Check sibling questions

Misc 6 - Prove sin 7x + sin 5x + sin 9x + sin 3x - Class 11

Misc 6 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Misc 6 - Chapter 3 Class 11 Trigonometric Functions - Part 3

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

Misc 6 Prove that ((sin⁑〖7π‘₯ + sin⁑〖5π‘₯) + (sin⁑〖9π‘₯ + sin⁑〖3π‘₯)γ€— γ€— γ€— γ€—)/((cos⁑〖7π‘₯ + π‘π‘œπ‘  5π‘₯) + (cos⁑〖9π‘₯ + cos⁑〖3π‘₯)γ€— γ€— γ€— ) = tan 6x Solving L.H.S ((sin⁑〖7π‘₯ + sin⁑〖5π‘₯) + (sin⁑〖9π‘₯ + sin⁑〖3π‘₯)γ€— γ€— γ€— γ€—)/((cos⁑〖7π‘₯ + π‘π‘œπ‘  5π‘₯) + (cos⁑〖9π‘₯ + cos⁑〖3π‘₯)γ€— γ€— γ€— ) Lets solve numerator and Denominator separately Solving numerator (sin 7x + sin 5x) + ( sin 9x + sin 3x) = 2 sin ((7π‘₯ + 5π‘₯)/2) cos ((7π‘₯ βˆ’ 5π‘₯)/2) + 2sin ((9π‘₯ +3π‘₯)/2) cos ((9π‘₯ βˆ’3π‘₯)/2) = 2 sin (12π‘₯/2) . cos (2π‘₯/2) + 2sin (12π‘₯/2) cos (6π‘₯/2) = 2 sin 6x . cos x + 2 sin 6x . cos 3x = 2sin 6x (cos x + cos 3x) Now solving Denominator (cos 7x + cos 5x) + (cos 9x + cos 3x) = 2 cos ((7π‘₯ + 5π‘₯)/2) cos ((7π‘₯ βˆ’ 5π‘₯)/2) + 2 cos ((9π‘₯ +3π‘₯)/2) cos ((9π‘₯ βˆ’3π‘₯)/2) = 2 cos (12π‘₯/2) . cos (2π‘₯/2) + 2 cos (12π‘₯/2) cos (6π‘₯/2) = 2 cos 6x . cos x + 2 cos 6x . cos 3x = 2 cos 6x (cos x + cos 3x) Now solving L.H.S ((sin⁑〖7π‘₯ + sin⁑〖5π‘₯) + (sin⁑〖9π‘₯ + sin⁑〖3π‘₯)γ€— γ€— γ€— γ€—)/((cos⁑〖7π‘₯ + π‘π‘œπ‘  5π‘₯) + (cos⁑〖9π‘₯ + cos⁑〖3π‘₯)γ€— γ€— γ€— ) = (2 sin⁑6π‘₯ (cos⁑π‘₯ + cos⁑3π‘₯))/(2 cos⁑6π‘₯ (cos⁑π‘₯ + cos⁑3π‘₯)) = (sin⁑6π‘₯ )/(cos⁑6π‘₯ ) = (sin⁑6π‘₯ )/(cos⁑6π‘₯ ) = tan 6x = R.H.S Hence L.H.S = R.H.S Hence proved

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.