1. ## nid Help please... sum and difference

Use the cosine , sine and tangent sum and difference identities to simplify each of the following:
1.) cos (A + B) + cos (A - B)
2.) cos (A + B) - cos (A - B)
3.) sin (A + B) + sin (A - B)
4.) sin (A + B) - sin (A - B)
5.) tan20degrees + tan32degrees / (1- tan 20degrees X tan 32 degrees)
6.) tan 35 degrees - tan 12 degrees / ( 1 + tan 35 degrees X tan 12 degrees)

2. Do you know the identities? (I would like to learn them)

3. here it is:
cos ( A + B ) = cos A cos B + sin A sin B
cos ( A + B ) = cos A cos B - sin A sin B
sin ( A + B ) = sin A cos B + cos A sin B
sin ( A - B) = sin A cos B - cos A sin B
tan ( A - B ) = tan A - tan B / ( 1 + tan A tan B)
tan ( A + B) = tan A + tan B / (1 - tan A tan B)

im not sure w/ my answers... but it says simplify

my answer in number 1 is 2cosAcosB am i correct?

my answer in number 1 is 2cosAcosB am i correct?
assuming the conversions in your post are correct, yes.

2)2sinAsinB

3)2sinAcosB

4)2cosAsinB

tan ( A - B ) = tan A - tan B / ( 1 + tan A tan B)
tan ( A + B) = tan A + tan B / (1 - tan A tan B)
do you mean: $\tan(A-B)=\tan A-\frac{\tan B}{1+\tan A\tan B}$

or: $\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

5. ^
^
the 2nd one of course....
btw how old are u?

btw how old are u?
I'm 14 (15 in september)

5)tan(20+32)=tan50

6)tan(35-12)=tan23

Do you need to see the steps?
I'll do a few of them . . .

Use the cosine, sine and tangent sum/difference identities to simplify:

$1)\;\cos(A + B) + \cos(A - B)$

We have: . . $\cos(A + B) \qquad\quad\;\;+ \qquad\quad\cos(A - B)$

Then: . $\overbrace{(\cos A\cos B - \sin A\sin B)} + \overbrace{(\cos A\cos B + \sin A\sin B)}$

. . . . . . . . . . . . . . . . $= \;\;\;2\cos A\cos B$

$3)\;\sin(A + B) +\sin (A - B)$

We have: . . $\sin(A + B) \qquad\quad\;\;+ \qquad\quad\sin(A - B)$

Then: . $\overbrace{(\sin A\cos B + \sin B\cos A)} + \overbrace{(\sin A\cos B - \sin B\cos A)}$

. . . . . . . . . . . . . . . . $= \;\;\;2\sin A\cos B$

$5)\;\frac{\tan20^o + \tan32^o}{1- \tan 20^o\cdot\tan32^o}$

We're exprected to recognize the form: . $\frac{\tan A + \tan B}{1 - \tan A\!\cdot\tan B}\:=\:\tan(A + B)$

Therefore: . $\frac{\tan20^o + \tan32^o}{1 - \tan20^o\!\cdot\tan32^o} \:=\:\tan(20^o + 32^o) \:=\:\tan52^o$