How would i solve questions such as sin(x)=$\displaystyle \frac{1}{2}$ or cos(3x)=$\displaystyle \frac{-\sqrt{3}}{2}$ without using a calculator.
Thanks in advance
$\displaystyle \sin x =-\frac{1}{2}$
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 $\displaystyle \cos 3x=-\frac{\sqrt{3}}{2}$
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
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 ?
Hello RajThe 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 $\displaystyle \sin\tfrac{\pi}{6} = \tfrac12$ and that $\displaystyle \cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}$.
Then you need a technique that will enable you to handle negative signs: for instance, how to use the fact that $\displaystyle \cos\tfrac{\pi}{6} = \tfrac{\sqrt3}{2}$ to enable you to find an angle whose cosine is $\displaystyle -\tfrac{\sqrt3}{2}$. (One such angle is $\displaystyle \tfrac{5\pi}{6}$.)
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 $\displaystyle \sin\tfrac{\pi}{6} = \tfrac12$ means that the sine of all these angles will also equal one-half: $\displaystyle \tfrac{5\pi}{6},\tfrac{13\pi}{6},\tfrac{17\pi}{6}, ...$
Finally, you'll need a technique that will enable you to deal with mutliple angles. For example, if $\displaystyle \sin3x = \tfrac12$, we've just seen that the possible values of $\displaystyle 3x$ are $\displaystyle \tfrac{\pi}{6},\tfrac{5\pi}{6},\tfrac{13\pi}{6},\t frac{17\pi}{6},...$. So we'd divide each of these by $\displaystyle 3$ to get the values of $\displaystyle x$: $\displaystyle \tfrac{\pi}{18},\tfrac{5\pi}{18},\tfrac{13\pi}{18} ,\tfrac{17\pi}{18},...$
Practice makes perfect!
Grandad