Ex 3.3, 16 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Dec. 13, 2024 by Teachoo
Last updated at Dec. 13, 2024 by Teachoo
Ex 3.3, 16 Prove that πππ β‘γ9π₯ βγ πππ γβ‘5π₯ γ/(π ππ 17π₯ β π ππβ‘3π₯ ) =βπ ππβ‘γ2π₯ γ/πππ β‘10π₯ Solving L.H.S πππ β‘γ9π₯ βγ πππ γβ‘5π₯ γ/(π ππ 17π₯ β π ππβ‘3π₯ ) We solve cos 9x β cos 5x & sin 17x β sin 3x seperately cos 9x β cos 5x = β 2 sin ((9x+5x)/2) sin((9xβ5x)/2) = β 2 sin (14π₯/2) sin (4π₯/2) = β 2 sin 7x sin (2x) sin 17x β sin 3x = 2 cos ((17x+3x)/2) sin((17xβ3x)/2) = 2 cos (20π₯/2) sin (14π₯/2) = 2 cos 10x sin 7x Now, πππ β‘γ9π₯ βγ πππ γβ‘5π₯ γ/(π ππ 17π₯ β π ππβ‘3π₯ ) = (βπ γπ¬π’π§ γβ‘γ(ππ±)γπ¬π’π§ γβ‘γ(ππ±)γ γ)/(π πππβ‘γ(πππ±)π¬π’π§β‘γ (ππ±)γ γ ) = γβsinγβ‘γ(2x)γ/πππ β‘γ(10x)γ = R.H.S So, L.H.S. = R.H.S. Hence proved
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo