Ex 3.3

Chapter 3 Class 11 Trigonometric Functions
Serial order wise

### Transcript

Ex 3.3, 6 Prove that: cos (Ο/4βπ₯) cos (Ο/4βπ¦) β sin (Ο/4βπ₯) sin (Ο/4βπ¦) = sinβ‘(π₯ + π¦) Solving L.H.S We know that cos (A + B) = cos A cos B β sin A sin B The equation given in Question is of this form Where A = (π/4 βπ₯) B = (π/4 βπ¦) Hence cos (Ο/4βπ₯) cos (Ο/4βπ¦) β sin (Ο/4βπ₯) sin (Ο/4βπ¦) = cos [(π/πβπ)" " +(π/π βπ)] = cos [Ο/4βπ₯+Ο/4 βπ¦] = cos [Ο/4+Ο/4βπ₯βπ¦] = cos [Ο/4+Ο/4βπ₯βπ¦] = cos [π/π " " β(π+π)] Putting Ο = 180Β° = cos [(180Β°)/2β(π₯+π¦)] = cos [90Β° β(π₯+π¦)] = sin (π+π) = R.H.S Hence proved = cos [(π/πβπ)" " +(π/π βπ)] = cos [Ο/4βπ₯+Ο/4 βπ¦] = cos [Ο/4+Ο/4βπ₯βπ¦] = cos [Ο/4+Ο/4βπ₯βπ¦] = cos [π/π " " β(π+π)] Putting Ο = 180Β° = cos [(180Β°)/2β(π₯+π¦)] = cos [90Β° β(π₯+π¦)] = sin (π+π) = R.H.S Hence proved