Trigonometry Laws

The subject of trigonometry has a bearing on triangles. In fact the name trigonometry is derived combining the terms triangles geometry. Each of the basic trigonometric ratios (Sine, Cosine and Tangents) of the interior angles of any triangle have some established relations with the sides of the triangle. Some important relations out of those are framed as trigonometry laws of a triangle or in short as trigonometry laws. Let us study those important laws one by one.

Law of Sines

Let us consider a triangle general in nature, that is a scalene triangle which has all interior angle different and all measures of the sides are different as well. Such a triangle is shown below.
Law of Sines
In any scalene triangle ABC , as shown above, the law of sine in the form of formula is,
$[\frac{(a)}{(\sin A)}]$ = $[\frac{(b)}{(\sin B)}]$ = $[\frac{(c)}{\sin C}]$The above formula is known as law of sines and let us show how this law is established.

Law of Sines Triangle
In the same triangle draw the altitude h which divides a into two parts as d and e.

Now, h = c*sin B from left side of A and h = b*sin C

Therefore, b*sin C = c*sin B or, $\frac{(b)}{(sin B)}$ = $\frac{(c)}{(sin C)}$

In the same way, by drawing a perpendicular from B on AC, it can be shown that
 
$\frac{(a)}{(sin A)}$ = $\frac{(c)}{(sin C)}$

Therefore,  $\frac{(a)}{(sin A)}$ = $\frac{(b)}{(sin B)}$ = $\frac{(c)}{(sin C)}$

Thus the law of sine is proved.

Law of Cosines

The law of cosine in the form of formula is,
a2 = b2 + c2 2bc cos A,   b2 = c2 + a2 2ca cos B   and    c2 = a2 + b2 2ab cos CThe proof of law of cosine is as follows. Let us refer the earlier diagram again.

Law of Cosines
In the same triangle draw the altitude h which divides a into two parts as d and e.

Let us prove the law of cosine with the same diagram

a = d + e = c cos B + b cos C. Multiplying by a both sides,

a2 = ac cos B + ab cos C . Same way we can establish by drawing other altitudes,

b2 = bc cos A+ ab cos C

c2 = ac cos B + bc cos A.  Therefore,

b2 + c2 - a2 = bc cos A+ ab cos C + ac cos B + bc cos A - ac cos B - ab cos C =  2bc cos A

or, a2 = b2 + c2 - 2bc cos A. Same way we can prove,

b2 = c2 + a2 - 2ca cos B   and    c2 = a2 + b2 - 2ab cos C

Thus the law of cosine is proved.

Law of Sines and Cosines

It may appear that it is easier to work with law of sine but the law of sine can not be used when the measures of all three sides are known but none of the angles are known. In such a case, the law of cosine is the only source of help, till at least finding one of the angles.

Suppose we only know the measures of two sides and one angle which is not an included angle. In this case, the entire triangle can be solved by law of sine alone. But in some cases, there is a possibility of getting two solutions or ambiguous solutions.  Both the  solutions may be correct or one of them may be extraneous depending upon the application. Let us explain cases where two possible solutions can occur.

Consider the case when the measures of sides b, c and measure of angle B is known. There will be two solutions for angle A correspondingly two solutions for the measures of side a, if,

1) angle B is acute
2) b < c
3) b > c*sin B

Similarly law of cosine is ineffective if we only know the measures of two sides and one angle which is not an included angle. In this case, the entire triangle can be solved by law of sine alone.

The law of cosine is also helpful to find the type of triangle when you know only the measures of all the sides. The method is, label the greatest side as a, thereby, the greatest angle is A. Now rewriting the formula  a2 = b2 + c2 – 2bc cos A as,
$\cos A$ = $\frac{[(b^{2}+c^{2})-(a^{2})]}{[(2bc )]}$If cos A is positive, the angle A is acute, and hence the triangle is an acute triangle as A is the greatest angle. If cos A is negative, the angle A is obtuse, and hence the triangle is an obtuse triangle as A is the greatest angle. If cos A is 0, the angle A is right angle, and hence the triangle is a right triangle.

Law of Tangents

The law of tangents is used in a triangle when two sides and the included angle is known or any two angles and any one side is known. If a, b and c are the sides of any triangle, corresponding to angles A, B and C, then as per law of tangents,
$\frac{[(a-b)]}{[(a+b)]}$$[\frac{\tan \frac{1}{2}(A-B)}{\tan \frac{1}{2}(A+B)}]$The law of tangents is proved as follows.

As per law of sines,

$\frac{a}{\sin A}$ = $\frac{b}{\sin B}$. Let this ratio be k.

Then, a = k*sin (A) and b = k*sin (B)

Therefore,

$\left [ \frac{(a-b)}{(a+b)} \right ]$ = $\frac{[k*\sin (A)-k*\sin (B)]}{[k*\sin (A)+k*\sin (B)]}$ = $\frac{[\sin (A)-\sin (B)]}{[\sin (A)+\sin (B)]}$

As per the formula for sum and difference of two sine ratios,
sin (A) -  sin (B) = 2 sin $\frac{1}{2}$ (A – B)*cos $\frac{1}{2}$ (A + B) and  sin (A) +  sin (B) = 2 sin $\frac{1}{2}$ (A + B)*cos $\frac{1}{2}$ (A - B)

Therefore,

$\left [ \frac{(a-b)}{(a+b)} \right ]$ = $\frac{[2\sin \frac{1}{2}(A-B)*\cos \frac{1}{2}(A+B)]}{2\sin \frac{1}{2}(A+B)*\cos \frac{1}{2}(A-B)}$

=$\frac{[\tan \frac{1}{2}(A-B)]}{[\tan \frac{1}{2}(A+B)]}$

Thus the law of tangent is proved.

This law is more convenient than the law of cosine to solve a triangle when only two sides and the included angle is known.  It is explained below. The above formula can be rewritten as,
$\tan$ $\frac{1}{2}$ (A – B) = $[\frac{(a – b)}{(a + b)}]$ * $\tan$ $\frac{1}{2}$ (A + B) = $[\frac{(a – b)}{(a + b)}]$ * $\cot$ $\frac{1}{2}$ (C)Since we know all the items on the right, (A – B) can be figured out.

We also know (A + B), since (A + B) = 180o – (C)

By solving we can find the angles A and B. The remaining side c can now be found by law of sine.