Question

Math

Posted 5 months ago

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Given constants $C$ and $D$, suppose that $\tan A$ and $\tan B$ are the solutions to $x^{2}+C x+D=0$, where $\tan (A+B)$ is defined.
Simplify $\sin ^{2}(A+B)+C \sin (A+B) \cos (A+B)+D \cos ^{2}(A+B)$.
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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

Use the identity $\sin^2 x + \cos^2 x = 1$ to simplify the expression

step 2

Substitute $\sin^2(A+B) + \cos^2(A+B)$ with 1 in the given expression

step 3

The expression simplifies to $1 + C \sin(A+B) \cos(A+B) + D \cos^2(A+B)$

step 4

Use the identity $\sin(2x) = 2 \sin x \cos x$ to simplify $C \sin(A+B) \cos(A+B)$

step 5

Substitute $C \sin(A+B) \cos(A+B)$ with $\frac{C}{2} \sin(2(A+B))$ in the expression

step 6

The expression now is $1 + \frac{C}{2} \sin(2(A+B)) + D \cos^2(A+B)$

step 7

Use the quadratic equation $x^2 + Cx + D = 0$ to express $\tan(A+B)$ in terms of $C$ and $D$

step 8

Since $\tan A$ and $\tan B$ are roots, use Vieta's formulas: $\tan A + \tan B = -\frac{C}{1}$ and $\tan A \cdot \tan B = \frac{D}{1}$

step 9

Use the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ to find $\tan(A+B)$

step 10

Substitute $\tan A + \tan B$ and $\tan A \cdot \tan B$ into the identity to get $\tan(A+B) = \frac{-C}{1-D}$

step 11

Use the identity $\tan^2 x + 1 = \sec^2 x$ to relate $\tan(A+B)$ and $\cos^2(A+B)$

step 12

Substitute $\cos^2(A+B)$ with $\frac{1}{\sec^2(A+B)}$ in the expression

step 13

The expression now is $1 + \frac{C}{2} \sin(2(A+B)) + D \left(\frac{1}{\sec^2(A+B)}\right)$

step 14

Substitute $\sec^2(A+B)$ with $1 + \tan^2(A+B)$ and then with $1 + \left(\frac{-C}{1-D}\right)^2$

step 15

Simplify the expression to find the final simplified form

Answer

The final simplified form of the expression is $1 + \frac{C}{2} \sin(2(A+B)) + D \left(\frac{1}{1 + \left(\frac{-C}{1-D}\right)^2}\right)$.

Key Concept

Trigonometric identities and Vieta's formulas

Explanation

The solution involves using trigonometric identities to simplify the given expression and applying Vieta's formulas to relate the sum and product of the roots of the quadratic equation to the coefficients.

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