Class 11MathematicsTrigonometrical Ratios, Functions and IdentitiesTrigonometrical ratios of sum and difference of two and three angles
MediumProve that $\sin^{2} 6x - \sin^{2} 4x = \sin 2x \sin 10x$.
(N/A) We use the identity $\sin^{2} A - \sin^{2} B = \sin(A+B) \sin(A-B)$.
$L.H.S. = \sin^{2} 6x - \sin^{2} 4x$
$= \sin(6x + 4x) \sin(6x - 4x)$
$= \sin(10x) \sin(2x)$
$= \sin 2x \sin 10x$
$= R.H.S.$
$L.H.S. = \sin^{2} 6x - \sin^{2} 4x$
$= \sin(6x + 4x) \sin(6x - 4x)$
$= \sin(10x) \sin(2x)$
$= \sin 2x \sin 10x$
$= R.H.S.$
Explore More
Similar Questions
If $A = \frac{\pi}{24}$,then $\frac{\cos A + \cos 3A + \cos 5A + \cos 7A}{\sin A + \sin 3A + \sin 5A + \sin 7A} = $
Prove that $\cos 2x \cos \frac{x}{2} - \cos 3x \cos \frac{9x}{2} = \sin 5x \sin \frac{5x}{2}$
Medium
View SolutionIf $\cos (\alpha + \beta) = \frac{3}{5}$,$\sin (\alpha - \beta) = \frac{5}{13}$ and $0 < \alpha, \beta < \frac{\pi}{4}$,then $\tan (2\alpha)$ is equal to
DifficultJEE MAIN 2019
View SolutionIf $A$ and $B$ $(A > B)$ are acute angles,$\sin (A-B)=\frac{16}{65}$ and $\sin B=\frac{5}{13}$,then $\tan A+\cot A=$
$(\cos \alpha + \cos \beta )^2 + (\sin \alpha + \sin \beta )^2 = $
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