# Thread: Help with solving these Trig Identities...

1. ## Help with solving these Trig Identities...

How do I solve:

Using Sum & Difference Identities

1. tan ( - pi/12)
2. tan (alpha + beta); alpha (-3/5) ends QIII; Beta (-1/4) ends QII
3. cos (-1/5 + -1/4)
4. (sin A - sin B)^2 + (cos A - cos B)^2

1. -2 + sq.root(3)
2.[ 192- 25 sq.root(15) ] / -119
3.[ 1 + 6 sq.root(10) ] / 20
4. -2 cos (A - B)

Thank you very much for your time! =)

2. Originally Posted by coolacu
How do I solve:

Using Sum & Difference Identities

1. tan ( - pi/12)
2. tan (alpha + beta); alpha (-3/5) ends QIII; Beta (-1/4) ends QII
3. cos (-1/5 + -1/4)
4. (sin A - sin B)^2 + (cos A - cos B)^2

1. -2 + sq.root(3)
2.[ 192- 25 sq.root(15) ] / -119
3.[ 1 + 6 sq.root(10) ] / 20
4. -2 cos (A - B)

Thank you very much for your time! =)
Check out the trigonometric sum formulas (google them for a list). These questions seem like simply picking the correct one and then working through the gears. For example here's the first one

$\tan(\frac{- \pi}{12}) = \tan(\frac{\pi}{4} - \frac{\pi}{3})$

Looking at the sum and difference formulas for tan we see that this is equal to

$\frac{\tan(\frac{\pi}{4})-\tan(\frac{\pi}{3})}{1+\tan{\frac{\pi}{4}}\tan{\fr ac{\pi}{3}}}$

$=\frac{1-\sqrt{3}}{1+\sqrt{3}}$

$=\frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})} = \frac{1-2\sqrt{3}+3}{-2}$

$= -2 + \sqrt{3}$

3. Hello, coolacu!

There must be a typo in #3.
Also in #4 . . .

$(4)\;\;(\sin A - \sin B)^2 + (\cos A - \cos B)^2$

Answer: . $-2\cos (A - B)$ . . . . no

$\text{Multiply out: }\;\sin^2\!A - 2\sin A\sin B + \sin^2\!B + \cos^2\!A-2\cos A\cos B + \cos^2\!B$

$\text{Rearrange: }\;\underbrace{\sin^2\!A + \cos^2\!A}_{\text{This is 1}} \;+\; \underbrace{\sin^2\!B + \cos^2\!B}_{\text{This is 1}} \;-\; 2\underbrace{(\cos A\cos B + \sin A\sin B)}_{\text{This is }\cos(A-B)}$

. . $\text{and we have: }\;{\color{blue}2 - 2\cos(A - B)}$

4. Hello coolacu
Originally Posted by coolacu
How do I solve:

Using Sum & Difference Identities

1. tan ( - pi/12)
2. tan (alpha + beta); alpha (-3/5) ends QIII; Beta (-1/4) ends QII
3. cos (-1/5 + -1/4)
4. (sin A - sin B)^2 + (cos A - cos B)^2

1. -2 + sq.root(3)
2.[ 192- 25 sq.root(15) ] / -119
3.[ 1 + 6 sq.root(10) ] / 20
4. -2 cos (A - B)

Thank you very much for your time! =)
I don't understand what questions 2 and 3 mean, and I think the answer to 4 is wrong. Here's what I get for that one:

$(\sin A - \sin B)^2 + (\cos A - \cos B)^2$

$= \sin^2A - 2\sin A\sin B + \sin^2B + \cos^2A - 2\cos A\cos B + \cos^2 B$

$= \sin^2 A + \cos^2A + \sin^2B +\cos^2 B -2(\cos A\cos B + \sin A \sin B)$

$= 2 - 2\cos(A-B)$

Can you explain what you mean by questions 2 and 3?