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General Solution of Trigonometric Equation

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Question

(i) Show that $t a n 9^{o} - t a n 27^{o} - t a n 63^{o} + t a n 81^{o} = 4$
(ii) Show that $c o s 42^{o} + c o s 78^{o} + c o s 162^{o} = 0$

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Solution

(i) We know that $t a n θ = c o t (90 - θ)$
=> $t a n 81 = c o t 9$
and $t a n 63 = c o t 27$

We need to simplify

$t a n 9 + c o t 9 - t a n 27 - c o t 27$
=> $\frac{s i n 9}{c o s 9} + \frac{c o s 9}{s i n 9} - \frac{s i n 27}{c o s 27} - \frac{c o s 27}{s i n 27}$
=> $\frac{1}{s i n 9. c o s 9} - \frac{1}{s i n 27. c o s 27}$
=> $\frac{2}{2 s i n 9. c o s 9} - \frac{2}{2 s i n 27. c o s 27}$
=> $\frac{2}{s i n 18} - \frac{2}{s i n 54}$

Putting values of $s i n 18 = \frac{\sqrt{5} - 1}{4}, s i n 54 = \frac{\sqrt{5} + 1}{4}$ and simplying, we get the expression equal to 4.

(ii) $c o s 42 + c o s 78 + c o s 162$
=> $c o s (60 - 18) + c o s (60 + 18) + c o s 162$
=> $2 c o s 60. c o s 18 + c o s (180 - 18)$
=> $2 \frac{1}{2} c o s 18 - c o s 18$
=> $0$

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Similar questions

Q. Without using tables, show that

tan 9^{o} - tan 27^{o} - tan 63^{o} + tan 81^{o} = 4

tan 9^{o} - tan 27^{o} - tan 63^{o} + tan 81^{o} =

tan 9^{o} \times tan 27^{o} \times tan 63^{o} \times tan 81^{o} =

Q. Find the value of

tan 9^{o} - tan 27^{o} - tan 63^{o} + tan 81^{o}

.

Q. Show that (i) (ii)

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General Solution of Trigonometric Equation

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