greatest value of 9cos^2x+4sin^2x is

• 9cos^2x+4sin^2x
  9(1 - sin^2x) + 4sin^2x
  9 - 9sin^2x + 4sin^2x
  9 - 5sin^2x
  Sin^2nx ranges from 0 < Sin^2nx < 1
  (ie) Min value = 0, Max value = 1
  
  So max value = 9 - 5*0 = 9
  
• 10 years agoHelpfull: Yes(15) No(1)
• If Y = 9cos^2x + 4sin^2x
  = 5cos^2x + 4cos^2x + 4sin^2x
  = 5cos^2x + 4(cos^2x + sin^2x)
  = 5cos^2x + 4
  
  Y = (4 + 5cos^2x)
  maxm value of cos^2x is 1
  so, Y max = 4 + 5*1 = 9
• 10 years agoHelpfull: Yes(11) No(0)
• 9cos^2x+4sin^2x
  9(1 - sin^2x) + 4sin^2x
  9 - 9sin^2x + 4sin^2x
  9 - 5sin^2x
  Sin^2nx ranges from 0 < Sin^2nx < 1
  (ie) Min value = 0, Max value = 1
  
  So 9 - 5(1) = 4
  
  Ans : 4
• 10 years agoHelpfull: Yes(2) No(6)
• 9cos^2x+4sin^2x=5cos^2x+4(sin^2x+cos^2x)=5cos^2x+4=5+4(since max value of cos=1)=9
• 10 years agoHelpfull: Yes(1) No(0)
• 9cos^2x+4sin^2x
  9(1 - sin^2x) + 4sin^2x
  9 - 9sin^2x + 4sin^2x
  9 - 5sin^2x
  Sin^2nx ranges from 0
• 10 years agoHelpfull: Yes(0) No(1)
• 9cos^2x+4sin^2x
  9(1 - sin^2x) + 4sin^2x
  9 - 9sin^2x + 4sin^2x
  9 - 5sin^2x
  Because Sin^2nx varies from 0 < Sin^2nx < 1
  hence min{sin^x}=0
  i.e. max value of given expression is 9
  
  
  
• 10 years agoHelpfull: Yes(0) No(0)
• 9cos^2x+4sin^2x
  =(5 cos^2x + 4cos^2x) + 4sin^2x
  =5cos^2x+ 4 ( cos^2x+sin^2x)
  =5cos^2x+ 4*1
  now, max value of cos@=1
  so max value= 5*1+4 = 9 ANS
• 10 years agoHelpfull: Yes(0) No(0)