#Trigo-1

Geometry Level 3

If cos ⁡ ( α + β ) = 0 \cos (\alpha + \beta) = 0 , then sin ⁡ ( α − β ) \sin (\alpha - \beta) can be reduced to

sin ⁡ 2 α \sin 2\alpha sin ⁡ α \sin \alpha cos ⁡ 2 β \cos 2\beta cos ⁡ β \cos \beta

Vilakshan Gupta by Vilakshan Gupta , 19, India

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1 solution

Solving the first equation gives:

α = 1 2 k π − β \alpha = \frac{1}{2} k\pi - \beta

Substituting this in the given sin leads to:

s i n ( 1 2 k π − β − β ) = s i n ( 1 2 k π − 2 β ) = c o s ( − 2 β ) = c o s ( 2 β ) sin( \frac{1}{2} k\pi - \beta - \beta) = sin( \frac{1}{2} k\pi - 2\beta) = cos(-2\beta) = cos(2\beta)

I have used the identities s i n ( A + 1 2 k π ) = c o s ( A ) sin(A + \frac{1}{2} k\pi) = cos(A) and c o s ( − A ) = c o s ( A ) cos (-A) = cos(A)

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