Well these are a teaser too..Trigonometry

Sinsane · #1 $Old$ December 4th, 2008, 06:46 AM

Solve the equation for solutions in the interval [0, 360). Round to the nearest degree.

sin2theta = -1/2 This I dont think I have it right..but I want to make sure I have it right on the test. Better safe than sorry losing 10 points

Oh and last one...if anyone can do this...

sqrt3 sec2theta = 2

I couldnt find that one at all.. I divided the sqrt 3, but after that..I was just confused.

Thanks!

running-gag · #2 $Old$ December 4th, 2008, 09:48 AM

Look at the trigonometric circle

$sin(2\theta) = -\frac{1}{2}$

gives $2\theta = -30 + 360k$ or $2\theta = -150 + 360k$ with k integer

Therefore $\theta = -15 + 180k$ or $\theta = -75 + 180k$

Solving the equation for solutions in the interval [0, 360[

$S = {105, 165, 285, 345}$

$\frac{\sqrt{3}}{cos(2\theta)} = 2$

$cos(2\theta) = \frac{\sqrt{3}}{2}$

$2\theta = -30 + 360k$ or $2\theta = 30 + 360k$

$\theta = -15 + 180k$ or $\theta = 15 + 180k$

Solving the equation for solutions in the interval [0, 360[

$S = {15, 165, 195, 345}$

Soroban · #3 $Old$ December 4th, 2008, 10:02 AM

Hello, Sinsane!

You're expected to know the trig values for certain angles.

. . For example: . $\sin30^o \:=\:\tfrac{1}{2}$ . . . and its variations.

If you dont, sayonara . . .

Quote:

Solve the equation for solutions in the interval [0°, 360°).

$\sin2\theta \:=\: \text{-}\tfrac{1}{2}$

This gives us: . $2\theta \:=\:210^o,\:330^o,\:570^o,\:690^o,\:\hdots$

Therefore: . $\theta \;=\;105^o,\:165^o,\:285^o,\:345^o$

Quote:

$\sqrt{3} \sec2\theta \:=\: 2$

So we have: . $\sec2\theta\;=\;\frac{2}{\sqrt{3}}$

Hence: . $2\theta \;=\;30^o,\:150^o,\:389^o,\:510^o,\: \hdots$

Therefore: . $\theta \;=\;15^o,\:75^o,\:195^o,\:255^o$