Question

Which of the following are identities?

Answer 1

\[A) \sin(x - \pi) = -sinx\]
\[B) \cos(x+y) + \cos(x-y) = 2cosxcosy\]
\[C) \cos(x+y) + \cos(x-y) = \cos^2x - \sin^2y\]
\[D) \sin(x+y) - \sin(x-y) = 2cosxsiny\]

Answer 2

Another one for the team! @zepdrix @agent0smith

Answer 3

I believe:
A) not an identity
B) yes!
C) no?
D) yes!

Answer 4

A) \[\sin(x-\pi) = \sin(x)\cos(\pi) - \cos(x)sen(\pi)\]

\[\cos(\pi) =-1\]
\[\sin(\pi) = 0\]

so, we have

\[\sin(x-\pi)=-\sin(x)\]

it's correct

Answer 5

lol

Answer 6

Thanks!!

Answer 7

So A is an identity too? Was I right on the others?

Answer 8

B) cos(x+y)+cos(x−y)=2cosxcosy

let's solve the first term from the left member
\[\cos(x+y)=\cos(x)\cos(y) - \sin(x)\sin(y)\]

the second one

\[\cos(x-y)= \cos(x)\cos(y) + \sin(x)\sin(y)\]

So, we have this

\[\cos(x)\cos(y) - \sin(x)\sin(y) + \cos(x)\cos(y) + \sin(x)\sin(y) = 2\cos(x)\cos(y)\]

simplifying, we have

\[2\cos(x)\cos(y)= 2\cos(x)\cos(y)\]

So, it's true :)

Answer 9

Right :) Is C the only one that's false?

Answer 10

it's obvious the C excercise is not an identidy ;)

Answer 11

let me check D excercise

Answer 12

how is it obvious? :P
That was the one I wasn't sure about...

Answer 13

For C, plug in x=y=0.

Answer 14

You mean plug in 0 for x and y?
1 + 1 = 1
I see...

Answer 15

D) sin(x+y)−sin(x−y)=2cosxsiny

Solving the first term from left member we have:

\[\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)\]

The second one:

\[\sin(x-y)= \sin(x)\cos(y) - \sin(y)\cos(x)\]

replacing we have

\[\sin(x)\cos(y) + \cos(x)\sin(y) - [\sin(x)\cos(y) - \cos(x)\sin(y)] = 2\cos(x)\sin(y)\]

be careful to the multiplication of the signs, we have
\[\sin(x)\cos(y) + \cos(x)\sin(y) - \sin(x)\cos(y) + \cos(x)\sin(y) = 2\cos(x)\sin(y)\]

Simplifying, we have_

\[2\cos(x)\sin)(y) = 2 \cos(x)\sin(y) \]

So, it's true :)

Answer 16

Thank you so much! You've been very helpful.

Answer 17

C) \[\cos(x+y) + \cos(x-y)= \cos^2(x) - \sin^2 (y)\]

Replacing we have

\[\cos(x)\cos(y) - \sin(x)\sin(y) + \cos(x)\cos(y) + \sin(x)\sin(y) = \cos^2(x) - \sin^2(y)\]

Simplifying we have

\[2\cos(x)\cos(y) = \cos^2(x) - \sin^2(y)\]

We sum and rest on the first member cos^2 (x) to not alterate , we have

\[2\cos(x)\cos(y) + \cos^2(x) - \cos^2(x) = \cos^2(x) - \sin^2(y)\]

We associate 2co(x)cos(y) - cos^2(y), like this:

\[\cos^2(x) + [2\cos(x)\cos(y) - \cos^2(x)] = \cos^2(x) - \sin^2(y)\]

Let's take out a common factor cos(x)

\[\cos^2(x) + \cos(x)[2\cos(y) - \cos(x)] = \cos^2(x) - \sin^2(y)\]

Conclusion

It's obvious than

\[\cos(x)[2\cos(y) - \cos(x)] = - \sin^2(y)\]

could you prove it?

Answer 18

@crisulcampo what is your goal with that last one? To prove algebraically that it isn't an identity...?

Answer 19

yes, it is hahahaha. I know that I could just replace x and y for any values, however I just wanted to demonstrate @abbles that is not equal by doing it algebraically... that's all