KCET · Maths · Trigonometric Ratios & Identities
The value of the \( \sin 1^{\circ}+\sin 2^{\circ}+\ldots \ldots+\sin 359^{\circ} \) is equal to
- A \( 00 \)
- B \( 11 \)
- C \( -1 \)
- D \( 180 \)
Answer & Solution
Correct Answer
(A) \( 00 \)
Step-by-step Solution
Detailed explanation
Given that, \(\sin 1^{\circ}+\sin 2^{\circ}+\cdots+\sin 359^{\circ}\)
We know that, \(\sin \theta+\sin \left(180^{\circ}+\theta\right)=0\)
So, \(\sin 1^{\circ}+\sin 181^{\circ}=0\)
Similarly, \(\sin 2^{\circ}+\sin 182^{\circ}=0\)
And so on, \(\sin 179^{\circ}+\sin 359^{\circ}=0\)
Therefore,
\(\sin 1^{\circ}+\sin 2^{\circ}+\ldots+\sin 359^{\circ}=\sin 180^{\circ}=0\)
We know that, \(\sin \theta+\sin \left(180^{\circ}+\theta\right)=0\)
So, \(\sin 1^{\circ}+\sin 181^{\circ}=0\)
Similarly, \(\sin 2^{\circ}+\sin 182^{\circ}=0\)
And so on, \(\sin 179^{\circ}+\sin 359^{\circ}=0\)
Therefore,
\(\sin 1^{\circ}+\sin 2^{\circ}+\ldots+\sin 359^{\circ}=\sin 180^{\circ}=0\)
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