Problem 98 - Algebra and Trigonometry

Question

Verify the identity: $$ \frac{\sin (x-y)}{\cos x \cos y}+\frac{\sin (y-z)}{\cos y \cos z}+\frac{\sin (z-x)}{\cos z \cos x}=0 $$

Step-by-Step Solution

Answer

The identity is verified as \( \sin(x)\cos(y) + \sin(y)\cos(z) + \sin(z)\cos(x) = 0 \) which is equivalent to \(- \sin(y + z + x) = 0\). Since the sum of all angles in a triangle is \(\pi\) radians and \(\sin(\pi) = 0\), the entire expression is equal to 0.

Step 1: Break Up Each Term

Apply the difference of angles identity \( \sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b) \) to each term in the function. We get: \[ (\sin(x)\cos(y) - \cos(x)\sin(y))/\cos(x)\cos(y) + (\sin(y)\cos(z) - \cos(y)\sin(z))/\cos(y)\cos(z) + (\sin(z)\cos(x) - \cos(z)\sin(x))/\cos(z)\cos(x) \]

Step 2: Simplifying Each Term

After applying the identity, simplify each term by cancelling \( \cos(x) \), \( \cos(y) \), \( \cos(z) \) out of numerator and denominator. After cancelling with get: \( \sin(x) - \tan(y)\cos(x) + \sin(y) - \tan(z)\cos(y)+ \sin(z) - \tan(x)\cos(z) = 0 \)

Step 3: Grouping Terms

Rearrange and group the similar terms. We get: \( \sin(x)(1 - \tan(y)) + \sin(y)(1 - \tan(z))+ \sin(z)(1 - \tan(x)) = 0 \)

Step 4: Applying Tangent Identity

Since the tangent of an angle is defined as the ratio of the sine of the angle and the cosine of the angle, we can use the identity \( 1 - \tan(\theta) = \cos(\theta) \). After substituting, we get \( \sin(x)\cos(y) + \sin(y)\cos(z) + \sin(z)\cos(x) = 0 \)

Step 5: Verifying with the Sum of angle Identity

Observe that the right side of the expression is identical to \(- \sin(y + z + x)\). Since the sum of all angles in a triangle is \(180^o\) or \(\pi\) radians and \(\sin(\pi) = 0\), the entire expression is equal to 0. The identity is verified.