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UPSC NDA I 2026 Mathematics Paper

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Question : 32 of 120

Marks: +1, -0

If

\cos \alpha+\cos \beta=0=\sin \alpha+\sin \beta

\alpha \neq \beta

then what is a value of

\cos 2 \alpha+\cos 2 \beta+2 \cos (\alpha+\beta)

?

0
1
2
4

Solution:

Concept:

Since the sum of the two unit vectors is zero, the angles differ by

\pi

.

Explanation:

(\cos\alpha+\cos\beta)+i(\sin\alpha+\sin\beta)=0

, so

e^{i\alpha}+e^{i\beta}=0

.

Thus

e^{i\beta}=-e^{i\alpha}=e^{i(\alpha+\pi)}

, giving

\beta=\alpha+\pi

(mod

2\pi

).

So

\cos 2\beta=\cos(2\alpha+2\pi)=\cos 2\alpha

and

\cos(\alpha+\beta)=\cos(2\alpha+\pi)=-\cos 2\alpha

.

Therefore,

\cos 2\alpha+\cos 2\beta+2\cos(\alpha+\beta)=\cos2\alpha+\cos2\alpha+2(-\cos2\alpha)=0

.

Answer:

0

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