Maximizing 3sin2x+8sin⁡xcos⁡x+9cos2x. What went wrong?Let f(x)=3sin2x+8sin⁡xcos⁡x+9cos2x. For some x∈[0,π2],f attains its maximum value, m. Compute m+100cos2xWhat I did was rewrite the equation as f(x)=6cos2x+8sin⁡xcos⁡x+3. Then I let a=&lt;6cos⁡x,8cos⁡x&gt; and b=&lt;cos⁡x,sin⁡x&gt;.Using Cauchy-Schwarz, I got that the maximum occurs when tan⁡x=43 and that the maximum value is 10cos⁡x. However, that produces a maximum of 9 for f(x), instead of the actual answer of 11.

Question

Maximizing 3sin2x+8sin⁡xcos⁡x+9cos2x. What went wrong?Let f(x)=3sin2x+8sin⁡xcos⁡x+9cos2x. For some x∈[0,π2],f attains its maximum value, m. Compute m+100cos2xWhat I did was rewrite the equation as f(x)=6cos2x+8sin⁡xcos⁡x+3. Then I let a=&amp;lt;6cos⁡x,8cos⁡x&amp;gt;  and  b=&amp;lt;cos⁡x,sin⁡x&amp;gt;.Using Cauchy-Schwarz, I got that the maximum occurs when tan⁡x=43 and that the maximum value is 10cos⁡x. However, that produces a maximum of 9 for f(x), instead of the actual answer of 11.

slaastro132z · Accepted Answer

f(x)=2sin2x+8sin⁡xcos⁡x+8cos2x+1=2(sin⁡x+2cos⁡x)2+1Let α∈[0,π2] and cos⁡α=25. Then sin⁡α=15 andsin⁡x+2cos⁡x=5(sin⁡xsin⁡α+cos⁡xcos⁡α)=5cos⁡(x−α)attaining its maximum when x=αSo, m=2(5)2+1=11 and m+100cos2α=11+100(45)=91

tralhavahr9c · Accepted Answer

Using the identitycos2x=1+cos⁡2x2and2sin⁡xcos⁡x=sin⁡2xthe question changes to finding minimum value of the function6+3cos⁡2x+4sin⁡2xAnd now using a standard result that the range of a function asin⁡α±bcos⁡α is [−a2+b2,a2+b2]Hence the range of the given expression becomes [1,11]

Maximizing \(\displaystyle{3}{{\sin}^{{2}}{x}}+{8}{\sin{{x}}}{\cos{{x}}}+{9}{{\cos}^{{2}}{x}}\). What went wrong? Let \(\displaystyle{f{{\left({x}\right)}}}={3}{{\sin}^{{2}}{x}}+{8}{\sin{{x}}}{\cos{{x}}}+{9}{{\cos}^{{2}}{x}}\).

Answered question

Answer & Explanation

New Questions in Precalculus