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Given constants CC and DD, suppose that tan⁡A\tan A and tan⁡B\tan B are the solution...
Mar 29, 2024
Given constants CC and DD, suppose that tan⁡A\tan A and tan⁡B\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 sin⁡2(A+B)+Csin⁡(A+B)cos⁡(A+B)+Dcos⁡2(A+B)\sin ^{2}(A+B)+C \sin (A+B) \cos (A+B)+D \cos ^{2}(A+B).
Solution by Steps
step 1
Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 to simplify the expression
step 2
Substitute sin⁡2(A+B)+cos⁡2(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)+Dcos⁡2(A+B)1 + C \sin(A+B) \cos(A+B) + D \cos^2(A+B)
step 4
Use the identity sin⁡(2x)=2sin⁡xcos⁡x\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))+Dcos⁡2(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 tan⁡A\tan A and tan⁡B\tan B are roots, use Vieta's formulas: tan⁡A+tan⁡B=−C1\tan A + \tan B = -\frac{C}{1} and tan⁡A⋅tan⁡B=D1\tan A \cdot \tan B = \frac{D}{1}
step 9
Use the identity tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} to find tan⁡(A+B)\tan(A+B)
step 10
Substitute tan⁡A+tan⁡B\tan A + \tan B and tan⁡A⋅tan⁡B\tan A \cdot \tan B into the identity to get tan⁡(A+B)=−C1−D\tan(A+B) = \frac{-C}{1-D}
step 11
Use the identity tan⁡2x+1=sec⁡2x\tan^2 x + 1 = \sec^2 x to relate tan⁡(A+B)\tan(A+B) and cos⁡2(A+B)\cos^2(A+B)
step 12
Substitute cos⁡2(A+B)\cos^2(A+B) with 1sec⁡2(A+B)\frac{1}{\sec^2(A+B)} in the expression
step 13
The expression now is 1+C2sin⁡(2(A+B))+D(1sec⁡2(A+B))1 + \frac{C}{2} \sin(2(A+B)) + D \left(\frac{1}{\sec^2(A+B)}\right)
step 14
Substitute sec⁡2(A+B)\sec^2(A+B) with 1+tan⁡2(A+B)1 + \tan^2(A+B) and then with 1+(−C1−D)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+(−C1−D)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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