As the name suggests an equation (or) Inequation In valuing one (or) more trigonometric ratios of unknown angles is called a trigonometric equation (or) inequation.

In this lesson we will learn how to affirm the general solution of trigonometric equations and inequations in valuing one (or) more of the six trigonometric ratios. We shall highlight the areas where a student must be careful while solving such equations and inequations we will also discuss the methods to solve equations / Inequations invaluing both trigonometric and non trigonometric functions.

(i) Basic Trigonometric equations:

Consider the following

cos q – sin q = 1                                                                   (1)

sin² q + cos² q = 1                                                                (3)

Equation (1) is satisfied if we put q = 0, 2Π, 4Π  …etc.

In it. Equation (2) is satisfied if we put q = 7Π/12 , 11Π/12, -Π/12

etc in it. But equation (3) is satisfied for any value of q. Equation (1) and (2) are called trigonometric equations while (3) is a trigonometric Identify. A trigonometric equation has three kinds of solutions.

(1)   Principal solution: Numerically smallest value of the equation unknown angle satisfying the given equation

(2)   Particular solution : Any value of angle satisfying the given equation.

(3)           General solution: Collection of all particular solution. For example the equation of (q – Π/6) = √3 is satisfied if q = -Π/6 (or) 7Π/6 (or) -5Π/6, etc

→q = Π/3 , 4Π/3 , -2Π/3 etc. Out of these the numerically smallest is q = Out of these the numerically smallest is q =/3.

This is the principal solution. It is a particular solution also.

Also the above values of q taken individually are particular solutions. For general solution we observe that since cotangent function is periodic with period p and it takes the value  only once in the Interval
[0, Π], (q , -Π/6) can be equal to any particular solution plus a multiple of Π. If we take the particular solution as the principal solution then q – Π/6 = nΠ + Π/6 , n ε l

q = nΠ+Π/3 , n ε l is the general solution. If in any equation we get two values of the unknown angle which are numerically equal but opposite in sign, the principal solution is assumed as the positive angle (e.g.) principal solution of sec q = 2 is q = Π/3, although q = –Π/3 also satisfies it.

Method to find principal value (numerically least angle)

(i)    First draw a trigonometric – circle and mark the quadrant in which the angle may be.

(ii)   Select anticlockwise direction for 1^st and 2^nd quadrants and select clockwise for 3^rd and 4^th quadrants

(iii)  Find the angle in the first rotation.

(iv)   Select the numerically least angle from these two values the angle thus found will be the principal value.

(v)    In case two angles (one with positive sign and the other with negative sign) qualify for the numerically least angle then we select the angle with positive sign as the principal value

The SHE is the universal reference for reporting relative half-cell potentials. It is a type of gas electrode and was widely used in early studies as a reference electrode, and as an indicator electrode for the determination of pH values. The SHE could be used as either an anode or cathode depending upon the nature of the half-cell it is used with. The SHE consists of a platinum electrode immersed in a solution with a hydrogen ion concentration of 1.00M and H₂ (g) at 1 atm pressure kept at 25^o C.

The platinum electrode is made of a small square of platinum foil which is platinized (known as platinum black). Hydrogen gas, at a pressure of 1 atmosphere, is bubbled around the platinum electrode. The platinum black serves as a large surface area for the reaction to take place, and the stream of hydrogen keeps the solution saturated at the electrode site with respect to the gas. It is interesting to note that even though the SHE is the universal reference standard, it exists only as a theoretical electrode which scientists use as the definition of an arbitrary reference electrode with a half-cell potential of 0.00 volts.

(Because half-cell potentials cannot be measured, this is the perfect electrode to allow scientists to perform theoretical research calculations.) The reason this electrode cannot be manufactured is due to the fact that no solution can be prepared that yields a hydrogen ion activity of 1.00M.

Hydrogen electrode is made by adding platinum black to platinum wire or a platinum plate. It is immersed in the test solution and an electric charge is applied to the solution and platinum black with hydrogen gas. The hydrogen-electrode method is a standard among the various methods for measuring pH. The values derived using other methods become trustworthy only when they match those measured using hydrogen electrode method. However, this method is not appropriate for daily use because of the effort and expense involved, with the inconvenience of handling hydrogen gas and great influence of highly oxidizing or reducing substances in the test solution.

E⁰ (H₂/H⁺) = 0 = E⁰ (H⁺/H₂)

Now the other half cell can be divided into two categories:

Þ            one which will act as anode

Þ            the others which will acts as cathode

Each type of cell arrangement will give an EMF value which will be actually the EMF value of unknown electrodes as EMF value of SHE is ‘0’ volts.

Example

Cu electrode (half cell) acts as cathode with SHE i.e., as Cu²⁺/Cu [Cu²⁺ (1.0 M) + 2e --> Cu]. The experimental measurement of EMF value for this cell arrangement give 0.34 volts. Since Cu electrode shows reduction with SHE, the given value of EMF represents the reduction potential of Cu half-cell.

E⁰ (Cu²⁺/Cu) = + 0.34 volts

The oxidation potential of Cu half-cell is just the negative of this value.

E⁰ (Cu/Cu²⁺) = – 0.34 volts

ü        There is no set of rules to distinguish between gravimetric and volumetric analysis and all the laws used in one are equally applicable to the other. For making our calculations simple, we divide stoichiometry into two parts

Þ      simple calculation based on mole concept (Gravimetric Analysis)

Þ      complex calculation based on volume of solutions and their concentrations (Volumetric analysis)

ü        While studying gravimetric analysis, we have three types of relationships in a chemical reaction.

Þ      Calculations involving mass – mass relations

Þ      Calculation involving mass – volume relations

Þ      Calculation involving volume – volume relations

Mass – Mass Relationship

ü        Following steps are followed while making necessary calculations

Þ      Write down balanced molecular equation for the chemical change

Þ      Write down the number of moles below the formula of each of the reactant and product

Þ      Write down the relative masses of the reactants and the products with the help of formula below the respective formula. These are the theoretical amounts of reactants and products.

Þ      By the applications of unitary method, mole concept or proportionality method, the unknown factor or factors can be determined.

Illustrations

1. How many gms of oxygen are required to burn completely 570 gm of octane.

Now balancing the chemical equation, we get

Then let us understand what this reaction means.

Does it mean,

Well, actually, the reaction does not mean either of the two. The first interpretation is wrong because the numbers (stoichiometric coefficients) appearing before substances in a reaction are not the number of atoms or molecules. If it were so, what do you think would be the interpretation of this reaction.

Limiting Reagent

ü        The reagent that gives the least no of moles of the product is the limiting reagent.

Illustrations

1.      1g of Mg is burnt in a closed vessel which contain 0.5g of O₂.

i)    Which reactant is left in excess?

ii)      Find the mass of excess reactant?

Percent Yield

Generally when a reaction is carried out in the laboratory we do not obtain the theoretical amount of the product. The amount of the product that is actually obtained is called the actual yield. Knowing the actual yield and theoretical yield the percentage yield can be calculated as

Mathematics – Trigonometric Identities

Now we shall discuss another example of two-dimensional motion that is motion of a particle on a circular path. This type of motion is called circular motion.

Consider a particle P is moving on circle of radius r on X – Y plane with origin O as centre.

The position of the particle at a given instant may be described by angle q, called angular position of the particle, measured in radian. As the particle moves on the path, its angular position (q) changes. The rate of change of angular position is called angular velocity (w) measured in radian per second.

The rate of change of angular velocity is called angular acceleration, measured in rad/s². Thus, the angular acceleration is

RELATION BETWEEN DIFFERENT PARAMETERS OF CIRCULAR MOTION

ü        It is easy to derive the equations of rotational kinematics for the case of constant angular acceleration with fixed axis of rotation. These equations are of the same form as those for one-dimensional translational motion.

w = w₀ +αt                                                             …(i)

Φ = Φ₀ + w₀t + αt²/2                                 …(ii)

w² = w₀² + 2α (Φ- Φ₀)                             …(iii)

Φ= Φ₀ + (w₀ +w)/(2t)                               …(iv)

Here,Φ is the initial angle and w₀ is the initial angular speed.

Illustrations

1.     (a) What is the angular velocity of the minute and hour hands of a clock?

(b) Suppose the clock starts malfunctioning at 7 AM which decelerates the minute hand at the rate of 4p radians/day. How much time would the clock loose by 7 AM next day?

Sol. Angular speed of

minute hand : wmh = 2Π rad/hr

= 48Π rad/day = (Π/1800) rad/sec

hour hand : whh = (Π/6) rad/hr

= 4Π rad/day = (Π/21600) rad/sec

(b)     Assume at t = 0, Φ₀ = 0, when the clock begins to malfunction. Use equation (ii) to get the angle covered by the minute hand in one day.

Hence the minute hand complete 23 revolutions, so the clock losses 1 hour.

Inverse of trigonometric ratios

We know that y = sin x means y is the value of sine of angle x if we consider domain and co-domain both as set R of real numbers. Sine ratio as seen from the fig. is many-one into function.

ü        But it is clear that if we restrict the domain to [-Π/2 , Π/2]  and range to [–1, 1], then. y = sin x is one-one onto and hence it is invertible.

So, y = sin x                             x ε [-Π/2 , Π/2] , y ε [–1, 1]

Þ  x = sin^–1 y                           y ε [–1, 1] , x ε [-Π/2 , Π/2]

ü        This value of x is called the principal value, i.e. belonging to [-Π/2 , Π/2]  and [-Π/2 , Π/2] range and it is called principal value range.

ü        The smallest numerical angle is called principal value.

ü        In general the inverse circular functions with their domain and range can be as given below:

Inverse Circular Function

Þ            sin^-1 x = θ        iff           sin θ = x, -Π/2 ≤ θ ≤ Π/2

Domain [-1,1]

Range [-Π/2 , Π/2]

Graph

Inverse Circular Function

Þ            cos^-1 x = θ   iff          cos θ = x, 0 ≤ θ ≤ Π

Domain [–1, 1]

Range [0, Π]

Graph

Inverse Circular Function

Þ            tan^-1 x = θ               iff   tan θ= x,  -Π/2< θ < Π/2

Domain (–∞, ∞)

Range (-Π/2 , Π/2)

Graph

along the vertical y-axis with a uniform downward acceleration ‘g’ and

• along the horizontal x-axis with a uniform velocity forward.

Consider a particle projected with an initial velocity u at an angle α with the horizontal x-axis as shown in figure 2.17. Velocity and accelerations can be resolved into two components:

Velocity along x-axis = u_x = u cos α

Acceleration along x-axis a_x = 0

Velocity along y-axis = u_y = u sin α

Acceleration along y-axis a_y = -g

Here we use different equation of motions of one dimension derived earlier to get the different parameters.

v² = v₀² – 2g (y – y₀)                                …(C)

Total Time of flight

When body returns to the same horizontal level, the resultant displacement in vertical y-direction is zero. Use equation B.

Therefore,             0 = (u sin α) t – (½)gt²

or                   t = 2uSinα / g

(as t cannot equal to 0)

Horizontal Range

Horizontal Range (OA)  = Horizontal velocity × Time of flight

=        u cos α × 2u sin α/g

= u² sin 2α/g

=

Maximum Height

At the highest point of the trajectory, vertical component of velocity is zero.

Therefore              0 = (u sin α)² – 2g H_max

or,           H_max = u² sin² α/ 2g

Equation of trajectory

Assuming the point of projection as the origin of co-ordinates and horizontal direction as the x-axis and vertical direction as the y-axis. Let P (x, y) be the position of the particle at instant after t second.

Then x = u cos α.t                and         y = u sin α.t – 1/2 gt²

Eliminating  ‘t’ form the above equations, we get,

y = x tan α – gx²/ 2u² cos² α

This is the equation of trajectory which is a parabola  (y = ax + bx²).

Illustrations

1.      A gun moving at a speed of 30m/sec fires at an angle 300 with a velocity 150m/s relative to the gun. Find the distance between the gun and the projectile when projectile hits the ground . (g = 10 m/sec)

Sol. Vertical component of velocity = 150 sin 300 = 75 m/sec

Horizontal component of velocity relative to gun = 150 cos 300

=                       =   75 √3 m/sec

Horizontal component of velocity relative to ground

=                              = 75 √3 + 30 ≈ 160 m/sec

Time of flight = (2 * 75 )/g = 15 sec

Range of projectile              = 160 × 15 = 2400 m

Distance moved by the gun and projectile = 2400 – 450 = 1950 m.

Consider a particle projected horizontally with a velocity u  from a point O as shown in figure 2.18.

Assuming the point of projection O as the origin of coordinates and horizontal direction as the X-axis and vertical direction as Y-axis. Let P (x, y) be the position of the particle after t seconds.

x = horizontal distance covered in time t = ut. …(1)

y = vertical distance covered in time t  = ½gt² ….(2)

Eliminate t from equations  (1) and (2) then

We get, y = (1/2 ) (g/u²) x²

This is the equation of parabola passing through the origin, with its vertex at the origin O. Hence the trajectory is a parabola.

Physics – Units and Measurements

Mathematics – Trigonometry