5.5 Trig identities

Trigonometric identities

Deriving values of trigonometric functions for ${30}^{\circ }$ , ${45}^{\circ }$ And ${60}^{\circ }$

Keeping in mind that trigonometric functions apply only to right-angled triangles, we can derive values of trigonometric functions for ${30}^{\circ }$ , ${45}^{\circ }$ and ${60}^{\circ }$ . We shall start with ${45}^{\circ }$ as this is the easiest.

Take any right-angled triangle with one angle ${45}^{\circ }$ . Then, because one angle is ${90}^{\circ }$ , the third angle is also ${45}^{\circ }$ . So we have an isosceles right-angled triangle as shown in [link] .

If the two equal sides are of length $a$ , then the hypotenuse, $h$ , can be calculated as:

So, we have:

We can try something similar for ${30}^{\circ }$ and ${60}^{\circ }$ . We start with an equilateral triangle and we bisect one angle as shown in [link] . This gives us the right-angled triangle that we need, with one angle of ${30}^{\circ }$ and one angle of ${60}^{\circ }$ .

If the equal sides are of length $a$ , then the base is $\frac{1}{2}a$ and the length of the vertical side, $v$ , can be calculated as:

So, we have:

You do not have to memorise these identities if you know how to work them out.

Two useful triangles to remember

Alternate definition for $tan\theta$

We know that $tan\theta$ is defined as: $tan\theta =\frac{\mathrm{opposite}}{\mathrm{adjacent}}$ This can be written as:

But, we also know that $sin\theta$ is defined as: $sin\theta =\frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$ and that $cos\theta$ is defined as: $cos\theta =\frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$

Therefore, we can write

A trigonometric identity

One of the most useful results of the trigonometric functions is that they are related to each other. We have seen that $tan\theta$ can be written in terms of $sin\theta$ and $cos\theta$ . Similarly, we shall show that: ${sin}^2\theta +{cos}^2\theta =1$

We shall start by considering $▵ABC$ ,

We see that: $sin\theta =\frac{AC}{BC}$ and $cos\theta =\frac{AB}{BC}.$

We also know from the Theorem of Pythagoras that: $A{B}^2+A{C}^2=B{C}^2.$

So we can write:

Trigonometric identities

1. Simplify the following using the fundamental trigonometric identities:
   1. $\frac{cos\theta }{tan\theta }$
   2. ${cos}^2\theta .{tan}^2\theta +{tan}^2\theta .{sin}^2\theta$
   3. $1-{tan}^2\theta .{sin}^2\theta$
   4. $1-sin\theta .cos\theta .tan\theta$
   5. $1-{sin}^2\theta$
   6. $\left(\frac{1-{cos}^2\theta }{{cos}^2\theta }\right)-{cos}^2\theta$
2. Prove the following:
   1. $\frac{1+sin\theta }{cos\theta }=\frac{cos\theta }{1-sin\theta }$
   2. ${sin}^2\theta +(cos\theta -tan\theta )(cos\theta +tan\theta )=1-{tan}^2\theta$
   3. $\frac{(2{cos}^2\theta -1)}{1}+\frac{1}{(1+{tan}^2\theta )}=\frac{1-{tan}^2\theta }{1+{tan}^2\theta }$
   4. $\frac{1}{cos\theta }-\frac{cos\theta {tan}^2\theta }{1}=1$
   5. $\frac{2sin\theta cos\theta }{sin\theta +cos\theta }=sin\theta +cos\theta -\frac{1}{sin\theta +cos\theta }$
   6. $\left(\frac{cos\theta }{sin\theta },+,tan,\theta \right)·cos\theta =\frac{1}{sin\theta }$