MHB Simplifying a trigonometric expression

[Alexstrasuz1]

(tg 570* sin 1469) / (ctg 495 * cos 781) =
Its in degrees

 

[Prove It]

(tg 570* sin 1469) / (ctg 495 * cos 781) =
Its in degrees

I would first try to write everything in terms of sines and cosines, and then change all the angles to the corresponding angles in the first cycle (so $\displaystyle \begin{align*} 0^{\circ} \leq \theta < 360^{\circ} \end{align*}$...)

 

[Alexstrasuz1]

I did it and I got [(sin30 * sin29)/cos30]/[(cos135*cos61)/sin135]
My problem is this is for exam for university application and we can't use calculator or the sheets with trigonometric table like value of sin30 etc, I am ok with algebra since my university is focusing more on chemistry.

 

[Prove It]

I did it and I got [(sin30 * sin29)/cos30]/[(cos135*cos61)/sin135]
My problem is this is for exam for university application and we can't use calculator or the sheets with trigonometric table like value of sin30 etc, I am ok with algebra since my university is focusing more on chemistry.

Some of those values can be simplified straight away - sin(30), cos(30), cos(135) and sin(135)...

 

[MarkFL]

When I consider that the period of the tangent and cotangent functions is $180^{\circ}$ and the period of the sine and cosine functions is $360^{\circ}$, and the fact that cotangent is an odd function, I can then rewrite the given expression as:

$$-\frac{\tan\left(30^{\circ}\right)\sin\left(29^{\circ}\right)}{\cot\left(45^{\circ}\right)\cos\left(61^{\circ}\right)}$$

Now, to deal with the sine and cosine function, we see the angles $29^{\circ}$ and $61^{\circ}$ are complementary...which means the sine of one is the cosine of the other and vice versa.

As for the tangent and cotangent functions, those are special angles for which we should know the values of those functions at those angles.