Very quick question please (1 Viewer)

[Existential]

Find the exact value of tan5pi/6.

Forgive me for asking such an easy question. I must be having a memory lapse as I've looked in the textbook and I just can't remember how to do these types of questions. Try not to bag me out too much. XD

I just need an explanation, as I know the answer is -1/sqrt3. I know where 1/sqrt3 comes from (the triangle), but how do you get from tan 5pi/6 to -tanpi/6? (What do you do with ASTC etc.?)

 

[Mature Lamb]

Note that 5pi/6 is in the 2nd quadrant (since 5pi/6 in radians is equivalent to 150 degrees). 150 degrees is the same as 180-150, or 30 degrees in the first quadrant. Using ASTC, you can see tan is negative in the 2nd quadrant. Thus the exact value of tan(5pi/6) = -tan(30) = -1/sqrt(3)

 

[SpiralFlex]

Let's convert it into degrees to make it easier for you.

So we now need to find the exact value of . Clearly is in the SECOND QUADRANT and tangent in that quadrant is a NEGATIVE. Also the second quadrant is

So, or another way of saying this is how far is it from the horizontal axis? .

So that's how we get or .

Which equals to your answer of .

 

[Existential]

ahhh thanks. also if i wanted to keep it in radians i could just start with pi - 5pi/6 = pi/6 than go from there.

another question please (last one i think)

Find the exact value of cosec 2pi/3. I'm confused with this one at the part when deciding whether its ive or +ve. Is cosec +ve in the second quadrant (sin)?

thanks.

 

[mitchy_boy]

Graph it, and it's below the x axis.

 

[mitchy_boy]

Learn all square roots of numbers in decimal form. Saves so much time. Just sit down and do, like sqrt(2), sqrt(3), 1/sqrt(2), etc.

 

[SpiralFlex]

ahhh thanks. also if i wanted to keep it in radians i could just start with pi - 5pi/6 = pi/6 than go from there.

another question please (last one i think)

Find the exact value of cosec 2pi/3. I'm confused with this one at the part when deciding whether its ive or +ve. Is cosec +ve in the second quadrant (sin)?
thanks.

Let's consider it's reciprocal, has reciprocity to


So we can note that,

Let's find

To make it easier, covert it into degree form.

So we need to find,

As you can obviously see it is in the SECOND QUADRANT, so sine in the SECOND QUADRANT is POSITIVE. It is from the horizontal axis. So

Remember what we said, , so therefore

Rationalise at will,

 

[Existential]

Thanks for that, it was a good refreshment.
I know i promised that was the last one, but here is one more. Sorry, but do it if you want (i like the full working and proper script you use). Also, anyone else is welcome too lol.
Okay I know its simple but again I just need an explanation -

"A point of intersection of the curves y=asinx and y=cosx occurs when x=pi/6. Evaluate a.

 

[AAEldar]

Thanks for that, it was a good refreshment.
I know i promised that was the last one, but here is one more. Sorry, but do it if you want (i like the full working and proper script you use). Also, anyone else is welcome too lol.
Okay I know its simple but again I just need an explanation -

"A point of intersection of the curves y=asinx and y=cosx occurs when x=pi/6. Evaluate a.