Prove that: 10^ 50^ 60^ 70^ = 3 16 - Vidyadip

Question

Prove that:
$\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ=\frac{\sqrt3}{16}$

✓

Answer

$\text{LHS}=\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ$
$\Big(\sin10^\circ\sin50^\circ\sin70^\circ\frac{\sqrt3}{2}\Big)$$\Big[\because\ \sin60^\circ=\frac{\sqrt3}{2}\Big]$
$=\ \sin(90^\circ-80^\circ)\sin(90^\circ-40^\circ)\sin(90^\circ-20^\circ)\frac{\sqrt3}{2}$
$=\ \cos80^\circ\cos40^\circ\cos20^\circ\frac{\sqrt3}{2}$
$=\ \frac{\sqrt3}{2\times2}(2\cos40^\circ\cos20^\circ)\cos80^\circ$$[\because\ 2\cos\text{A}\cos\text{B}=(\cos\text{A+B})+\cos(\text{A}-\text{B})]$
$=\ \frac{\sqrt3}{2\times2}[\cos(40^\circ+20^\circ)+\cos(40^\circ-20^\circ)]\cos80^\circ$
$=\ \frac{\sqrt3}{2\times2}[\cos60^\circ+\cos20^\circ]\cos80^\circ$
$=\ \frac{\sqrt3}{2\times2}\Big[\frac{1}{2}+\cos20^\circ\Big]\cos80^\circ$
$=\ \frac{\sqrt3}{4}\Big[\frac{1}{2}\cos80^\circ+\cos20^\circ\cos80^\circ\Big]$
$=\ \frac{\sqrt3}{8}[\cos80^\circ+2\cos20^\circ\cos80^\circ]$
$=\ \frac{\sqrt3}{8}[\cos80^\circ+\cos(80^\circ+20^\circ)+\cos(80^\circ-20^\circ)]$
$=\ \frac{\sqrt3}{8}[\cos80^\circ+\cos100^\circ+\cos60^\circ]$
$=\ \frac{\sqrt3}{8}[\cos80^\circ+\cos(180^\circ-80^\circ)+\cos60^\circ]$
$=\ \frac{\sqrt3}{8}[\cos60^\circ]=\frac{\sqrt3}{16}=\text { RHS}$

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

Explore more

More "5 Marks Questions" from Transformation Formulae→Transformation Formulae — all question types→MATHS STD 11 Science question papers→Rajasthan Board STD 11 Science question papers→Rajasthan Board question paper generator→

Similar questions

Differentiate the following from first principle:

$(-\text{x})^{-1}$

Find the (i) lengths of major and minor axes, (ii) coordinate of the vertice, (iii) coordinate of the foci, (iv) eccentricity, and (v) length of the latus rectum of ellipe: $16 x^2+25 y^2=400$.

Calculate the mean deviation of the following income groups of five and seven members from their medians:

I Income in ₹	II Income in ₹
4000	3800
4200	4000
4400	4200
4600	4400
4800	4600
	4800
	5800

Find the equation to the ellipse in the following case:

Vertices $(0, \pm13),$ foci $(0, \pm5)$

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\text{a}}}\frac{\text{a}\sin\text{x}-\text{x}\sin\text{a}}{\text{ax}^2-\text{xa}^2}$

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has:

No girl?
At least one boy and one girl?
At least 3 girls?

Differentiate the following from the first principle

$\text{e}^{\sqrt{\text{ax}+\text{b}}}$

$\frac{\cos^2\text{B}-\cos^2\text{C}}{\text{b + c}}+\frac{\cos^2\text{C}-\cos^2\text{A}}{\text{c + a}}+\frac{\cos^2\text{A}-\cos^2\text{B}}{\text{a + b}}=0$

Solve the following equations:
$4\sin\text{x}\cos\text{x}+2\sin\text{x}+2\cos\text{x}+1=0$

Prove that:
$\cos20^\circ\cos100^\circ+\cos100^\circ\cos140^\circ-\cos140^\circ\cos200^\circ=-\frac{3}{4}$