If sinA+sinB+sinC=cosA+cosB+cosC=0, prove that sin2A+sin2B+sin2C=cos2A+cos2B+cos2C=3/2.
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Step-by-Step Solution
Step 1: Set up the given equations
We are given two equations. Let's label them as equation (1) and equation (2) for easy reference. These equations state that the sum of the sines of angles A, B, and C is 0, and similarly, the sum of their cosines is also 0.
Step 2: Square and add the equations
To relate the given sums to sums of squares, we square both equation (1) and equation (2). Then, we add the squared equations together. This step is crucial because it will allow us to use the identity sin2x+cos2x=1.
Step 3: Expand and simplify using trigonometric identity
We expand the squared terms using the formula (x+y+z)2=x2+y2+z2+2xy+2yz+2zx. Then, we group the sin2θ+cos2θ terms, which simplify to 1. We also use the cosine addition formula cos(X−Y)=cosXcosY+sinXsinY to simplify the cross-product terms.
Step 4: Simplify the equation
After simplifying the sum of the sin2θ+cos2θ terms to 3, we rearrange the equation to isolate the sum of the cosine difference terms. This gives us a key intermediate result.
Step 5: Express sin2A+sin2B+sin2C in terms of cosines
We use the identity sin2x=1−cos2x for each term in the expression sin2A+sin2B+sin2C. This allows us to express the sum of sine squares in terms of the sum of cosine squares.
Step 6: Calculate sin2A+sin2B+sin2C
From equation (1), we have sinA+sinB=−sinC. Squaring both sides gives sin2A+sin2B+2sinAsinB=sin2C. Rearranging, we get sin2A+sin2B−sin2C=−2sinAsinB. Similarly, from equation (2), we get cos2A+cos2B−cos2C=−2cosAcosB. Adding these two results and using sin2x+cos2x=1 and cos(A−B)=cosAcosB+sinAsinB, we find cos(A−B)=−1/2. By symmetry, cos(B−C)=−1/2 and cos(C−A)=−1/2.
Step 7: Substitute back and find the final result
We use the property that if x+y+z=0 and ∣x∣=∣y∣=∣z∣=1, then x2+y2+z2=0. By letting x=eiA, y=eiB, z=eiC, we have x+y+z=(cosA+cosB+cosC)+i(sinA+sinB+sinC)=0+i(0)=0. Thus, ei2A+ei2B+ei2C=0. Expanding this, we get (cos2A+cos2B+cos2C)+i(sin2A+sin2B+sin2C)=0. This implies cos2A+cos2B+cos2C=0. Using the double angle formula cos2θ=2cos2θ−1, we substitute and solve for cos2A+cos2B+cos2C. Similarly, using cos2θ=1−2sin2θ, we solve for sin2A+sin2B+sin2C.