 10

\[Cos^2\theta-\sin^2\theta(\cos^2\theta+\sin^\theta)? \]or \[(Cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^\theta) ?\]

Answer 11

latter

Answer 12

So how (when expanding the bracket) does it simplify to cos^2(theta)-sin^2(theta)?

Answer 13

Never mind. stupid question

Answer 14

Thanks for helping! I understand now

Answer 15

Haha, there are no stupid questions, please feel free to ask.

Answer 16

Good job!

Answer 17

I just realized! I would not have thought of it though... need to think more outside the box!

Answer 18

Sorry again, but again, It wasn't a stupid question... How does it simplify to that answer?

Answer 19

That is good. Discussing stimulates thought. Also, sometimes not doubting your own good capability helps too.

Answer 20

I understand the (a-b)(a+b) rule, but how does that simplify it to the second answer... wouldn't it just come back to the first answer?

Answer 21

So you realize that
\[\cos^4\theta - \sin^2\theta = (\cos^2\theta + \sin^2\theta)(\cos^2\theta-\sin^2\theta)\]
But since
\[\cos^2\theta + \sin^2\theta = 1\]
we get
\[\cos^4\theta - \sin^2\theta = (1)(\cos^2\theta-\sin^2\theta)\]

Answer 22

er typo... i meant sin^4 theta in the left side

Answer 23

Not completely no.
I realize the first and second identity, but I'm not sure how it links up!? to
the last equation?

Answer 24

You take the first equation, and put the value of cos^2 + sin^2 from the second equation in it.

Answer 25

Ohhhhh!

Answer 26

Thanks. Registration has finally hit!

Answer 27

That's great! I'm glad I was able to help