Sin 2x Formula

What is the formula for Sin 2x ?

Sin 2x is equal to 2.sin x.cos x.

Proof of Sin 2x's Formula

Way 1

In the above isosceles triangle ABC;

$AB=AC=1$

$BD=DC=\sin x$

$AD=\cos x$

$BE=\sin 2x$

Area of ​​triangle ABC;

$A(ABC)=\frac{AC.BE}{2}$

$A(ABC)=\frac{BC.AD}{2}$

$\frac{AC.BE}{\cancel{2}}=\frac{BC.AD}{\cancel{2}}$

$1.\sin 2x=2\sin x.\cos x$

$\sin 2x=2.\sin x.\cos x$

Way 2

In the right triangle ABC above;

$AC=1$

$AB=\sin x$

$BC=\cos x$

$AD=DC=a$

$BD=\cos x-a$

For triangle ABC;

$(AB{)}^2+(BC{)}^2=(AC{)}^2$

${\sin }^{2}x+{\cos }^{2}x=1^2$

${\sin }^{2}x+{\cos }^{2}x=1$

For triangle ABD;

$(AB{)}^2+(BD{)}^2=(AD{)}^2$

${\sin }^{2}x+(\cos x-a{)}^2=a^2$

${\sin }^{2}x+{\cos }^{2}x-2.\cos x.a+a^2=a^2$

${\sin }^{2}x+{\cos }^{2}x-2.\cos x.a+\cancel{a^2}-\cancel{a^2}=0$

$1-2.\cos x.a=0$

$-2.\cos x.a=-1$

$\frac{\cancel{-2.\cos x}.a}{\cancel{-2.\cos x}}=\frac{-1}{-2.\cos x}$

$a=\frac{1}{2.\cos x}$

In triangle ABC;

$\sin 2x=\frac{AB}{AD}=\frac{\sin x}{a}$

$\sin 2x=\frac{\sin x}{\frac{1}{2.\cos x}}$

$\sin 2x=\sin x.2.\cos x$

$\sin 2x=2.\sin x.\cos x$

Way 3

Formulas used to find the trigonometric value of the sum or difference of two angles with known trigonometric values ​​are called sum-difference formulas. We can find the value of the formula for sin 2x by using the following sum formula for sine.

$\boxed{\sin (a+b)=\sin a.\cos b+\sin b.\cos a}$

$\sin 2x=\sin (x+x)$

$\sin 2x=\sin x.\cos x+\sin x.\cos x$

$\sin 2x=2.\sin x.\cos x$

$Question:$

$\sin x-\cos x=\frac{1}{2}\Rightarrow \sin 2x=?$

$Answer:$

$\sin x-\cos x=\frac{1}{2}$

$(\sin x-\cos x{)}^2=(\frac{1}{2}{)}^2$

${\sin }^{2}x-2.\sin x.\cos x+{\cos }^{2}x=\frac{1}{4}$

${\sin }^{2}x+{\cos }^{2}x-2.\sin x.\cos x=\frac{1}{4}$

$1-\sin 2x=\frac{1}{4}$

$-\sin 2x=\frac{1}{4}-1$

$-\sin 2x=\frac{1-4}{4}$

$-\sin 2x=-\frac{3}{4}$

$\sin 2x=\frac{3}{4}$

$Question:$

$\cos 20°=x\Rightarrow \frac{\sin 80°}{\cos 40°.\sin 20°}=?$

$Answer:$

$\frac{\sin 80°}{\cos 40°.\sin 20°}=\frac{2.\sin 40°.\cancel{\cos 40°}}{\cancel{\cos 40°}.\sin 20°}$

$\frac{\sin 80°}{\cos 40°.\sin 20°}=\frac{\mathrm{2.2.}\cancel{\sin 20°}.\cos 20°}{\cancel{\sin 20°}}$

$\frac{\sin 80°}{\cos 40°.\sin 20°}=4.\cos 20°$

$\frac{\sin 80°}{\cos 40°.\sin 20°}=4x$