Chapter 3 Class 11 Trigonometric Functions
Ex 3.1, 2 (i)
Ex 3.2, 7 Important
Ex 3.2, 8
Ex 3.2, 9 Important
Ex 3.3, 4
Ex 3.3, 5 (i) Important
Ex 3.3, 8 Important
Ex 3.3, 11 Important
Ex 3.3, 18 Important
Ex 3.3, 23 Important
Ex 3.3, 21 Important
Question 7 Important Deleted for CBSE Board 2025 Exams
Question 4 Important Deleted for CBSE Board 2025 Exams
Question 8 Important Deleted for CBSE Board 2025 Exams
Question 9 Important Deleted for CBSE Board 2025 Exams
Example 20 Important
Example 21 Important
Misc 4 Important
Misc 7 Important You are here
Chapter 3 Class 11 Trigonometric Functions
Last updated at April 16, 2024 by Teachoo
Misc 7 Prove that: sin 3x + sin2x – sin x = 4 sin x cos 𝑥/2 cos 3𝑥/2 Solving L.H.S sin 3x + sin 2x − sin x = sin 3x + (sin 2x – sin x) = sin 3x + 2cos ((2𝑥 + 𝑥)/2) . sin ((2𝑥−𝑥)/2) = sin 3x + 2 cos (𝟑𝒙/𝟐) sin 𝒙/𝟐 We know that sin 2x = 2 sin x cos x Divide by x by x/2 sin 2x/2 = 2 sin x/2 cos x/2 sin x = 2 sin x/2 cos x/2 Now Replace x by 3x sin 3x = 2 sin 𝟑𝐱/𝟐 cos 𝟑𝐱/𝟐 = 2 sin 3𝑥/2 cos 3𝑥/2 + ["2 cos " 3𝑥/2 " sin " 𝑥/2] = 2 cos 3𝑥/2 ["sin " 𝟑𝒙/𝟐 " + sin " 𝒙/𝟐] Using sin x + sin y = 2 sin (𝑥 + 𝑦)/2 cos (𝑥 − 𝑦)/2 Putting x = 3𝑥/2 & y = 𝑥/2 , = 2 cos 3𝑥/2 ["2 sin " ((3𝑥/2 " + " 𝑥/2))/2 " . cos " ((3𝑥/2 " − " 𝑥/2))/2] = 2 cos 3𝑥/2 ["2 sin " (((3𝑥 + 𝑥)/2))/2 " . cos " (((3𝑥 − 𝑥)/2))/2] = 2 cos 3𝑥/2 ["2 sin " ((4𝑥/2))/2 " . cos " ((2𝑥/2))/2] = 2 cos 3𝑥/2 ["2 sin " ((2𝑥/1))/2 " . cos " ((𝑥/1))/2] = 2 cos 𝟑𝒙/𝟐 ["2 sin " 𝟐𝒙/𝟐 " . cos " 𝒙/𝟐] = 2 cos 3𝑥/2 ["2 sin " 𝑥" . cos " 𝑥/2] = 4 cos 𝟑𝒙/𝟐 sin 𝒙 cos 𝒙/𝟐 = R.H.S Hence L.H.S = R.H.S Hence proved