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Math
Posted 5 months ago
Given constants CC and DD, suppose that tanA\tan A and tanB\tan B are the solutions to x2+Cx+D=0x^{2}+C x+D=0, where tan(A+B)\tan (A+B) is defined.

Simplify sin2(A+B)+Csin(A+B)cos(A+B)+Dcos2(A+B)\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 sin2x+cos2x=1\sin^2 x + \cos^2 x = 1 to simplify the expression
step 2
Substitute sin2(A+B)+cos2(A+B)\sin^2(A+B) + \cos^2(A+B) with 1 in the given expression
step 3
The expression simplifies to 1+Csin(A+B)cos(A+B)+Dcos2(A+B)1 + C \sin(A+B) \cos(A+B) + D \cos^2(A+B)
step 4
Use the identity sin(2x)=2sinxcosx\sin(2x) = 2 \sin x \cos x to simplify Csin(A+B)cos(A+B)C \sin(A+B) \cos(A+B)
step 5
Substitute Csin(A+B)cos(A+B)C \sin(A+B) \cos(A+B) with C2sin(2(A+B))\frac{C}{2} \sin(2(A+B)) in the expression
step 6
The expression now is 1+C2sin(2(A+B))+Dcos2(A+B)1 + \frac{C}{2} \sin(2(A+B)) + D \cos^2(A+B)
step 7
Use the quadratic equation x2+Cx+D=0x^2 + Cx + D = 0 to express tan(A+B)\tan(A+B) in terms of CC and DD
step 8
Since tanA\tan A and tanB\tan B are roots, use Vieta's formulas: tanA+tanB=C1\tan A + \tan B = -\frac{C}{1} and tanAtanB=D1\tan A \cdot \tan B = \frac{D}{1}
step 9
Use the identity tan(A+B)=tanA+tanB1tanAtanB\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} to find tan(A+B)\tan(A+B)
step 10
Substitute tanA+tanB\tan A + \tan B and tanAtanB\tan A \cdot \tan B into the identity to get tan(A+B)=C1D\tan(A+B) = \frac{-C}{1-D}
step 11
Use the identity tan2x+1=sec2x\tan^2 x + 1 = \sec^2 x to relate tan(A+B)\tan(A+B) and cos2(A+B)\cos^2(A+B)
step 12
Substitute cos2(A+B)\cos^2(A+B) with 1sec2(A+B)\frac{1}{\sec^2(A+B)} in the expression
step 13
The expression now is 1+C2sin(2(A+B))+D(1sec2(A+B))1 + \frac{C}{2} \sin(2(A+B)) + D \left(\frac{1}{\sec^2(A+B)}\right)
step 14
Substitute sec2(A+B)\sec^2(A+B) with 1+tan2(A+B)1 + \tan^2(A+B) and then with 1+(C1D)21 + \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+C2sin(2(A+B))+D(11+(C1D)2)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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