Trigonometry Procedure...

[Orion1]

Because Trigonometry is a pre-requisite for Calculus, the purpose of this procedure is for use in 'calculator banned' Calculus courses, to eliminate the requirement of memorizing sixteen 'special angles' in Trigonometry to derive the correct solution.

Memorizing this procedure can produce all the correct solutions with knowledge of only the signs of four quadrants and three special angles in quadrant I.

This is the procedure that I developed in the 'rough', and I have not thoroughly tested it. Please let me know if there is a better 'symbolic' way to write this procedure or if this procedure produces any errors.


f_t( \theta_1 ) = \sin \theta_1, \cos \theta_1, \tan \theta_1 \; \text{etc.}

\theta_1 = \frac{n a}{b}
\theta_2 = \frac{\theta_1}{n} = \frac{a}{b}

\text{if} \; \theta_1 = \pi \; \text{then} \; \theta_2 = 0

f_q(\theta_1) = \text{sgn} [f_t (\theta_1)]

f_1(\theta_1) = f_q(\theta_1) \cdot f_t(\theta_2)

f_1(\theta_1) = \text{sgn} [f_t (\theta_1)] \cdot f_t(\theta_2)

\text{example:}
f_t( \theta_1 ) = \cos \frac{2 \pi}{3}
\theta_1 = \frac{2 \pi}{3} \; \; \; n = 2 \; \; \; a = \pi \; \; \; b = 3
\theta_2 = \frac{a}{b} = \frac{\pi}{3}
f_q \left( \frac{2 \pi}{3} \right) = \text{sgn} \left[ \cos \frac{2 \pi}{3} \right] = -1
f_1( \theta_1 ) = \cos \frac{2 \pi}{3} = \text{sgn} \left[ \cos \frac{2 \pi}{3} \right] \cdot \cos \frac{\pi}{3}
\cos \frac{2 \pi}{3} = - \cos \frac{\pi}{3} = - \frac{1}{2}
\boxed{\cos \frac{2 \pi}{3} = - \frac{1}{2}}

[VietDao29]

Firstly, you haven't given any proof your for "theorem". So you cannot claim that your theorem is correct.
\text{if} \; \theta_1 = \pi \; \text{then} \; \theta_2 = 0
???
Why must it be so?!?! :confused:
f_1(\theta_1) = \text{sgn} [f_t (\theta_1)] \cdot f_t(\theta_2)
Ok, so you are saying that:
\cos \left( \frac{\pi}{3} \right) = \cos \left( \frac{2 \pi}{6} \right) = \mbox{sgn} \left( \cos \left( \frac{2 \pi}{6} \right) \right) \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}? Right?

[Orion1]

\text{if} \; \theta_1 = \pi \; \text{then} \; \theta_2 = 0
Why must it be so?!?!

Because the area formed by the triangle with respect to origin and zero radians is zero. 0 and pi radians are the only two special angles where this is true and hence pi radians is the only exception to the formula.

A_t = \frac{a \cdot b}{2} \sin \pi = 0

This case would also be true for the formula:
\text{if} \; \theta_1 = \pi \; \text{then} \; \theta_2 = - \pi

A_t = \frac{a \cdot b}{2} \sin -\pi = 0

Right?
Negative.
\cos \left( \frac{\pi}{3} \right) \neq \frac{\sqrt{3}}{2}
\theta_2 \neq \frac{a}{2b}

Formula:
\cos \left( \frac{n a}{b} \right) = \text{sgn} \left[ \cos \left( \frac{n a}{b} \right) \right] \cos \left( \frac{a}{b} \right)

Proof:
\cos \left( \frac{n \pi}{3} \right) = \text{sgn} \left[ \cos \left( \frac{n \pi}{3} \right) \right] \cos \left( \frac{\pi}{3} \right) = \pm \frac{1}{2}

\text{if} \; \theta_1 = n \pi \; \text{then} \; \theta_2 = 0

\cos \left( n \pi \right) = \text{sgn} \left[ \cos \left( n \pi \right) \right] \cos \left( 0 \right) = \pm 1