Question
Prove that:
$\cos20^\circ\cos40^\circ\cos80^\circ=\frac{1}{8}$

Answer

$\text{LHS}=\cos20^\circ\cos40^\circ\cos80^\circ$
$=\ \frac{1}{2}(2\cos20^\circ\cos40^\circ)\cos80^\circ$
$=\ \frac{1}{2}[\cos(40^\circ+20^\circ)+\cos(40^\circ-20^\circ)]\cos80^\circ$$[\because\ 2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})]$
$=\ \frac{1}{2}[\cos60^\circ+\cos20^\circ]\cos80^\circ$
$=\ \frac{1}{2}\Big[\frac{1}{2}+\cos20^\circ\Big]\cos80^\circ$
$=\ \frac{1}{2}[\cos80^\circ+2\cos20^\circ\cos80^\circ]$
$=\ \frac{1}{4}[\cos80^\circ+\cos(80^\circ+20^\circ)+\cos(20^\circ-80^\circ)]$
$=\ \frac{1}{4}[\cos80^\circ+\cos100^\circ+\cos60^\circ]$
$=\ \frac{1}{4}[\cos80^\circ+\cos(180^\circ-80^\circ)+\cos60^\circ]$
$=\ \frac{1}{4}[\cos80^\circ-\cos80^\circ+\cos60^\circ]$
$= \frac{1}{4}\Big[\frac{1}{2}\Big]=\frac{1}{8}=\text{RHS}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If the image of the point (2, 1) with respect to the line mirror be (5, 2), find the equation of the mirror.
Find the mean deviation from the mean for following data:
$x_i$
$5$
$7$
$9$
$10$
$12$
$15$
$f_i$
$8$
$6$
$2$
$2$
$2$
$6$
Prove that the area of the parallelogram formed by the lines $a_1x + b_1y + c_1 = 0, a_1x + b_1y+ d_1 = 0, a_2x + b_2y + c_2 = 0, a_2x + b_2y + d_2 = 0$ is $\Big|\frac{(\text{d}_1-\text{c}_1)(\text{d}_2-\text{c}_2)}{\text{a}_1\text{b}_2-\text{a}_2\text{b}_1}\Big|$ sq.units.
Deduce the condition for these lines to form a rhombus.
If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, find
$(\text{A}\times\text{B})\cap(\text{A}\times\text{C})$
A team of medical students doing their internship have to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple or very simple are respectively, $0.15, 0.20, 0.31, 0.26, 08.$ Find the probabilities that a particular surgery will be rated:
Prove that:
$\frac{\sin\text{A}\sin2\text{A}+\sin3\text{A}\sin6\text{A}}{\sin\text{A}\cos2\text{A}+\sin3\text{A}\cos6\text{A}}=\tan5\text{A}$
Reduce each of the following expressions to the sine and cosin of a single expression:
$\cos\text{x}-\sin\text{x}$
Find the equation of the parabola, if
The focus is at $(0, -3)$ and the vertex is at $(0, 0).$
Prove that $\frac{(2\text{n})!}{2^{2\text{n}}(\text{n}!)^2}\leq\frac{1}{\sqrt{3\text{n}+1}}$ for all $\text{n}\in\text{N}.$
The circle $x^2 + y^2 - 2x - 2y + 1 = 0$ is rolled along the positive direction of $x-$axis and makes one complete roll. Find its equation in new-position.