NEET Physics Class 11 Chapter 10 Mathematical Tools – Trigonometry
Measurement Of Angle And Relationship Between Degrees And Radian
In navigation and astronomy, angles are measured in degrees, but in calculus, it is best to use units called radians because they simplify later calculations.
Let ACB be a central angle in a circle of radius r, as in the figure. Then the angle ACB or θ is defined in radius as –
θ = \(\frac{\text { Arc length }}{\text { Radius }}\)
θ = \(\frac{\widehat{A B}}{r}\)
If r = 1 then θ = AB
The radian measure for a circle of unit radius of angle ACB is defined as the length of the circular arc AB. Since the circumference of the circle is 2π and one complete revolution of a circle is 360º, the relation between radians and degrees is given by: π radians = 180º
Angle Conversion Formulas
1 degree = \(\frac{\pi}{180}\) (≈ 0.02) radian
Degrees to radians: multiply by \(\frac{\pi}{180}\)
1 radian ≈ 57 degrees
Radians to degrees : multiply by \(\frac{180}{\pi}\)
Question 1.
- Convert 45º to radians.
- Convert \(\frac{\pi}{6}\) rad to degrees.
Answer:
- \(\text { 45 – } \frac{\pi}{180}=\frac{\pi}{4} \mathrm{rad}\)
- \(\frac{\pi}{6} \cdot \frac{180}{\pi}=30^{\circ}\)
Question 2. Convert 30º to radians.
Answer:
⇒ \(30^{\circ} \times \frac{\pi}{180}\)
⇒ \(\frac{\pi}{6} \mathrm{rad}\)
Question 3. Convert \(\frac{\pi}{3}\) rad to degrees.
Answer:
⇒ \(\frac{\pi}{3} \times \frac{180}{\pi}\)= 60º
Standard values
- \(30^{\circ}=\frac{\pi}{6} \mathrm{rad}\)
- \(45^{\circ}=\frac{\pi}{4} \mathrm{rad}\)
- \(60^{\circ}=\frac{\pi}{3} \mathrm{rad}\)
- \(90^{\circ}=\frac{\pi}{2} \mathrm{rad}\)
- \(120^{\circ}=\frac{2 \pi}{3} \mathrm{rad}\)
- \(135^{\circ}=\frac{3 \pi}{4} \mathrm{rad}\)
- \(150^{\circ}=\frac{5 \pi}{6} \mathrm{rad}\)
- \(180^{\circ}=\pi \mathrm{rad}\)
- \(360^{\circ}=2 \pi \mathrm{rad}\)
(Check these values yourself to see that they satisfy the conversion formulae)
NEET Physics Class 11 Chapter 10 Mathematical Tools – Measurement Of Positive And Negative Angles
An angle in the xy-plane is said to be in standard position if its vertex lies at the origin and its initial ray lies along the positive x-axis.
Angles measured counterclockwise from the positive x-axis are assigned positive measures; angles measured clockwise are assigned negative measures.
NEET Physics Class 11 Chapter 10 Mathematical Tools – Six Basic Trigonometric Functions
The trigonometric function of a general angle θ is defined in terms of x, y, and r.
Sine: sinθ = \(\frac{\text { opp }}{\text { hyp }}=\frac{\mathrm{y}}{\mathrm{r}}\)
Cosecant: cosecθ = \(\frac{\text { hyp }}{\text { opp }}=\frac{\mathrm{r}}{\mathrm{y}}\)
Cosine: cosθ = \(\frac{\text { adj }}{\text { hyp }}=\frac{x}{r}\)
Secant: secθ = \(\frac{\text { hyp }}{\text { adj }}=\frac{r}{x}\)
Tangent: tanθ = \(\frac{\text { opp }}{\text { adj }}=\frac{y}{x}\)
Cotangent: cotθ = \(\frac{\text { adj }}{\text { opp }}=\frac{x}{y}\)
Values Of Trigonometric Functions
If the circle has radius r = 1, the equations defining sinθ and cos θ become Cos θ = x, sinθ = y
We can then calculate the values of the cosine and sine directly from the coordinates of P.
Question 1. Find the six trigonometric ratios from the given figure
Answer:
sinθ = \(\frac{\text { opp }}{\text { hyp }}=\frac{4}{5}\)
cosθ = \(\frac{\text { adj }}{\text { hyp }}=\frac{3}{5}\)
tan θ = \(\frac{\text { opp }}{\text { adj }}=\frac{4}{3}\)
cosec θ = \(\frac{\text { hyp }}{\text { opp }}=\frac{5}{4}\)
sec θ = \(\frac{\text { hyp }}{\text { adj }}=\frac{5}{3}\)
cotθ = \(\frac{\text { adj }}{\text { opp }}=\frac{3}{4}\)
Question 2. Find the sine and cosine of angle θ shown in the unit circle if the coordinates of point p are as shown.
Answer:
cos θ = x-coordinate of P = – \(\frac{1}{2}\)
2sin θ = y-coordinate of P = \(\frac{\sqrt{3}}{2}\)
NEET Physics Class 11 Chapter 10 Mathematical Tools – Rules For Finding Trigonometric Ratio Of Angles Greater Than 90°
Step 1 → Identify the quadrant in which the angle lies.
Step 2 → If angle = (nπ ± θ) where n is an integer. Then trigonometric function of (nπ ± θ)= same trigonometric function of θ and the sign will be decided by the CAST Rule.
If angle = \(\left[(2 n+1) \frac{\pi}{2} \pm \theta\right]\) where n is an integer. Then the trigonometric function of \(\left[(2 n+1) \frac{\pi}{2} \pm \theta\right]\)= complimentary trigonometric function of θ and sign will be decided by CAST Rule.
Values of sin θ, cos θ, and tan θ for some standard angles.
Question 1. Evaluate sin 120°
Answer:
sin 120°
= sin (90° + 30°)
= cos 30°
⇒ \(\frac{\sqrt{3}}{2}\)
Aliter sin 120°
= sin (180° – 60°)
= sin 60°
⇒ \(\frac{\sqrt{3}}{2}\)
Question 2. Evaluate cos 135°
Answer:
cos 135°
= cos (90° + 45°)
= – sin 45°
⇒ \(-\frac{1}{\sqrt{2}}\)
Question 3. Evaluate cos 210°
Answer:
cos 210°
= cos (180° + 30°)
= – cos30°
⇒ \(-\frac{\sqrt{3}}{2}\)
Question 4. Evaluate tan 210°
Answer:
tan 210°
= tan (180° + 30°)
= tan 30°
= \(\frac{1}{\sqrt{3}}\)
NEET Physics Class 11 Chapter 10 Mathematical Tools – General Trigonometric Formulas
Question 1. \(\cos ^2 \theta+\sin ^2 \theta=1\)
Answer:
⇒ \(1+\tan ^2 \theta=\sec ^2 \theta\)
⇒ \(1+\cot ^2 \theta={cosec}^2 \theta\)
Question 2. cos(A + B) = cos A cos B – sin A sin B
Answer:
sin( A + B) = sin A cos B + cos A sin B
⇒ \(\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}\)
Question 3. sin 2θ = 2 sin θ cos θ; cos 2θ = cos2 θ – sin2θ = 2cos2 θ – 1 = 1 – 2sin2 θ
Answer:
⇒ \(\cos ^2 \theta=\frac{1+\cos 2 \theta}{2}\)
⇒ \(\sin ^2 \theta=\frac{1-\cos 2 \theta}{2}\)
Question 4.
In Δ ABC, the sine rule
Answer:
ΔABC need not be right-angled, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)
Question 5. Cosine Rule:
Answer:
cosA = \(\frac{\mathrm{b}^2+\mathrm{C}^2-\mathrm{a}^2}{2 \mathrm{bc}}\)
cosB = \(\frac{a^2+C^2-b^2}{2 a c}\)
cosC = \(\frac{a^2+b^2-c^2}{2 a b}\)
NEET Physics Class 11 Chapter 10 Mathematical Tools – Coordinate Geometry
To specify the position of a point in space, we use a right-handed rectangular axes coordinate system. This system consists of
- Origin
- Axis or axes.
If a point is known to be on a given line or in a particular direction only one coordinate is necessary to specify its position, if it is in a plane, two coordinates are required, if it is in space three coordinates are needed.
Origin
This is any fixed point that is convenient to you. All measurements are taken w.r.t. this fixed point.
Axis or Axes
Any fixed direction passing through the origin and convenient to you can be taken as an axis.
- If the position of a point or the position of all the points under consideration always happens to be in a particular direction, then only one axis is required. This is generally called the x-axis.
- If the positions of all the points under consideration are always in a plane, two perpendicular axes are required.
- These are generally called the x and y-axis. If the points are distributed in space, three perpendicular axes are taken which are called the x, y, and z-axis.
Position Of A Point In xy Plane
The position of a point is specified by its distances from the origin along (or parallel to) the x and y-axis as shown in the figure.
Here x-coordinate and y-coordinate are called abscissa and ordinate respectively.
Distance Formula
The distance between two points (x1, y1) and (x2, y2) is given by
d = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
Note: In space d = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)
Slope Of A Line
The slope of a line joining two points A(x1, y1) and B(x2, y2) is denoted by
m and is given by
m = \(\frac{\Delta \mathrm{y}}{\Delta \mathrm{x}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\tan \theta\) [If both axes have identical scales]
Here θ is the angle made by a line with a positive x-axis. The slope of a line is a quantitative measure of inclination.
Question 1. For points (2, 14) find abscissa and ordinate. Also, find the distance from the y and x-axis.
Answer:
Abscissa = x-coordinate = 2 = distance from y-axis.
Ordinate = y-coordinate = 14 = distance from the x-axis.
Question 2. Find the value of a if distances between the points (–9 cm, a cm) and (3 cm, 3cm) is 13 cm.
Answer:
By using distance formula d = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
⇒ \(13 \sqrt{[3-(-9)]^2+[3-a]^2}\)
⇒ 132 = 122 + (3 – a)2
⇒ (3 – a)2 = 132 – 122 = 52
⇒ (3 – a) = ± 5
⇒ a = 2 cm or 8 cm
Question 3. A dog wants to catch a cat. The dog follows the path whose equation is y-x = 0 while the cat follows the path whose equation is x2 + y2 = 8. The coordinates of possible points for catching the cat are.
- (2, – 2)
- (2, 2)
- (–2, 2)
- (–2, 2)
- (2, 4)
Answer:
Let catching point be (x1, y1) then, y1 – x1 = 0 and x12 + y12 = 8
Therefore, 2x12 = 8
⇒ x12 = 4
⇒ x1 = ± 2;
so possible ae (2, 2) and (–2, –2).