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Misc 5 - Chapter 3 Class 11 Trigonometric Functions
Last updated at Dec. 16, 2024 by Teachoo
Chapter 3 Class 11 Trigonometric Functions
Serial order wise
- Ex 3.1
- Ex 3.2
- Ex 3.3
- Examples
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Miscellaneous
- Principal and General Solutions
Transcript
Misc 5 Prove that: sin π₯ + sin 3π₯ + sin5π₯ + sin 7π₯ = 4cos π₯ cos 2π₯ sin 4π₯ Solving LHS sin π₯ + sin 3π₯ + sin5π₯ + sin 7π₯ = (π¬π’π§β‘π+πππ ππ)+(π¬π’π§β‘ππ+πππβ‘ππ) = 2 sin ((π₯ + 5π₯)/2) .cos ((π₯ β 5π₯)/2) + 2sin ((3π₯ + 7π₯)/2) cos ((3π₯ β 7π₯)/2) = 2 sin 3π₯ cos (β2π₯) + 2sin 5π₯ cos (β2π₯) = 2 sin 3π₯ cos 2π₯ + 2sin 5π₯ cos 2π₯ = 2 cos 2π₯ [ sin 3π₯ + sin 5π₯] = 2cos 2π₯ ("2sin " ((3π₯ + 5π₯)/2)" . cos" ((3π₯ β 5π₯)/2)) = 2cosβ‘2π₯ [2 sinβ‘4π₯.cosβ‘γ(βπ₯)γ] = π ππ¨π¬β‘ππ π¬π’π§β‘ππ ππ¨π¬β‘π = RH.S. = 2 cos 2π₯ [ sin 3π₯ + sin 5π₯] = 2cos 2π₯ ("2sin " ((3π₯ + 5π₯)/2)" . cos" ((3π₯ β 5π₯)/2)) = 2cosβ‘2π₯ [2 sinβ‘4π₯.ππ¨π¬β‘γ(βπ)γ] = π ππ¨π¬β‘ππ π¬π’π§β‘ππ ππ¨π¬β‘π = RH.S. Hence , L.H.S.= R.H.S. Hence proved
