Trigonometric Ratios Of Multiples And Fractions Of Angles

In this section we will learn about multiples and fractions of angles.

We saw how to evaluate ratios for angles of the form $(A \pm B)$. We can use these to come with the formulas for angles of the form $nA$ using substitution.

I. Evaluating Ratios Of 2A & 3A

We know that $\sin (A + B) = \sin A \cos B + \cos A \sin B$

Now, substituting $B = A$, we get:

$\sin (A + A) = \sin A \cos A + \cos A \sin A$

$\Rightarrow$ $\sin 2A = 2 \sin A \cos A$

Similarly, using $\cos (A + B) = \cos A \cos B - \sin A \sin B$ and substituting $B = A$, we get:

$\cos (A + A) = \cos A \cos A - \sin A \sin A$

$\Rightarrow \cos 2A = \cos^2A - \sin^2 A = 2 \cos^2 A - 1$

$\Rightarrow$ $\cos 2A = 2 \cos^2 A - 1 = 1 - 2 \sin^2 A$ $...eqn(i)$

Similarly, for an angle of the form $3A$, we can assert that:

$\sin (3A) = \sin (2A + A)$

$= \sin 2A \cos A + \cos 2A \sin A$

$= 2 \sin A \cos A \cos A + (2 \cos^2 A - 1) \sin A$

$= 2 \sin A \cos^2 A + (1 - 2\sin^2 A) \sin A$

$= 2 \sin A (1 - \sin^2 A) + \sin A - 2sin^3 A$

$= 2 \sin A - 2 \sin^3 A + \sin A - 2sin^3 A$

-----------book page break-----------

$= 3 \sin A - 4 \sin^3 A$

Similarly,

$\cos 3A = \cos (2A + A)$

$= \cos 2A \cos A - \sin 2A \sin A$

$= (2 \cos^2 A - 1) \cos A - (2 \sin A \cos A) \sin A$

$= 2 \cos^3 A - \cos A - 2 \sin^2 A \cos A$

$= 2 \cos^3 A - \cos A - 2 (1 - \cos^2 A) \cos A$

$= 2 \cos^3 A - \cos A - 2 \cos A + 2 \cos^3 A$

$= 4 \cos^3 A - 3 \cos A$

For $\tan 2A$,

$\tan 2A = \tan (A + A)$

$= \dfrac{\tan A + \tan A}{1 - \tan A \cdot \tan A}$

$= \dfrac{2 \tan A}{1 - \tan^2 A}$

For $\tan 3A$,

$\tan 3A = \tan (2A + A)$

$= \dfrac{\tan 2A + \tan A}{1 - \tan 2A \cdot \tan A}$

-----------book page break-----------

$= \dfrac{\dfrac{2 \tan A}{1 - \tan^2 A} + \tan A}{1 - \left(\dfrac{2 \tan A}{1 - \tan^2 A}\right) \times \tan A}$

$= \dfrac{\dfrac{2 \tan A + \tan A - \tan^3 A}{1 - \tan^2 A}}{\dfrac{1 - \tan^2 A - 2\tan^2 A}{1 - \tan^2 A}}$

$= \dfrac{3 \tan A - \tan^3 A}{1 - 3\tan^2 A}$

Likewise, it is possible to evaluate ratios for angles $nA$ for larger values of $n$, but the steps grow more and more tedious as the value of $n$ increases.

II. Evaluating Ratios Of A/2

Finding the ratios of $\dfrac{A}{2}$ should be fairly simple, given all the formulae that we have derived in the previous section.

Substituting $A$ with $\dfrac{A}{2}$ in $eqn(i)$, we get:

$\cos A = 1 - 2 \sin^2 \dfrac{A}{2}$

$\Rightarrow 2 \sin^2 \dfrac{A}{2} = 1 - \cos A$

-----------book page break-----------

$\Rightarrow \sin \dfrac{A}{2} = \pm \sqrt{\dfrac{1 - \cos A}{2}}$

Again, using the same substitution as before, we can also write:

$\cos A = 2 \cos^2 \dfrac{A}{2} - 1$

$\Rightarrow 2 \cos^2 \dfrac{A}{2} = 1 + \cos A$

$\Rightarrow \cos \dfrac{A}{2} = \pm \sqrt{\dfrac{1 + \cos A}{2}}$

Therefore,

$\tan \dfrac{A}{2} = \dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}}$

$= \sqrt{\dfrac{1 - \cos A}{1 + \cos A}}$