+0

# Trigonometry - were did I go wrong?

+1
285
2
+183
 Q 5f no 6   If Sin Ø + Sin 2Ø = a, & Cos Ø + Cos 2Ø = b Prove that (a2 + b2) (a2 + b2-3) =2 b Sin Ø + Sin 2Ø = a Let c = Ø+2Ø = 3Ø and d = Ø-2Ø = -Ø c + d = 2Ø c - d = 4Ø   Therefore a = 2 Sin Ø. Cos 2Ø & a2 = 4 Sin2 Ø. Cos22 Ø Cos Ø + Cos 2Ø = b Let c = Ø+2Ø = 3Ø and d = Ø-2Ø = -Ø c + d = 2Ø c - d = 4Ø   Therefore b = 2 Cos Ø. Cos 2Ø & b2 = 4 Cos2 Ø. Cos22 Ø a2 + b2 = 4 Sin2 Ø. Cos22 Ø + 4 Cos2 Ø. Cos22 Ø =   4 Cos22 Ø (Sin2 Ø + Cos2 Ø) = 4 Cos22 Ø Therefore (a2 + b2) (a2 + b2-3) = 4 Cos22Ø (4 Cos22 Ø-3) = 16 Cos4 2Ø – 12 Cos2 2Ø =4 Cos2Ø (4Cos32Ø – 3Cos 2Ø) =4 Cos 2Ø. Cos 6Ø 2b = 4 Cos Ø. Cos 2Ø Conclusion: 4 Cos Ø. Cos 2Ø = 4 Cos 2Ø. Cos 6Ø   CosØ =  Cos 6Ø  (??????)
Feb 24, 2019

#1
+30086
+4

Prove as follows:

(Note: your expressions for a and b are incorrect).

Feb 24, 2019

#1
+30086
+4

Prove as follows:

(Note: your expressions for a and b are incorrect).

Alan Feb 24, 2019
#2
+183
+1

Thanks for that Alan.  I had been following this factor theory ... Need to trace the fundamental mistake!

Regards and I re-assert that you do a great job !

Feb 24, 2019