Question

How do you prove the formula $\operatorname{sin}3A=3sinA-4{{\sin }^{3}}A$ using only the identity $\sin (A+B)=\sin A\cos B+\cos A\sin B$

Answer 1

We are asked to prove that $\operatorname{sin}3A=3sinA-4{{\sin }^{3}}A$. To do so we will start considering the left hand side. We use $\sin (A+B)=\sin A\cos B+\cos A\sin B$. We first split $3A \text{ as } (A+2A)$ and use $A=A$ and $B=2A$ on $\sin (A+B)$, we will split, then we will split $2A$ or $A+A$, we will also use other splitting of sin function to obtain the different identity we will be needing during the solution.

Complete step by step solution:
We are given $\operatorname{sin}3A=3sinA-4{{\sin }^{3}}A$ we have to prove the above defined identity. We start our process by considering the left side, we will process it and reach to the right. Now we have $\operatorname{sin}3A$. To solve our problem, we can use only one identity.
We start our process by considering the left side, we will process it and reach to the right. Now we have $\sin 3A$to solve our problem we can use only one identity $\sin (A+B)=\sin A\cos B+\cos A\sin B$
So to solve our problem, we rearrange terms to get more identity which can be used to solve our problem.
We have $\sin (3A)$
As we are given, $\sin (A+B)=\sin A\cos B+\cos A\sin B$
Changing $B$ to $-B$, we get,
$\sin (A-B)=\sin A\cos (-B)+\cos A(\sin (-B))........(i)$
As, $\cos (-\theta )=\cos \theta \,\,and\,\,\,\sin (-\theta )=-\sin \theta$, so we have
$\sin (A-B)=\sin A\cos B - \cos A\sin B.......(ii)$
Adding (ii) and (i) we get,
$\sin (A+B)+\sin (A-B)=2\sin A\cos B...........(iii)$
Now we have $3A$ so we split as $3A=A+2A$.
So, considering $A=A$ and $B=2A$ in equation (iii) we get,
$\sin (A+2A)+\sin (A-2A)=2\sin (A)(\cos (2A))$
Simplifying we get,
$\sin 3A-\sin A=2\sin A\cos 2A$
Adding $\sin A$ both sides, we get,
$\sin 3A=2\sin A\cos 2A+\sin A.........(iv)$
Now we put $A=A$ and $B=90-A$ in eq (iii), we will get,
$\sin (A+90-A)+\sin (A-(90-A)=2\sin A\cos (90-A)$
Simplifying we get,
$\sin (90)+\sin (2A-90)=2\sin A\cos (90-A)$
As $\cos (90-\theta )=\sin \theta \,\,\,and\,\,\sin (-\theta )=\sin \theta$, we get,
$\begin{aligned} & \sin (2A-90)=-\sin (90-2A) \\\ & \cos (90-A)=\sin A \\\ \end{aligned}$
So our equation is,
$\sin 90-\sin (90-2A)=2{{\sin }^{2}}A$
Again as $\sin (90-\theta )=\cos \theta$ , so we have $\sin (90-2A)=\cos 2A$
So our equation becomes
$\begin{aligned} & \sin (90{}^\circ )-\cos (2A)=2{{\sin }^{2}}A \\\ & \Rightarrow 1-\cos 2A=2{{\sin }^{2}}A \\\ \end{aligned}$
So, we get
$\cos 2A=1-2{{\sin }^{2}}A........(iv)$
Using this in equation (iv) we get
$\sin 3A=2\sin A(1-2{{\sin }^{2}}A)+\sin A$
Opening brackets, we get
$\Rightarrow 2\sin A-4{{\sin }^{3}}A+\sin A$
Adding like terms we get,
$=3\sin A-4{{\sin }^{3}}A$
So, we get
$\sin (3A)=3\sin A-4{{\sin }^{3}}A$
Hence proved.

Note: To start solving identity we should know in which quadrant, when ratio is positive or negative, which point ratio changes their behaviour, we must know which ratio is odd and which is even. $\cos x$ is an even relation while the $\sin x$ is an odd fraction, $\sin ,\cos$ change to one another at an odd multiple of $90{}^\circ$ .