Trigonometry Problem: Finding A - B
We are given the equation: $ \sin(3A - 20^\circ) = \cos(20^\circ - 3B) $ We need to find the value of $A - B$.
Using Trigonometric Identities
Recall the trigonometric identity: $ \sin(\theta) = \cos(90^\circ - \theta) $ Applying this identity to the left side of the given equation, we can rewrite $\sin(3A - 20^\circ)$ as $\cos(90^\circ - (3A - 20^\circ))$. $ \sin(3A - 20^\circ) = \cos(90^\circ - 3A + 20^\circ) = \cos(110^\circ - 3A) $
Now, substitute this back into the original equation: $ \cos(110^\circ - 3A) = \cos(20^\circ - 3B) $
Solving for A - B
If $\cos(x) = \cos(y)$, then a general relationship is $x = y$ or $x = -y$ (considering the principal values or basic angle relationships for simplicity in typical exam contexts). A more direct relationship derived from $\sin(X) = \cos(Y)$ is $X + Y = 90^\circ$. Let $X = 3A - 20^\circ$ and $Y = 20^\circ - 3B$. Therefore, we have: $ (3A - 20^\circ) + (20^\circ - 3B) = 90^\circ $
Simplify the equation: $ 3A - 20^\circ + 20^\circ - 3B = 90^\circ $ $ 3A - 3B = 90^\circ $ Factor out 3: $ 3(A - B) = 90^\circ $ Divide by 3 to find the value of $A - B$: $ A - B = \frac{90^\circ}{3} $ $ A - B = 30^\circ $
The value of $A - B$ is $30^\circ$.
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