# Math Help - easy trig question

1. ## easy trig question

How would i solve questions such as sin(x)= $\frac{1}{2}$ or cos(3x)= $\frac{-\sqrt{3}}{2}$ without using a calculator.

2. Originally Posted by Raj
How would i solve questions such as sin(x)= $\frac{1}{2}$ or cos(3x)= $\frac{-\sqrt{3}}{2}$ without using a calculator.

$\sin x =-\frac{1}{2}$

first of all , the reference angle is 30 degrees . But sin is negative so it would be in the 3rd and 4th quadrant ie

x=210 , 330

for $\cos 3x=-\frac{\sqrt{3}}{2}$

reference angle = 30 degrees and since cos is negative , it will be in the 2nd and 3rd quadrant .

ie 3x=150 , 210 , 510 , 570 , 870 , 930

x=50 , 70 , 170 , 190 , 290 , 310

Assuming that 0<x<360 , then 0<3x<1080

3. Originally Posted by mathaddict
$\sin x =-\frac{1}{2}$

first of all , the reference angle is 30 degrees . But sin is negative so it would be in the 3rd and 4th quadrant ie

x=210 , 330

for $\cos 3x=-\frac{\sqrt{3}}{2}$

reference angle = 30 degrees and since cos is negative , it will be in the 2nd and 3rd quadrant .

ie 3x=150 , 210 , 510 , 570 , 870 , 930

x=50 , 70 , 170 , 190 , 290 , 310

Assuming that 0<x<360 , then 0<3x<1080
thanks but i really have no idea how you got the reference angle

4. Originally Posted by Raj
thanks but i really have no idea how you got the reference angle
lets say for sin x = -1/2

we just ignore the negative sign first , ie sin x =1/2

So by recalling these special angles , x=30 degree , then from here we only the negative sign by deciding which quadrant should it be in .

Another example ,

cos x = - 0.445

same thing , ignore the sign first , then use the calculator to find the reference angle (63.58) , then now since its negative so cos will be in 2nd and 3rd quadrant ..

Clear ?

5. Hello Raj
Originally Posted by Raj
How would i solve questions such as sin(x)= $\frac{1}{2}$ or cos(3x)= $\frac{-\sqrt{3}}{2}$ without using a calculator.

The answer is: using a combination of experience - you simply recognise certain trig ratios - and technique.

For instance, you'll need experience - and the ability to memorise certain facts - to know that $\sin\tfrac{\pi}{6} = \tfrac12$ and that $\cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}$.

Then you need a technique that will enable you to handle negative signs: for instance, how to use the fact that $\cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}$ to enable you to find an angle whose cosine is $-\tfrac{\sqrt3}{2}$. (One such angle is $\tfrac{5\pi}{6}$.)

Then you'll need a technique that will enable you to find other angles with the same sine or cosine. For example, the fact that $\sin\tfrac{\pi}{6} = \tfrac12$ means that the sine of all these angles will also equal one-half: $\tfrac{5\pi}{6},\tfrac{13\pi}{6},\tfrac{17\pi}{6}, ...$

Finally, you'll need a technique that will enable you to deal with mutliple angles. For example, if $\sin3x = \tfrac12$, we've just seen that the possible values of $3x$ are $\tfrac{\pi}{6},\tfrac{5\pi}{6},\tfrac{13\pi}{6},\t frac{17\pi}{6},...$. So we'd divide each of these by $3$ to get the values of $x$: $\tfrac{\pi}{18},\tfrac{5\pi}{18},\tfrac{13\pi}{18} ,\tfrac{17\pi}{18},...$

Practice makes perfect!

For instance, you'll need experience - and the ability to memorise certain facts - to know that $\sin\tfrac{\pi}{6} = \tfrac12$ and that $\cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}$.