The product ∏ k = 1 45 c o s k ∘ c o s k ∘ + s i n k ∘ equals a b , where a , b are primes and a < b . Find a + b .

Pair sin k ∘ with cos ( 90 − k ) ∘ and use sum-to-product. The product is ( 2 ) 44 ⋅ 2 = 2 23 , so a + b = 25 .

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MathematicsTrigonometric Ratios & IdentitiesSummation of SeriesHard3 minQB

MathematicsHardsingle choice

The product $\prod_{k = 1}^{45} \frac{c o s k ^{\circ} + s i n k ^{\circ}}{c o s k ^{\circ}}$ equals $a^{b}$ , where $a, b$ are primes and $a < b$ . Find $a + b$ .

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Answer:

Solution:

Pair $sin k^{\circ}$ with $cos (90 - k)^{\circ}$ and use sum-to-product. The product is $(2)^{44} \cdot 2 = 2^{23}$ , so $a + b = 25$ .

Stream:JEESubject:MathematicsTopic:Trigonometric Ratios & IdentitiesSubtopic:Summation of Series

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