1. ## trigonometrical qn

Given that cot q=-t, t>0 and 0° q180° , find, in terms of t,

(a) cosec q
(b) tan(-q)
(c) sin(90°-q )

for(b) is it -(-1/t) = -(1/-t) ?? im kinda confused....

also can anyone tell me how to derive the following formulas from the four quadrants mtd....(as i dun want to memorise the formulas)

1) cos(-x)=cos x
2) tan(-x) = -tan x
3)sin(90°-x)=cos x

2. Originally Posted by maybeline9216
Given that cot q=-t, t>0° ≤q≤180° , find, in terms of t,
(a) cosec q
(b) tan(-q)
(c) sin(90°-q)

for(b) is it -(-1/t) = -(1/-t) ?? im kinda confused....
for (b), -(-1/t)=1/t.

For (a), remember the formula $\cot^2(x)+1=\text{cosec}^2(x)$

For (c), note that $\sin(90^\circ-x)=\cos(x).$ Then, $\cot(x)=\frac{\cos(x)}{\sin(x)}=\cos(x) \cdot \text{cosec}(x) \implies \cos(x)=\frac{\cot(x)}{\text{cosec}(x)}$

also can anyone tell me how to derive the following formulas from the four quadrants mtd....(as i dun want to memorise the formulas)
These formula are correct in whatever quadrant x is.

1) cos(-x)=cos x
2) tan(-x) = -tan x
3)sin(90°-x)=cos x
Draw a circle. Take a point x of the circle in quadrant 1 (it's in general easier). Remember that its abscissa is cos(x) and its ordinate is sin(x).
After that, find on the circle where -x is, that is to say the point forming an opposite angle comparing to x.
Find where 90-x is, etc... It's the way to remember that (yes, because you'll have to remember these formula, no matter what you want )

3. Opps...sorry i forgot to mention how to derive the following formulas from the four quadrant

For any angle x(whether acute or obtuse,i just dun understand the obtuse part)

For sin(-x) = -sin x (i understand because for the 3rd and 4th quadrant,sine is negative)

But i dun understand why
1) cos(-x)=cos x
2) tan(-x) = -tan x