3.12 Trigonometric values, equations and identities  (Page 2/3)

Reflection in 0

There is no change in function form. Function takes sign in accordance with sign rule.

$$\sin \left(0-x\right)=-\sin x;\cos \left(0-x\right)=\cos x;\tan \left(0-x\right)=-\tan x;$$ $$\mathrm{cosec}\left(0-x\right)=-\mathrm{cosec}x;\sec \left(0-x\right)=\sec x;\cot \left(0-x\right)=-\cot x$$

Reflection in π/2

Reflection in π/2 is also known as co-function identities. Functions are called co-functions when their compliments have same value. As such, sine and cosine are co-functions. In this case, there is change in function form as combination of angle involves π/2. Function takes sign in accordance with sign rule.

$$\sin \left(\frac{\pi }{2}-x\right)=\cos x;\cos \left(\frac{\pi }{2}-x\right)=\sin x;\tan \left(\frac{\pi }{2}-x\right)=\cot x;$$ $$\mathrm{cosec}\left(\frac{\pi }{2}-x\right)=\sec x;\sec \left(\frac{\pi }{2}-x\right)=\mathrm{cosec}x;\cot \left(\frac{\pi }{2}-x\right)=\tan x$$

Reflection in π

In this case, there is no change in function form. Function takes sign in accordance with sign rule.

$$\sin \left(\pi -x\right)=\sin x;\cos \left(\pi -x\right)=-\cos x;\tan \left(\pi -x\right)=-\tan x;$$ $$\mathrm{cosec}\left(\pi -x\right)=\mathrm{cosec}x;\sec \left(\pi -x\right)=-\sec x;\cot \left(\pi -x\right)=-\cot x$$

Shift by π/2

Shift refers to horizontal shift of graph. We shall explore this aspect of trigonometric function in detail in a separate module. From transformation point of view, there is change in function form as combination of angle involves π/2. Function takes sign in accordance with sign rule.

$$\sin \left(\frac{\pi }{2}+x\right)=\cos x;\cos \left(\frac{\pi }{2}+x\right)=-\sin x;\tan \left(\frac{\pi }{2}+x\right)=-\cot x;$$ $$\mathrm{cosec}\left(\frac{\pi }{2}+x\right)=\sec x;\sec \left(\frac{\pi }{2}+x\right)=-\mathrm{cosec}x;\cot \left(\frac{\pi }{2}+x\right)=-\tan x$$

Shift by π

In this case, there is no change in function form. Function takes sign in accordance with sign rule.

$$\sin \left(\pi +x\right)=-\sin x;\cos \left(\pi +x\right)=-\cos x;\tan \left(\pi +x\right)=\tan x;$$ $$\mathrm{cosec}\left(\pi +x\right)=-\mathrm{cosec}x;\sec \left(\pi +x\right)=-\sec x;\cot \left(\pi +x\right)=\cot x$$

Shift by 2π

In this case, there is no change in function form. Function takes sign in accordance with sign rule.

$$\sin \left(2\pi +x\right)=\sin x;\cos \left(2\pi +x\right)=\cos x;\tan \left(2\pi +x\right)=\tan x;$$ $$\mathrm{cosec}\left(2\pi +x\right)=\mathrm{cosec}x;\sec \left(2\pi +x\right)=\sec x;\cot \left(2\pi +x\right)=\cot x$$

Finding angles

Trigonometric functions are many-one relation. We are required to find angles corresponding to a given trigonometric value. For example, what are angles corresponding to sine value of -√3/2. In other words, we need to find angles whose sine evaluates to this value. Note that these values corresponds to intersection of parallel line y=-√3/2 with the graph of sine curve.

For the time being, let us concentrate the interval [0,2π], which corresponds to one cycle of four quadrants. We follow an algorithm as given here to find angles in this interval :

1 : Consider only numerical magnitude of the given value. Find acute angle whose trigonometric function value corresponds to the numerical magnitude of the given value.

2 : Use sign rule and identify quadrants in which trigonometric function has the sign that of given value.

3 : Use value diagram and determine the angles as required.

To see the working of the algorithm, let us consider sinx = -√3/2. Considering only the magnitude of numerical value, we have :

$$\sin \theta =\frac{\sqrt{3}}{2}=\sin \frac{\pi }{3}$$

Thus, required acute angle is π/3. Now, sine function is negative in third and fourth quadrants. Looking at the value diagram, the angle in third quadrant is :

$$\Rightarrow x=\pi +\theta =\pi +\frac{\pi }{3}=\frac{4\pi }{3}$$

Similarly, angle in fourth quadrant is :