Question

Help please

Answer 1

To find out if it is an identity, you have to start with one side and ignore the other side after the equal sign.

So let us start with the left hand side or the LHS.

\(\sf cos(\alpha+\beta )-cos(\alpha - \beta )\)

Do you know what

\(\sf cos(\alpha + \beta ) =?\)

and what

\(\sf cos(\alpha - \beta ) =?\)

Answer 2

If you don't, then say so. Don't feel embarrased or anything. After all we have to learn.

You should memorize them too, your teacher expects you to know them by heart. You can forget it after the course is done (unless you are next taking pre-calc, and calc). (:

Answer 3

I tried looking it up, but I don't know it

Answer 4

Ok, so you should memorize this as you will need it later in the course and to prove other identities and stuff.

\(\sf cos(\alpha + \beta ) =cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)\)

\(\sf cos(\alpha - \beta ) =cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta)\)

Answer 5

Alright, what do I do with those?

Answer 6

Subtract them.

\(\sf cos(\alpha+\beta )-cos(\alpha - \beta )\)

so substitute the values for each equation that I gave you above and subtract. (:

Answer 7

So I would end up with just sin(α)sin(β)?

Answer 8

\(\sf cos(\alpha+\beta )-cos(\alpha - \beta )\)

Substitute the formula in and this is what you should get.

\(\sf cos(\alpha)cos(\beta)-sin(\alpha)sin- [cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta)]\)

Can you simplify it any further, remember there is a minus that you have to distribute into the parenthesis.

Answer 9

The cos(α)cos(β) would cancel out, right?

Answer 10

yes they should

Answer 11

Then it would be (−sin(α)sin) -(sin(α)sin(β)])?

Answer 12

@TheSmartOne

Answer 13

\(\bf cos(\alpha + \beta)-cos(\alpha - \beta)
\\ \quad \\
cos({\color{brown}{ \alpha}} + {\color{blue}{ \beta}})= cos({\color{brown}{ \alpha}})cos({\color{blue}{ \beta}})- sin({\color{brown}{ \alpha}})sin({\color{blue}{ \beta}})
\\ \quad \\

cos({\color{brown}{ \alpha}} - {\color{blue}{ \beta}})= cos({\color{brown}{ \alpha}})cos({\color{blue}{ \beta}}) + sin({\color{brown}{ \alpha}})sin({\color{blue}{ \beta}})\qquad thus
\\ \quad \\

[cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)]-[cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta)]
\\ \quad \\
\cancel{cos(\alpha)cos(\beta)}-sin(\alpha)sin(\beta)\cancel{-cos(\alpha)cos(\beta)}-sin(\alpha)sin(\beta)
\\ \quad \\
-2sin(\alpha)sin(\beta)\)

Answer 14

Is that my answer to the left side? −2sin(α)sin(β)?

Answer 15

sure

Answer 16

notice the two remaining terms are the "same"
-same - same = -2same

Answer 17

So it's an identity

Answer 18

yeap

Answer 19

Thank you! Can you help me with another one?

Answer 20

Sorry was helping other people and didn't get a notification until rn.

Thanks for helping in the mean while jdoe :)