Differential Equations 1 Question 6

6. The value of k=1131sinπ4+(k1)π6sinπ4+kπ6 is equal to

(2016 Adv.)

(a) 33

(b) 2(33)

(c) 2(31)

(d) 2(2+3)

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Answer:

Correct Answer: 6. (c)

Solution:

  1. Here, k=1131sinπ4+(k1)π6sinπ4+kπ6

Converting into differences, by multiplying and dividing by sinπ4+kπ6π4+(k1)π6, i.e. sinπ6.

k=113sinπ4+kπ6π4+(k1)π6sinπ6sinπ4+(k1)π6sinπ4+kπ6

=2k=113sinπ4+kπ6cosπ4+(k1)π6sinπ4+(k1)π6sinπ4+kπ6=2k=113cotπ4+(k1)π6cotπ4+kπ6=2cotπ4cotπ4+π6++cotπ4+π6cotπ4+2π6+2+cotπ4+12π6cotπ4+13π6=2cotπ4cotπ4+13π6=21cot29π12=21cot2π+5π12=2(12+3)=2(31)



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