# A few questions

• Jan 11th 2010, 03:05 PM
Paymemoney
A few questions
Hi
Here are some of the questions I'm stuck on.
1)Simply the expression: $\displaystyle cos(y)cos(-2y)-sin(y)sin(-2y)$
this is what i have done and i cannot see where i have gone wrong.
$\displaystyle =cos((y)+(-2y))$
$\displaystyle =cos(y)$ 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 $\displaystyle sin(x)cos(2x)+cos(x)sin(2x)$.

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

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

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

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

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

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

book's answer is$\displaystyle 2+\sqrt3$
• Jan 11th 2010, 03:12 PM
Moo
Quote:

Originally Posted by Paymemoney
Hi
Here are some of the questions I'm stuck on.
1)Simply the expression: $\displaystyle cos(y)cos(-2y)-sin(y)sin(-2y)$
this is what i have done and i cannot see where i have gone wrong.
$\displaystyle =cos((y)+(-2y))$
$\displaystyle =cos(y)$ 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 $\displaystyle sin(x)cos(2x)+cos(x)sin(2x)$.
Yes, but you can write cos(2x)=1-2sinČ(x) and sin(2x)=2sin(x)cos(x)
and then use cosČ(x)=1-sinČ(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 $\displaystyle tan(\frac{5\pi}{12})$.

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

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

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

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

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

book's answer is$\displaystyle 2+\sqrt3$
Multiply your answer by $\displaystyle \frac{1-\sqrt{3}}{1-\sqrt{3}}$ (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)=aČ-bČ and (a-b)Č=aČ+bČ-2ab
• Jan 11th 2010, 03:23 PM
Paymemoney
ok thanks
• Jan 11th 2010, 04:00 PM
Paymemoney
So in question 2 does that mean i need to substitute values into?

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

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

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

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

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

$\displaystyle =4-2\sqrt3$
• Jan 11th 2010, 11:27 PM
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)

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