double angle/half angle formulas

• Dec 4th 2008, 04:05 PM
ipreferhistory
double angle/half angle formulas
I have two problems for homework tonight that I can't get.
they are over double and half angles.
if someone can get any of these, and show how they did it or give a short explanation how, that would be awesome.

1. sin2x[4cos^4(x) - 4cos^2(x)]^2
I started by changing sin2x to 2sinxcosx & squared the parenthetical part
then I multiplied it all out but I don't see any way to simplify.

2. sin(C + D) = 7/8, (C+D) is in quadrant I, and sin C = 2/8, C is in quadrant I
find sin D
this one I'm rather lost. I know the formula for sin(C + D), so I can make it
(sinC)(cosD) + (cosC)(sinD) = 7/8
change the sin C to 2/8... but can't get it all figured out.

• Dec 4th 2008, 04:33 PM
nzmathman
For (Q1),

$\displaystyle \sin(2x) \times (4 \cos^4x - 4 \cos^2x)^2$

$\displaystyle = 2 \sin{x} \cos{x} \times [4 \cos^2x ( \cos^2x - 1)]^2$

We know $\displaystyle 1 - \cos^2x = \sin^2x$:

So $\displaystyle 2 \sin{x} \cos{x} \times [4 \cos^2x ( \cos^2x - 1)]^2$

$\displaystyle = 2 \sin{x} \cos{x} \times (4 \cos^2x \times - \sin^2x)^2$

$\displaystyle = 2 \sin{x} \cos{x} \times (16 \cos^4x \, \sin^4x)$

$\displaystyle = 32 \sin^5x \, \cos^5x$
• Dec 4th 2008, 04:58 PM
nzmathman
For (Q2),

$\displaystyle \sin(C + D) = \frac{7}{8}$

We know $\displaystyle \sin{C} = \frac{2}{8} = \frac{1}{4}$

$\displaystyle C = \arcsin{\left( \frac{1}{4} \right)} = 14.478_{}^{o}$ (Quadrant I)
$\displaystyle \sin{(C + D)} = \sin{(14.478_{}^{o} + D)} = \frac{7}{8}$
$\displaystyle {14.478_{}^{o} + D} = \arcsin{\left(\frac{7}{8} \right)} = 61.045_{}^{o}$ (Quadrant I)

$\displaystyle D = 61.045_{}^{o} - 14.478_{}^{o} = 46.567_{}^{o}$

Now you can find $\displaystyle \sin{D}$ (Wink)
• Dec 4th 2008, 05:19 PM
ipreferhistory
thank you! I did not think to use arcsin for question 2.