MHB Simplifying a trigonometric expression

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The discussion focuses on simplifying the trigonometric expression (tg 570° sin 1469°) / (ctg 495° cos 781°) by converting everything into sines and cosines and reducing the angles to their equivalents within the first cycle. The user successfully rewrites the expression to include sine and cosine values for specific angles, noting that some can be simplified directly. They emphasize the importance of knowing the values for special angles, as calculators and trigonometric tables are not allowed in their exam. The relationship between complementary angles is also highlighted, which aids in further simplification. Understanding these principles is crucial for solving the expression effectively.
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(tg 570* sin 1469) / (ctg 495 * cos 781) =
Its in degrees
 
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Alexstrasuz said:
(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 $\displaystyle \begin{align*} 0^{\circ} \leq \theta < 360^{\circ} \end{align*}$...)
 
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.
 
Alexstrasuz said:
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)...
 
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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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