jee main 2021 crash course - Amazon AWS

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JEE MAIN 2021 CRASH COURSE


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Observations

1. (a) cos(2n𝜋) =

(b) cos((2n + 1)𝜋) =

(c) sin(n𝜋) =

2. (a) cos(-θ) =

(b) sin(-θ)=

(cos θ, sin θ)

(0, 1)

(1, 0)

(0, -1)

(-1, 0)

O X

Y

P

θ

1. Sine of supplementary angles are same.

2. Cosines of supplementary angles are negative of each other.

3. sin(-θ) = -sinθ and cos(-θ) = cosθ

Remarks

Compound Angles and Multiple Angle Formulae

1. sin(A + B) = sinA cosB + cosA sinB

2. sin(A - B) = sinA cosB - cosA sinB

3. cos(A + B) = cosA cosB - sinA sinB

4. cos(A - B) = cosA cosB + sinA sinB

Special Cases

These two special cases are very important.

Remark

1. sin 2θ and cos 2θ can be expressed in terms of tanθ as

2. sin(A + B) × sin(A - B) = sin2A - sin2B

cos(A + B) × cos(A - B) = cos2A - sin2B = cos2B - sin2A

List of most commonly used formulae and expressions.

If m tan(θ - 30°) = n tan (θ + 120°), then prove that

Solution :

Prove that cos A cos(60° - A) cos(60° + A)

Solution :

Prove the following results.(a) tan3A tan2A tanA = tan3A - tan2A - tanA(b) If A + B + C = 𝜋, then prove that tanA tanB tanC = tanA + tanB + tanC

Prove the following results.(a) tan3A tan2A tanA = tan3A - tan2A - tanA

Solution :

Prove the following results.(b) If A + B + C = 𝜋, then prove that tanA tanB tanC = tanA + tanB + tanC

Solution :

Prove that tan 71° tan 64° = 1 + tan 71° + tan 64°

Solution :

a sin θ + b cos θ

a sin θ + b cos θ

Expressing in terms of cosine only.

Expressing in terms of sine only.

2 4A B C D

IIT 1988

Solution :

Transformation of product into sum and difference

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

2 cos A sin B = 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)

Transformation Formulae

Transformation of sum and difference into product

Transformation Formulae

Show that

Solution :

If sinx + sin y = a and cos x + cos y = b, then evaluate

Solution :

Recall:

If sin θ = n sin(θ + 2⍺), then tan(θ + ⍺) = _____.

A B C D

Solution :

Result

cos12° cos24° cos36° cos48° cos72° cos84° =_______.

A B C D

Solution :

Prove that sin 10° sin 50° sin 70°

Solution :

Alternate Solution - ISolution :

Solution : Alternate Solution - II

Two Trigonometric Series

1. sin(a) + sin(a + d) + sin(a + 2d) +...+ sin(a + (n - 1)d)

2. cos(a) + cos(a + d) + cos(a + 2d) +...+ cos(a +(n - 1)d)

A B C D

Solution :

There are some identities which hold true only under some conditions.

Conditional Identities

If A + B + C = 𝜋, then :

(a) sin2A + sin2B + sin2C = 4sinA sinB sinC

(b) cos2A + cos2B + cos2C = -1 -4 cosA cosB cosC

(d) tanA + tanB + tanC = tanA tanB tanC

Result :

Trigonometric Equations


Observations(a) sinθ = 0 ⇒ θ =

(b) cosθ = 0 ⇒ θ =

(c) cosθ = 1 ⇒ θ =

(d) cosθ = -1 ⇒ θ =


Results(a) (i) sinθ = sin⍺ ⇒ θ = n𝜋 + (-1)n ⍺

(ii) cosθ = cos⍺ ⇒ θ = 2n𝜋 ± ⍺

(iii) tanθ = tan⍺ ⇒ θ = n𝜋 + ⍺

(b) (i) sin2θ = sin2⍺

(ii) cos2θ = cos2⍺ ⇒ θ = n𝜋 ± ⍺

(iii) tan2θ = tan2⍺

Solve the following equations.(a) cosx + cos2x + cos4x = 0, x ∈ [0, 𝜋]

Solution :

Solve the following equations.(b) cos4x = sinx

Solution :

Find the general solution of the following equation.cosx - sinx = 1

Solution :

Find the number of integral values of k for which the following equation has solutions. 7 cosx + 5sinx = 2k + 1

Solution :

Solve for x if sin2 x - 7 sin x cos x + 12 cos2 x = 0

Solution :

Lets Chakk de

y(1 - x) = 1 y(1 + x) = 1 x(1 - y) = 1x(1 + y) = 1A B C D

09-01-2020-Evening Shift

Solution :

Let ⍺ and β be two real roots of the equation (K + 1)tan2 tan x = (1 - K), where K ≠ 1 and λ are real numbers. If tan2 (⍺ + β) = 50, then value of λ is:

510A B C D

07-01-2020-Morning Shift

510A B C D


Let ⍺ and β be two real roots of the equation (K + 1)tan2 tan x = (1 - K), where K ≠ 1 and λ are real numbers. If tan2 (⍺ + β) = 50, then value of λ is:

Solution :

The value of sin10° sin30° sin50° sin70° is:

A B C D


Solution :

The maximum value of for any real value of θ is:

A B C D


A B C D


The maximum value of for any real value of θ is:

Solution :

A B C D


Solution :

Let S = {θ ∈ [-2𝜋, 2𝜋] : 2 cos2 θ + 3 sinθ = 0} then the sum of the elements of S is:

𝜋2𝜋A B C D


Solution :

If then the number of values of x for which

sinx - sin2x + sin 3x = 0, is

3 1 24A B C D


3 1 24A B C D


If then the number of values of x for which

sinx - sin2x + sin 3x = 0, is

Solution :

11th class Leaderboard

Simran Mondal Madhav Lata

Ahmad nayeem 23 Shivay

Harshit kumar Divyansh Srivastava

Sudhanshu Pandey Zapros WB

Himanshu

❖ Cover entire JEE Main Syllabus (as per the new pattern) with

India’s Best Teachers in 45 Sessions

❖ Solve unlimited doubts with Doubt experts on our Doubt App

from till Exam 8 AM to 11 PM

❖ 1 Batch Classes will be over in 10 Weeks

❖ 10 full and 10 Part syllabus test to make you exam ready

❖ Amazing tricks & tips to crack JEE Main 2021 Questions in

power-packed 90 Min sessions



Lightning Deal: ₹10000 ₹8000/-

Batch Starts From : Every Monday

*Crash Course Link Available in Description

Apply Coupon Code: AVKCC

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