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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: dinosaur93 on March 15, 2012, 08:50:01 pm

Title: Circular Function Question Thread
Post by: dinosaur93 on March 15, 2012, 08:50:01 pm: COMPOUND AND DOUBLE ANGLES

1. $\sin \frac{\pi}{12}$

$\sin 2u = 2\cos u \times \sin u$

$\sin 2 (\frac {\pi}{6}) = 2\sin (\frac {\pi}{6}) \times \cos (\frac {\pi}{6})$

$2 \times \frac {\sqrt3}{2} \times \frac {1}{2}$

$= \frac {\sqrt3}{2}$

Isn't this right? If not, which but did I go wrong? :(

The answer says $\frac {\sqrt 2 (\sqrt 3 + 1)}{4}$

2. How do you simplify the following? WITHOUT THE CALCULATOR AND USING THE COMPUND AND DOUBLE ANGLE FORMULA

a.) $\sin ( \theta - \frac {\pi}{6})$

b.) $\cos ( \theta - \frac {\pi}{3})$

c.) $\tan ( 2\theta - \frac {\pi}{4})$

d.) $\cos ( 2\theta - \frac {\pi}{2})$

e.) $\sin ( 2\theta + \frac {\pi}{4})$

f.) $\tan ( -\theta + \frac {3\pi}{4})$

SYMMETRY AND EXACT VALUES

Also, Could anyone explain the properties of the complementary angles, I don't seem to get the concept, thank you!

2. If $\sin (x)$ - 0.3, $\cos (a)$ = 0.5, $\tan (b)$ = 2.4 and $x, a,$ and $b$ are in the first quadrant. Find the value of the following:

a.) $\cos (\pi - x)$

b.) $\sin (\pi - a)$

c.) $\cos (2\pi - a)$

d.) $\tan (\pi - b)$

e.) $\cos (2\pi + x)$

f.) $\sin (2\pi - a)$

3.) If $\sin (\theta) = 0.3, \cos (x) = 0.7,$ and $\tan (\alpha) = 0.4$ , find the value of the following:

a.) $\sin (\frac {\pi}{3} - x)$

b.) $\cos (\frac {3\pi}{2} + \theta)$

c.) $\cos (\frac {\pi}{2} + \theta)$

d.) $\sin (\frac {3\pi}{2} - x)$

e.) $\sin (\frac {\pi}{2} + x)$

f.) $\cos (\frac {3\pi}{2} - \theta)$

g.) $\cos (\frac {3\pi}{2} + \theta)$

h.) $\sin (\frac {3\pi}{2} - x)$
Title: Re: Circular Function Question Thread
Post by: Stick on March 15, 2012, 09:07:10 pm: Without doing your homework for you, I strongly suggest you draw out the unit circle every time you do these problems and it's also helpful to draw out a CAST diagram (please give me a shout if you don't know what this is). As you start to learn more on exact values, there's two helpful triangles which can be drawn as well. If you don't learn them, feel free to contact me and I'll draw them for you. :)
Title: Re: Circular Function Question Thread
Post by: Greatness on March 15, 2012, 09:33:22 pm: For the first question i think you should use the compound angle formula :)
Title: Re: Circular Function Question Thread
Post by: Planck's constant on March 15, 2012, 09:33:54 pm: Quote from: elvin.lam1 on March 15, 2012, 08:50:01 pm
COMPOUND AND DOUBLE ANGLES

1. $\sin \frac{\pi}{12}$

$\sin 2u = 2\cos u \times \sin u$

$\sin 2 (\frac {\pi}{6}) = 2\sin (\frac {\pi}{6}) \times \cos (\frac {\pi}{6})$

$2 \times \frac {\sqrt3}{2} \times \frac {1}{2}$

$= \frac {\sqrt3}{2}$

Isn't this right? If not, which but did I go wrong? :(

The answer says $\frac {\sqrt 2 (\sqrt 3 + 1)}{4}$

mate you did nothing wrong. You gave a brilliant proof of :

$\sin \frac{\pi}{3}=\frac {\sqrt3}{2}$

The only problem is that the question asked for $\sin \frac{\pi}{12}$ :)
Title: Re: Circular Function Question Thread
Post by: Planck's constant on March 15, 2012, 09:50:23 pm: Quote from: elvin.lam1 on March 15, 2012, 08:50:01 pm

Also, Could anyone explain the properties of the complementary angles, I don't seem to get the concept, thank you!

When 2 angles add up to 90 degrees,

a + b = 90

Then,

sina = cosb
tana = cotb
seca = cosecb

Easiest way to visualise it, is through a right-angled triangle
The two pointy angles (a,b) add up to 90

Which two sides of the triangle have ratio sina ?
The same as those which have ratio cosb