Express in functions of theta:

Question

a. sin ( 810- theta)
b. cot ( theta - 360)
c cos (-180- theta)

Answer 1

sin ( 810- theta)
I am sure you meant: sin(180 - θ)
which is sinθ

cot(θ - 360) ....

recall that tan(-x) = -tanx
and tan(360-x) = -tanx
and of course the cot behaves just like the tan

cot(θ-360)
= -cot(360-θ)
= -(-cotθ)
= cotθ

cos(-180 - θ)
= cos(-(180 + θ))
= cos(180+θ) , since cos(-x) = cosx
= -cosθ , (180+θ) is in the 3rd quad

The above can be easily seen by the CAST rule.
You should also verify each answer you get for these type of questions by picking some angle and using your calculator

Answer 2

a.

810 ° = 720 ° + 90 ° = 2 * 360 ° + 90 °

So :

sin ( 810 ° - theta ) = sin ( 90 ° - theta )

sin ( A - B ) = sin A * cos B - cos A * sin B

In this case A = 90 ° , B = theta.

sin 90 ° = 1

cos 90 ° = 0

sin ( 90 ° - theta ) = sin 90 ° * cos theta - cos 90 ° * sin theta = 1 * cos theta - 0 * sin theta = cos theta

So :

sin ( 810 ° - theta ) = cos theta

2.

cot ( theta - 360 ° ) = cos ( theta - 360 ° ) / sin ( theta - 360 ° )

sin ( theta - 360 ° ) = sin theta

cos ( theta - 360 ° ) = cos theta

So :

cot ( theta - 360 ° ) = cos theta / sin theta = cot theta

cot ( theta - 360 ° ) = cot theta

3.

cos ( - 180 ° - theta )

All trigonometric functions of - 180 ° is same like trigonometric functions of 180 °

sin - 180 ° = sin 180 °

cos - 180 ° = cos 180 °

sin 180 ° = 0

cos 180 ° = - 1

cos ( A - B ) = sin A * sin B + cos A * cos B

In this case A = - 180 ° , B = theta

cos ( - 180 ° - theta ) = sin ( - 180 ) ° * sin theta + cos ( - 180 ° ) * cos theta =

sin 180 ° * sin theta + cos 180 ° * cos theta = 0 * sin theta + ( - 1 ) * cos theta =

0 - cos theta = - cos theta

cos ( - 180 ° - theta ) = - cos theta