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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Think about the following two identities: \[a^2 - b^2 = (a+b) (a-b)\] and \[\cos^2\theta + \sin^2\theta = 1\]
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.
Um Ok, and how would you show that?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

How would you show sin^ + cos^ = 1?
It's just a rule, isn;t it? I mean, how would you show the workings? Not sure I completely get this topic yet.
Yes. It is a rule. But, the proof has to do with the right triangle and pythagorean theorem.
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?
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".
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
\[Cos^2\theta-\sin^2\theta(\cos^2\theta+\sin^\theta)? \]or \[(Cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^\theta) ?\]
latter
So how (when expanding the bracket) does it simplify to cos^2(theta)-sin^2(theta)?
Never mind. stupid question
Thanks for helping! I understand now
Haha, there are no stupid questions, please feel free to ask.
Good job!
I just realized! I would not have thought of it though... need to think more outside the box!
Sorry again, but again, It wasn't a stupid question... How does it simplify to that answer?
That is good. Discussing stimulates thought. Also, sometimes not doubting your own good capability helps too.
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?
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 \[\cos^4\theta - \sin^2\theta = (1)(\cos^2\theta-\sin^2\theta)\]
er typo... i meant sin^4 theta in the left side
Not completely no. I realize the first and second identity, but I'm not sure how it links up!? to the last equation?
You take the first equation, and put the value of cos^2 + sin^2 from the second equation in it.
Ohhhhh!
Thanks. Registration has finally hit!
That's great! I'm glad I was able to help

Not the answer you are looking for?

Search for more explanations.

Ask your own question