Trigonometric Identity Application
The question asks us to find the value of $x + y$ given the equation $\sin(5x - 40^\circ) = \cos(5y + 40^\circ)$. To solve this, we need to use a fundamental trigonometric identity that relates the sine and cosine functions.
Using the Sine-Cosine Relationship
We know that for any two angles $A$ and $B$, if $\sin(A) = \cos(B)$, then the angles are complementary (or related in a way that their sum is $90^\circ$, possibly plus multiples of $360^\circ$). A common form of this identity is:
$$ \sin(A) = \cos(B) \implies A + B = 90^\circ $$
This identity holds true when $A$ and $B$ are complementary angles.
Applying the Identity to the Given Equation
In our problem, we have:
- $A = 5x - 40^\circ$
- $B = 5y + 40^\circ$
Substitute these into the identity $A + B = 90^\circ$:
$$ (5x - 40^\circ) + (5y + 40^\circ) = 90^\circ $$
Solving for x + y
Now, let's simplify the equation:
- Combine the terms involving $x$ and $y$: $$ 5x + 5y - 40^\circ + 40^\circ = 90^\circ $$
- The constant terms cancel out: $$ 5x + 5y + 0^\circ = 90^\circ $$ $$ 5x + 5y = 90^\circ $$
- Factor out 5 from the left side: $$ 5(x + y) = 90^\circ $$
- To find the value of $x + y$, divide both sides by 5: $$ x + y = \frac{90^\circ}{5} $$ $$ x + y = 18^\circ $$
Conclusion
Therefore, the value of $x + y$ is $18^\circ$. This corresponds to the first option provided.
Similar Questions
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