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Trigonometric Ratios for Sum of Two Angles

If α and β ar...

Question

If $α$ and $β$ are distinct roots of $a c o s θ + b s i n θ = c$ , Prove that $s i n (α + β) = \frac{2 a b}{a^{2} + b^{2}}$ .
or
Prove that $c o s 20^{\circ} c o s 40^{\circ} c o s 60^{\circ} c o s 80^{\circ} = \frac{1}{16}$

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Solution

It is given that, $α$ and $β$ are the distinct roots of $a c o s θ + b s i n θ = c$

$∴ a c o s α + b s i n α = c$ ...(i)

and $a c o s β + b s i n β = c$ ...(ii)

From Eqs. (i) and (ii), we get

$a (c o s α - c o s β) + b (s i n α - s i n β) = 0$

$\Rightarrow - 2 a s i n (\frac{α + β}{2}) s i n (\frac{α - β}{2}) + 2 b c o s (\frac{α + β}{2}) s i n (\frac{α - β}{2}) = 0$ $[∵ c o s C - c o s D = - 2 s i n (\frac{C + D}{2}) s i n (\frac{C_{D}}{2}) and s i n C - s i n D = 2 c o s (\frac{C + D}{2}) s i n (\frac{C - D}{2})]$

$\Rightarrow t a n \frac{α + β}{2} = \frac{b}{a}$

Now, $s i n (α + β) = \frac{2 t a n (\frac{α + β}{2})}{1 + t a n^{2} (\frac{α + β}{2})} = \frac{2 \frac{b}{a}}{1 + \frac{b^{2}}{a^{2}}} = \frac{2 a b}{a^{2} + b^{2}}$

We have, $L H S = c o s 20^{\circ} c o s 40^{\circ} c o s 60^{\circ} c o s 80^{\circ} = c o s 60^{\circ} c o s 20^{\circ} c o s 40^{\circ} c o s 80^{\circ}$

$= \frac{1}{2} (c o s 20^{\circ} c o s 40^{\circ}) c o s 80^{\circ}$

$= \frac{1}{4} (2 c o s 20^{\circ} c o s 40^{\circ}) c o s 80^{\circ}$ $[∵ 2 c o s A c o s B = c o s (A + B) + c o s (A - B)]$

$= \frac{1}{4} (c o s 60^{\circ} + c o s 20^{\circ}) c o s 80^{\circ} = \frac{1}{4} (c o s 60^{\circ} c o s 80^{\circ} + c o s 20^{\circ} c o s 80^{\circ})$

$= \frac{1}{4} (\frac{1}{2} c o s 80^{\circ} + \frac{1}{2} .2 c o s 20^{\circ} . c o s 80^{\circ})$

$= \frac{1}{4} [\frac{1}{2} c o s 80^{\circ} + \frac{1}{2} c o s (20^{\circ} + 80^{\circ}) + c o s (20^{\circ} - 80^{\circ})]$

$= \frac{1}{4} [\frac{1}{2} c o s 80^{\circ} + \frac{1}{2} (c o s 100^{\circ} + c o s 60^{\circ})] = \frac{1}{4} [\frac{1}{2} c o s 80^{\circ} + \frac{1}{2} {c o s (180 - 80^{\circ}) + \frac{1}{2}}]$

$= \frac{1}{4} (\frac{1}{2} c o s 80^{\circ} - \frac{1}{2} c o s 80^{\circ} + \frac{1}{4}) = \frac{1}{16} = R H S$

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c o s 20^{\circ} c o s 40^{\circ} c o s 60^{\circ} c o s 80^{\circ} = \frac{1}{16}

Q. Prove that --
cos20°cos40° cos60° cos80° =1

α

β

are the solution of

a

cos θ + b sin θ = c

, then show that :

cos (α - β) = \frac{2 c^{2} - (a^{2} + b^{2})}{a^{2} + b^{2}}

{c o s 20}^{\circ} c o s 40^{\circ} {c o s 60}^{\circ} c o s 80^{o} = \frac{1}{16}

If $s i n α + s i n β = a a n d c o s α + c o s β = b$ , prove that

(i) $s i n (α + β) = \frac{2 a b}{a^{2} + b^{2}}$

(ii) $c o s (α - β) = \frac{a^{2} + b^{2} - 2}{2}$

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Trigonometric Ratios for Sum of Two Angles

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