Trig Equation

Geometry Level 1

sin ⁡ 2 α − sin ⁡ 2 β = ? \large \sin^2\alpha -\sin^2\beta = \ ? \

sin ⁡ α 2 \sin\alpha^2 sin ⁡ ( α + β ) sin ⁡ ( α − β ) \sin(\alpha+\beta)\sin(\alpha-\beta) sin ⁡ ( α − β ) 2 \sin(\alpha-\beta)^2 sin ⁡ β 2 \sin\beta^2

Anish Harsha by Anish Harsha , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rishabh Jain
Jan 29, 2016

Using sin ⁡ 2 A = 1 − cos ⁡ 2 A 2 sin ⁡ 2 x − sin ⁡ 2 y = ( 1 − cos ⁡ 2 x ) 2 − ( 1 − cos ⁡ 2 y ) 2 = 1 2 ( cos ⁡ 2 y − cos ⁡ 2 x ) Using cos ⁡ C − cos ⁡ D = 2 sin ⁡ ( D − C 2 ) sin ⁡ ( C + D 2 ) = sin ⁡ ( 2 x + 2 y 2 ) sin ⁡ ( 2 x − 2 y 2 ) = sin ⁡ ( x + y ) sin ⁡ ( x − y ) \color{darkviolet}{\text{Using}~\sin^2A=\dfrac{1-\cos 2A}{2}} \\ \sin^2 x-\sin^2 y= \dfrac{(1- \cos 2x)}{2}-\dfrac{(1-\cos 2y)}{2}\\ = \dfrac{1}{2}(\cos2y-\cos2x)~~\\ \color{darkviolet}{\text{Using}~\cos C-\cos D=2 \sin (\dfrac{D-C}{2}) \sin (\dfrac{C+D}{2})} \\ = \sin (\dfrac{2x+2y}{2})\sin (\dfrac{2x-2y}{2}) \\ =\large \sin (x+y)\sin (x-y)

9 Helpful 0 Interesting 0 Brilliant 0 Confused

B o n u s : cos ⁡ 2 α − cos ⁡ 2 β = ? ? \Large Bonus: \color{#D61F06}{\cos^2 \alpha- \cos^2 \beta}=?? = ( 1 − sin ⁡ 2 α ) − ( 1 − s i n 2 β ) =(1-\sin^2 \alpha) - (1-sin^2 \beta) = − ( sin ⁡ 2 α − sin ⁡ 2 β ) =-(\sin^2 \alpha - \sin^2 \beta ) = sin ⁡ ( α + β ) sin ⁡ ( β − α ) =\sin (\alpha+\beta) \sin(\beta-\alpha)

Rishabh Jain - 5 years, 4 months ago

Same method. Clone of Calvin sirs problem!

Aareyan Manzoor - 5 years, 4 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...