Class 11MathematicsTrigonometrical Ratios, Functions and IdentitiesTrigonometrical ratios of multiple and sub-multiple angles
MediumProve that: $\cos 4x = 1 - 8 \sin^2 x \cos^2 x$
(N/A) $L.H.S. = \cos 4x$
$= \cos 2(2x)$
$= 1 - 2 \sin^2 2x$ (using $\cos 2A = 1 - 2 \sin^2 A$)
$= 1 - 2(2 \sin x \cos x)^2$ (using $\sin 2A = 2 \sin A \cos A$)
$= 1 - 2(4 \sin^2 x \cos^2 x)$
$= 1 - 8 \sin^2 x \cos^2 x$
$= R.H.S.$
$= \cos 2(2x)$
$= 1 - 2 \sin^2 2x$ (using $\cos 2A = 1 - 2 \sin^2 A$)
$= 1 - 2(2 \sin x \cos x)^2$ (using $\sin 2A = 2 \sin A \cos A$)
$= 1 - 2(4 \sin^2 x \cos^2 x)$
$= 1 - 8 \sin^2 x \cos^2 x$
$= R.H.S.$
Explore More
Similar Questions
Evaluate: $\cos^2 A(3 - 4\cos^2 A)^2 + \sin^2 A(3 - 4\sin^2 A)^2$
Easy
View SolutionIf $\sin 3\alpha = 4\sin \alpha \sin (x + \alpha )\sin (x - \alpha ),$ then $x = $
Medium
View Solution$\frac{\sin 5 \theta}{\sin \theta}$ is equal to
If $\cos ^3 x \sin 4 x = \sum_{r=0}^{n} a_{r} \sin rx$ for all $x \in R$,then $a_3+a_5 : a_1+a_7 = $
$2\cos x - \cos 3x - \cos 5x = $
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo