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if n is an integer, ( n cdot 180^circ ) represents an integer multiple …

Question

if n is an integer, ( n cdot 180^circ ) represents an integer multiple of ( 180^circ ) and ( (2n + 1) cdot 90^circ ) represents an odd integer multiple of ( 90^circ ), and so on. determine whether the expression is equal to 0, 1, −1, or is undefined. ( sinleft(4n + 3) cdot 90^circ
ight ) ( sinleft(4n + 3) cdot 90^circ
ight ) with options: = 1, is undefined, = −1, = 0

Explanation:

Step1: Expand the angle

First, expand the angle inside the sine function: \((4n + 3)\cdot90^{\circ}=4n\cdot90^{\circ}+3\cdot90^{\circ}=360^{\circ}n + 270^{\circ}\).

Step2: Use the periodicity of sine

The sine function has a period of \(360^{\circ}\), so \(\sin(360^{\circ}n + 270^{\circ})=\sin(270^{\circ})\) because \(\sin(\theta + 360^{\circ}k)=\sin\theta\) for any integer \(k\) (here \(k = n\)).

Step3: Evaluate \(\sin(270^{\circ})\)

We know that \(\sin(270^{\circ})=\sin(180^{\circ}+ 90^{\circ})\). Using the identity \(\sin(A + B)=\sin A\cos B+\cos A\sin B\), with \(A = 180^{\circ}\) and \(B = 90^{\circ}\), we get \(\sin(180^{\circ})\cos(90^{\circ})+\cos(180^{\circ})\sin(90^{\circ})\). Since \(\sin(180^{\circ}) = 0\), \(\cos(90^{\circ}) = 0\), \(\cos(180^{\circ})=- 1\), and \(\sin(90^{\circ}) = 1\), this simplifies to \(0\times0+(-1)\times1=-1\).

Answer:

\(=-1\)