Trigonometric Expression Simplification 2
Given constants $ C $ and $ D, $ suppose that $ \tan A $ and $ \tan B $ are the solutions to
\[x^2 + Cx + D = 0,\]where $ \tan (A + B) $ is defined.
Simplify
\[\sin^2 (A + B) + C \sin (A + B) \cos (A + B) + D \cos^2 (A + B).\]Your expression should contain only one of the variables $ A, $ $ B, $ $ C, $ and $ D $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
By Vieta's formulas, $ \tan A + \tan B = -C $ and $ \tan A \tan B = D $. Then from the angle addition formula,
\[\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = -\frac{C}{1 - D}.\]We write the expression we are interested in, in terms of $ \tan (A + B) $:
\begin{align*}
&\sin^2 (A + B) + C \sin (A + B) \cos (A + B) + D \cos^2 (A + B) \\
&= \cos^2 (A + B) \tan^2 (A + B) + C \cos^2 (A + B) \tan (A + B) + D \cos^2 (A + B) \\
&= \cos^2 (A + B) (\tan^2 (A + B) + C \tan (A + B) + D) \\
&= \frac{\cos^2 (A + B)}{\sin^2 (A + B) + \cos^2 (A + B)} (\tan^2 (A + B) + C \tan (A + B) + D) \\
&= \frac{1}{\tan^2 (A + B) + 1} \cdot (\tan^2 (A + B) + C \tan (A + B) + D).\end{align*}
Then
\begin{align*}
&\frac{1}{\tan^2 (A + B) + 1} \cdot (\tan^2 (A + B) + C \tan (A + B) + D) \\
&= \frac{1}{(-\frac{C}{1 - D})^2 + 1} \cdot \left( \left( -\frac{C}{1 - D} \right)^2 - C \cdot \frac{C}{1 - D} + D \right) \\
&= \frac{(1 - D)^2}{(1 - D)^2 + C^2} \cdot \frac{D (C^2 + (1 - D)^2)}{(1 - D)^2} \\
&= \boxed{D}.\end{align*}
Freeflows