If sin A + sin B + sin C=cosA+cosB+cosC=0 prove that sin²A + sin²B + sin² C=cos2A+cos2B+cos2C=3/2.
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Step-by-Step Solution
Step 1: Square the given equations
We are given two equations: sinA+sinB+sinC=0 and cosA+cosB+cosC=0. To introduce squared terms, we square both of these equations.
Step 2: Expand the squared equations
Expanding the squared sums (x+y+z)2=x2+y2+z2+2(xy+yz+zx), we get two new equations. Let's call them equation (1) and equation (2).
Step 3: Add the expanded equations
Now, we add equation (1) and equation (2). We group the terms sin2X+cos2X and the product terms.
Step 4: Apply trigonometric identities
Using the identity sin2X+cos2X=1 and the cosine addition formula cos(X−Y)=cosXcosY+sinXsinY, we simplify the sum. This gives us equation (3).
Step 5: Isolate the sum of cosines
From equation (3), we can isolate the sum of the cosine difference terms.
Step 6: Substitute back into expanded equations
From equations (1) and (2), we can express sin2A+sin2B+sin2C and cos2A+cos2B+cos2C in terms of the product sums.
Step 7: Final Proof
We know that 2(sinAsinB+sinBsinC+sinCsinA)=(sinA+sinB+sinC)2−(sin2A+sin2B+sin2C). Since (sinA+sinB+sinC)2=0, we have 2(sinAsinB+sinBsinC+sinCsinA)=−(sin2A+sin2B+sin2C). Substituting this into equation (1) gives 2(sin2A+sin2B+sin2C)=3. Thus, sin2A+sin2B+sin2C=23. Similarly, for the cosine terms, we can use the identity 2(cosAcosB+cosBcosC+cosCcosA)=(cosA+cosB+cosC)2−(cos2A+cos2B+cos2C). Since (cosA+cosB+cosC)2=0, we have 2(cosAcosB+cosBcosC+cosCcosA)=−(cos2A+cos2B+cos2C). Substituting this into equation (2) gives 2(cos2A+cos2B+cos2C)=3. Thus, cos2A+cos2B+cos2C=23.