An Easy Way To Learn The Basic Trigonometric Identities

First learn the structure. Learn the positions of the six trigonometry functions. First comes the $\sin$ function, underneath it comes the $\cos$ function.

Then in the next column comes the $\csc$ function, underneath it comes the $\sec$ function. And in the last column comes the $\cot$ function and underneath it comes the $\tan$ function. Learning this position is important and just one time. The six trig functions are specially placed in the above given places, so as to serve our need.  Continue reading

The Trigonometric Ratio Hexagon

I found a book this morning containing a hexagonal figure of trigonometric ratios.  It reminded me of the mnemonics I created when I was still in high school in order to remember the formulas during exams. Unlike in other countries, we were not allowed to have formula sheets during exams, so we have to memorize them all.

In the figure, any trigonometric ratio is a product of its immediate neighbors. Therefore,  Continue reading

The Proof of the Tangent Half-Angle Formula

In this post, we prove the following trigonometric identity:

$\displaystyle \tan \frac{\theta}{2} = \frac{\sin\theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta}$.

Proof

Consider a semi-circle with “center” $O$ and diameter $AB$ and radius equal to 1 unit as shown below.  If we let $\angle BOC =\theta$, then by the Inscribed Angle Theorem, $\angle CAB = \frac{\theta}{2}$.

Draw $CD$ perpendicular to $OB$ as shown in the second figure. We can compute for the sine and cosine of $\theta$ which equal to the lengths of $CD$ and $OD$, respectively. In effect, $BD = 1 - \cos \theta$ and $AD = 1 + \cos \theta$. Continue reading

The Complements Theorem

In this post, we discuss the proof behind one of the most commonly used identities in trigonometry. We examine the equations below and  show why the relationships always hold.

$\sin \theta = \cos (90^{\circ} - \theta)$
$\cos \theta = \sin(90^{\circ} - \theta)$

To students who have taken trigonometry, I’m sure that you have met these equation before.  The proof of these equations are as follows.

Consider triangle $ABC$ right angled at $C$. From the definitions, we know that

$\sin A = \displaystyle\frac{a}{c}$
$\cos A = \displaystyle\frac{b}{c}$
$\sin B = \displaystyle\frac{b}{c}$
$\cos B = \displaystyle\frac{a}{c}$

Therefore, (1) $\sin A = \cos B$ and (2) $\cos A = \sin B$.

Now, If we let $A = \theta$, then $B = 90^{\circ} - \theta$, then substituting the values of $A$ and $B$ in (1) and (2), we have

$\sin \theta = \cos (90^{\circ} - \theta)$
$\cos \theta = \sin (90^{\circ} - \theta)$

and these are what we want to show.

As exercises, use the strategy above, or any strategy you want to prove the following identities.

1.) $\cot\theta = \tan (90^{\circ} - \theta)$
2.) $\sec \theta = \csc(90^{\circ} - \theta)$

An elementary proof of the cosine law

The cosine law states that in triangle $ABC$ with side lengths $a, b,$ and $c$, the following equations are satisfied:

$a ^2 = b^2 + c^2 - 2bc$ $\cos A$*

$b ^2 = a^2 + c^2 - 2ac$ $\cos B$

$c ^2 = a^2 + b^2 - 2ab$ $\cos C$

The discussion below shows how these equations were derived. Continue reading