Ex 3.3, 16 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Dec. 13, 2024 by Teachoo
Ex 3.3
Ex 3.3, 1
Important
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Chapter 3 Class 11 Trigonometric Functions
Serial order wise
Transcript
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
