If cos(β - γ) + cos(γ - α) + cos(α - β) = -3/2, then (A) A is false and B is true (B) both A and B a

Q: Why did we multiply by 2? It seems like an extra step.

It clears the fraction, but more importantly, it creates the coefficient $2$ needed for the algebraic identity $2ab$. Without that $2$, we wouldn't be able to complete the square.

Q: How was I supposed to guess that I needed to add 3?

It is the 'Magic 3' trick. When you see a sum of cosines equal to a constant, look for ways to complete the square. Since we have three angles, we need three pairs of $\cos^2 \theta + \sin^2 \theta$ to balance the equation.

Q: Why does $X^2 + Y^2 = 0$ imply $X=0$ and $Y=0$?

Because we are in the domain of real numbers. Squares of real numbers are always non-negative. If the sum of two non-negative numbers is zero, each must individually be zero.

Question

If $\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = -3/2$, then

Options:

(A) A is false and B is true

(B) both A and B are true

(C) both A and B are false

(D) A is true and B is false

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the trigonometric equation:

$$\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = -\frac{3}{2}$$

To simplify the expression, we first clear the fraction by multiplying the entire equation by $2$:

$$2[\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta)] = -3$$

### The Expansion Strategy

Next, we invoke the compound angle identity, $\cos(A - B) = \cos A \cos B + \sin A \sin B$. Expanding each term in the equation yields:

$$2[\cos \beta \cos \gamma + \sin \beta \sin \gamma + \cos \gamma \cos \alpha + \sin \gamma \sin \alpha + \cos \alpha \cos \beta + \sin \alpha \sin \beta] = -3$$

We then move the $-3$ to the left side of the equation to set it to zero:

$$2(\cos \beta \cos \gamma + \cos \gamma \cos \alpha + \cos \alpha \cos \beta) + 2(\sin \beta \sin \gamma + \sin \gamma \sin \alpha + \sin \alpha \sin \beta) + 3 = 0$$

### The Magic 3 Transformation

To reveal the hidden structure, we split the constant $3$ into $1 + 1 + 1$. We then substitute each $1$ with the Pythagorean identity $\cos^2 	heta + \sin^2 	heta$ for each respective angle $\alpha, \beta,$ and $\gamma$:

$$(\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) + (\cos^2 \gamma + \sin^2 \gamma) + 2(\cos \beta \cos \gamma + \cos \gamma \cos \alpha + \cos \alpha \cos \beta) + 2(\sin \beta \sin \gamma + \sin \gamma \sin \alpha + \sin \alpha \sin \beta) = 0$$

By grouping the cosine and sine terms separately, we recognize the algebraic identity $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$. The equation collapses into a perfect square form:

$$(\cos \alpha + \cos \beta + \cos \gamma)^2 + (\sin \alpha + \sin \beta + \sin \gamma)^2 = 0$$

### Final Conclusion

Since we are working with real numbers, the sum of two squares can only equal zero if each individual square is zero. This leads us to the following system of equations:

$$\cos \alpha + \cos \beta + \cos \gamma = 0$$

$$\sin \alpha + \sin \beta + \sin \gamma = 0$$

These conditions represent the fundamental solution to the problem, demonstrating that the sum of the vectors $(\cos \alpha, \sin \alpha)$, $(\cos \beta, \sin \beta)$, and $(\cos \gamma, \sin \gamma)$ must be the **zero vector**.

Animated Solution for Mathematics - Trigonometry: If $cos (β - γ) + cos (γ - α) + cos (α - β) = - 3/2$ , then

Select Answer:

Visualized Solution

The Given Equation

Clearing the Fraction

Expanding the Cosine Differences

The Expanded Form

Rearranging the Equation

The Magic Three Trick

Substituting the Ones

Grouping Cosine Terms

Grouping Sine Terms

Recognizing the Perfect Square

Condensing the Equation

The Zero Sum Condition

Final Answer

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Expansion Strategy

The Magic 3 Transformation

Final Conclusion

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If cos(β−γ)+cos(γ−α)+cos(α−β)=−3/2, then