# A few questions

• January 11th 2010, 04:05 PM
Paymemoney
A few questions
Hi
Here are some of the questions I'm stuck on.
1)Simply the expression: $cos(y)cos(-2y)-sin(y)sin(-2y)$
this is what i have done and i cannot see where i have gone wrong.
$=cos((y)+(-2y))$
$=cos(y)$ answers says its cos(-y).why?

2)express sin(3x) in terms of sin(x). This questions follows on from the previous question which was asking to expand sin(x+2x) answer to this is $sin(x)cos(2x)+cos(x)sin(2x)$.

3)Use the compound angle formulas and appropriate angles to find the exact value of $tan(\frac{5\pi}{12})$.

This is what i have done:
$tan(\frac{5\pi}{12}=tan(\pi-\frac{7\pi}{12})$

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

$=\frac{tan(\pi)-tan(\frac{\pi}{3})-tan(\frac{\pi}{4})-tan(\pi)tan(\frac{\pi}{3})tan(\frac{\pi}{4})}{1+ta n(\pi)tan(\frac{\pi}{3})+tan(\pi)tan(\frac{\pi}{4} )+tan(\frac{\pi}{3})tan(\frac{\pi}{4})}$

$=\frac{0-\sqrt3-1-0*\sqrt3*1}{1+0*\sqrt3+0*1+\sqrt3*1}$

$=\frac{\sqrt3-1}{1+\sqrt3}$

book's answer is $2+\sqrt3$
• January 11th 2010, 04:12 PM
Moo
Quote:

Originally Posted by Paymemoney
Hi
Here are some of the questions I'm stuck on.
1)Simply the expression: $cos(y)cos(-2y)-sin(y)sin(-2y)$
this is what i have done and i cannot see where i have gone wrong.
$=cos((y)+(-2y))$
$=cos(y)$ answers says its cos(-y).why?

You're correct, and the answer's correct too, because cos(-x)=cos(x)

Quote:

2)express sin(3x) in terms of sin(x). This questions follows on from the previous question which was asking to expand sin(x+2x) answer to this is $sin(x)cos(2x)+cos(x)sin(2x)$.
Yes, but you can write cos(2x)=1-2sin²(x) and sin(2x)=2sin(x)cos(x)
and then use cos²(x)=1-sin²(x) to simplify and get something with respect to sin(x) only.

Quote:

3)Use the compound angle formulas and appropriate angles to find the exact value of $tan(\frac{5\pi}{12})$.

This is what i have done:
$tan(\frac{5\pi}{12}=tan(\pi-\frac{7\pi}{12})$

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

$=\frac{tan(\pi)-tan(\frac{\pi}{3})-tan(\frac{\pi}{4})-tan(\pi)tan(\frac{\pi}{3})tan(\frac{\pi}{4})}{1+ta n(\pi)tan(\frac{\pi}{3})+tan(\pi)tan(\frac{\pi}{4} )+tan(\frac{\pi}{3})tan(\frac{\pi}{4})}$

$=\frac{0-\sqrt3-1-0*\sqrt3*1}{1+0*\sqrt3+0*1+\sqrt3*1}$

$=\frac{\sqrt3-1}{1+\sqrt3}$

book's answer is $2+\sqrt3$
Multiply your answer by $\frac{1-\sqrt{3}}{1-\sqrt{3}}$ (multiplying the numerator and the denominator by the conjugate of the denominator - it's similar to 'rationalizing the denominator')
and use the identities : (a-b)(a+b)=a²-b² and (a-b)²=a²+b²-2ab
• January 11th 2010, 04:23 PM
Paymemoney
ok thanks
• January 11th 2010, 05:00 PM
Paymemoney
So in question 2 does that mean i need to substitute values into?

$cos^2(x)=1-sin^2(x)$

This is what i done for question 3 but i still don't get the answer, what have a done wrong.

$=\frac{(\sqrt3-1)(1+\sqrt3)}{(1+\sqrt3)(1+\sqrt3)}$

$=\sqrt3^2-1^2$

$=\sqrt3^2+1^2-2\sqrt3$

$=4-2\sqrt3$
• January 12th 2010, 12:27 AM
Moo
For question 2, no. Follow the steps and use this identity to get the final form.

For the other one... you're not multiplying by what I said (Tongueout)

$\frac{\sqrt3-1}{1+\sqrt3}=\frac{\sqrt3-1}{1+\sqrt3}\cdot\frac{1-\sqrt3}{1-\sqrt3}=\frac{-(1-\sqrt3)^2}{(1+\sqrt3)(1-\sqrt3)}=\frac{-\dots}{1-3}=\dots$
• January 12th 2010, 01:16 AM
Paymemoney
if you don;t mind could you do question 2 for me coz i still can't get the answer -_-.
• January 12th 2010, 01:21 AM
differentiate
$sin(3x) = sin (2x +x)$
$= sin2xcosx + cos2xsinx$
$= cosx(2sinxcosx) + sinx(cos^2x - sin^2x)$
$= 2sinx cos^2x + sinx(1- 2sin^2x)$
$= 2sinxcos^2x + sinx - 2sin^3x$
$= 2sinx(1-sin^2x) + sinx - 2sin^3x$
$= 2sinx - 2sin^3x + sinx - 2sin^3x$
$= 3sinx - 4sin^3x$