With reference to the right-angled triangle shown, what is the value of $\sin(A)\cos(B) + \cos(A)\sin(B)$?
To solve the given problem, we need to evaluate the expression: \(\sin(A)\cos(B) + \cos(A)\sin(B)\).
This expression resembles the formula for the sine of the sum of two angles. According to trigonometric identities:
\(\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)\)
Thus, the given expression is simply \(\sin(A + B)\).
In a right-angled triangle, the sum of angles \(A\) and \(B\) equals 90 degrees since the third angle is 90°.
\(A + B = 90^\circ\)
Therefore, \(\sin(A + B) = \sin(90^\circ)\).
We know from trigonometric values that:
\(\sin(90^\circ) = 1\)
Thus, the value of \(\sin(A)\cos(B) + \cos(A)\sin(B)\) is 1.
Therefore, the correct answer is 1.
Important Questions from Trigonometry (Notes)
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