Trigonometry 

Trigonometry is that branch of mathematics that relates to the sides and angles of triangles, especially in a right-angled triangle. In solving a vast array of problems, one has to learn the basic principles of trigonometric functions and identities. Below, we discuss the major topics in trigonometry IIT JEE maths and how they help in JEE preparation.

1.0Key Concepts of Trigonometry 

1. Trigonometric Ratios and identities 

In a right-angled triangle, there exist six fundamental trigonometric ratios and identities defining the relationship between the angles and the sides:

$\text{ Sine }(\sin ):\sin \theta =\frac{\text{ Perpendicular }}{\text{ Hypotenuse }}$

$\text{ Cosine }(\cos ):\cos \theta =\frac{\text{ Base }}{\text{ Hypoternuse }}$

$\text{ Tangent }(\tan ):\tan \theta =\frac{\text{ Perpendicular }}{\text{ Base }}$

$\text{ Cotangent }(\cot ):\cot \theta =\frac{1}{\tan \theta }=\frac{\text{ Base }}{\text{ Perpendicular }}$

$\text{ Secant (sec): }\sec \theta =\frac{1}{\cos \theta }=\frac{\text{ Hypotenuse }}{\text{ Base }}$

$\text{ Cosecant (csc): }\mathrm{cosec}\theta =\frac{1}{\sin \theta }=\frac{\text{ Hypotenuse }}{\text{ Perpendicular }}$

2. Trigonometric Identities

Trigonometric identities play a crucial role in simplifying trigonometric expressions and solving equations. Some important identities include:

Pythagorean Identity

${\sin }^2\theta +{\cos }^2\theta =1$

$1+{\tan }^2\theta ={\sec }^2\theta$

$1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta$

Reciprocal Identities

$\sec \theta =\frac{1}{\cos \theta }\cdot \mathrm{cosec}\theta =\frac{1}{\sin \theta },\cot \theta =\frac{1}{\tan \theta }$

Sum and Difference Identities

$\sin (A\pm B)=\sin A\cos B\pm \cos A\sin B$

$\cos (A\pm B)=\cos A\cos B\mp \sin A\sin B$

$\tan (A\pm B)=\frac{\tan A\pm \tan B}{1\mp \tan A\tan B}$

$\cot (A\pm B)=\frac{\cot A\cdot \cot B\mp 1}{\cot B\pm \cot A}$

$\sin (A+B)\sin (A-B)={\sin }^{2}A-{\sin }^{2}B={\cos }^{2}B-{\cos }^{2}A$

$\cos (A+B)\cos (A-B)={\cos }^{2}A-{\sin }^{2}B={\cos }^{2}B-{\sin }^{2}A$

\sin (A+B+C)=\sinA \cos B \cos C+\sin B \cos A \cos C+\sin C \cos A \cos B-\sin A \sin B \sin C

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

$\tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}$

Transformation Formula

$2\sin A\cos B=\sin (A+B)+\sin (A-B)$

$2\sin B\cos A=\sin (A+B)-\sin (A-B)$

$2\cos A\cos B=\cos (A+B)+\cos (A-B)$

$2\sin A\sin B=\cos (A-B)-\cos (A+B)$

$\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

$\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

$\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

$\cos A-\cos B=2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

Multiple Angle Ratios

$\sin 2A=2\sin A\cos A=\frac{2\tan A}{1+{\tan }^{2}A}$

$\cos 2A=2{\cos }^{2}A-1=1-2{\sin }^{2}A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}={\cos }^{2}A-{\sin }^{2}A$

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

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

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

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

3. Trigonometric Functions of Angles

The trigonometric functions are further defined on the unit circle. This defines the concept of sine, cosine, tangent, and other related functions for all angles and just right triangles. The periodicity of the trigonometric circular function is deeper with a unit circle.

Unit Circle Definition: For an angle, , the sine and cosine are defined as:

$\sin \theta =y\text{-coordinate },\cos \theta =\mathrm{x}\text{ - coordinate, }$

and

$\tan \theta =\frac{\sin \theta }{\cos \theta }$

This can be visualized by seeing the behaviour of the values of the trigonometric functions for angles larger than 90°, that is, in the second, third, and fourth quadrants.

4. Trigonometric Equations

The Trigonometric equations often involve using the basic trigonometric ratios and identities for determining the measures of unknown angles. For instance, general equations for these trigonometric functions are:

Sinθ = Sinα, and the general solution is θ = nπ + (-1)nα, where n ∈ Z

Cosθ = Cosα, and the general solution is θ = 2nπ + α, where n ∈ Z

Tanθ = Tanα, and the general solution is θ = nπ + α, where n ∈ Z

2.0Graphs of Trigonometric Functions

Trigonometric Function	Graphs
Sine  y=sinx Domain is R Range (-1,1)
Cos y=cosx Domain is R Range (-1,1)
Tan y=tanx Domain = R - (2n+1)π/2 Range = R
Cot y= cotx Domain = R - nπ Range = R
Sec y= secx Domain: R - {(2x+1)π/2} Range: (-\infty,-1) \cup(1, \infty)
Cosec y = cosecx Domain: R - {nπ} Range : (-\infty,-1) \cup(1, \infty)

3.0Solved Problems 

Question 1: If

$\mathrm{Sin}x=-\frac{3}{5},\text{ where }\pi <x<\frac{3\pi }{2},\text{ then }80\left({\tan }^{2}x-\cos x\right)$is equal to

Ans: 109

Explanation

$\sin x=-\frac{3}{5}\text{ where }\pi <x<\frac{3\pi }{2}$

$\tan x=\frac{3}{4},\cos x=\frac{-4}{5}$

$80\left({\tan }^{2}x-\cos x\right)$

$=80\left(\frac{9}{16}+\frac{4}{5}\right)$

$=\left(\frac{45+64}{80}\right)$

=109

Question2: Suppose

$\theta \in \left[0,\frac{\pi }{4}\right]\text{ is a solution of }4\cos \theta -3\sin \theta =1\text{. }$ 

Then

$\cos \theta$ is equal to:

Ans:

$\frac{4}{(3\sqrt{6}-2)}$

Explanation

$4\cos \theta -3\sin \theta =1$

$4\cos \theta -1=3\sin \theta$

Squaring Both sides

$16{\cos }^2\theta -8\cos \theta +1=9\left(1-{\cos }^2\theta \right)$

$16{\cos }^2\theta -8\cos \theta -8=-9{\cos }^2\theta$

$\Rightarrow 25{\cos }^2\theta -8\cos \theta -8=0$

$\Rightarrow \cos \theta =\frac{8\pm \sqrt{64+4\times 25\times 8}}{2.25}$

$=\frac{8\pm 4\sqrt{4+50}}{2.25}$

$=\frac{4\pm 2\sqrt{54}}{25}$

$\text{ As }\theta \in \left[0,\frac{\pi }{4}\right]$

$\Rightarrow \cos \theta =\frac{4+6\sqrt{6}}{25}=\frac{4}{3\sqrt{6-2}}$

Question 3: Show that 2 sin²β + 4 cos (α + β) sin α sin β + cos 2 (α + β) = cos 2α 

Solution: LHS 

= 2 sin²β + 4 cos (α + β) sin α sin β + cos 2(α + β) 

= 2 sin²β + 4 (cos α cos β – sin α sin β) sin α sin β + (cos 2α cos 2β – sin 2α sin 2β) 

= 2 sin²β + 4 sin α cos α sin β cos β – 4 sin²α sin²β + cos 2α cos 2β – sin 2α sin 2β 

(4 sin α cos α sin β cos β = 2 sin α cos α .2sin β cos β)

(2 sin α cos α =sin 2α and 2sin β cos β=sin 2β) 

= 2 sin²β + sin 2α sin 2β – 4 sin²α sin²β + cos 2α cos 2β – sin 2α sin 2β 

(cos 2β = 1–2 sin²β)

= (1 – cos 2β) – (2 sin²α) (2 sin²β) + cos 2α cos 2β 

= (1 – cos 2β) – (1 – cos 2α) (1 – cos 2β) + cos 2α cos 2β 

= (1 – cos 2β)(1–1+cos 2α) + cos 2α cos 2β

= (1 – cos 2β)(cos 2α) + cos 2α cos 2β

= cos 2α – cos 2α cos 2β + cos 2α cos 2β

= cos 2α 

Problem 4: Prove that Sec8-1sec4-1=tan8tan2

Solution: LHS

Sec8-1sec4-1=1/cos8-11/cos4-1

=(1-cos8)cos4cos8(1-cos4)

(cos2 = 1 – sin²; sin² = 1 – cos2; sin²4 = 1 – cos8)

=2sin24cos4cos8sin22

=sin4(2sin4cos4)2cos8sin22

(sin2 = 2sincos; sin8 = 2sin4cos4)

=sin4sin82cos8sin22

=2sin2cos2sin8sin22cos8

(sincos=tan)

= tan8tan2