1. Sin 2x = Sin 2x = sin(2x)=2sin(x). cos(x)
Sin(2x) = 2 * sin(x)cos(x)
Proof:
To express Sine, the formula of “Angle Addition” can be used.
sin(2x) = sin(x+x)
Since Sin (a + b) = Sin(a). Sin(b) + Cos(a).Cos(b)
Therefore, sin(x+x) = sin(x)cos(x) + cos(x)sin(x) = 2. sin(x). cos(x)
Also, Sin 2x = $\frac{2tanx}{1+\tan 2x}$
To Prove Sin2x in the form of tanx x which is equal to $\frac{2tanx}{1+\tan 2x}$
Now let us start the proof from the right-hand side and hence, prove it as LHS = RHS
RHS = $\frac{2tanx}{1+\tan 2x}$
⇒ 2.$\frac{sinx}{cosx}$ / Sec²x
⇒ 2.$\frac{sinx}{cosx}$ / $\frac{1}{\cos 2x}$
⇒ 2.$\frac{sinx}{cosx}$ . $\frac{\cos 2x}{1}$
⇒ 2sinxcosx
⇒ sin2x
Hence Proved LHS = RHS
2. cos2x
Cos 2x = $\frac{(1 – tan2x)}{(1+ tan2x)}$
Proof: To prove LHS = RHS
We are solving RHS which is equal to $\frac{(1 – tan2x)}{(1+ tan2x)}$
⇒ $\frac{1 – \frac{\sin 2x}{\cos 2x}}{1+ \frac{\sin 2x}{\cos 2x}}$
⇒ $\frac{ \cos 2x- \sin 2x}{\cos 2x}$ / $\frac{\cos 2x+ \sin 2x}{\cos 2x}$
⇒ $\frac{ \cos 2x- \sin 2x}{\cos 2x} \frac{\cos 2x}{\cos 2x+ \sin 2x}$ (Since $\cos 2x+ \sin 2x=1)$
⇒ $\frac{ \cos 2x- \sin 2x}{\cos 2x} \frac{\cos 2x}{\cos 2x+ \sin 2x}$
⇒ Hence, $\cos 2x- \sin 2x= $Cos2x
Another Method of proving – cos2x = cos²x – sin²x
Now, $\cos 2x – sin2x \frac{\cos 2x}{\cos 2x}$
⇒ $\frac{\cos 2x (1-\sin ^{2}x)}{ cos2x}$
⇒ cos²x (cos²x – sin²x /cos²x)
⇒ cos²x (1 – tan²x )
⇒ $\frac{( 1 – tan2x)}{\sec 2x }$
⇒ $\frac{(1 – tan2x )}{(1+ tan2x )}$
Hence, RHS is proved
3. Tan 2x
Proof:
As we know, tan(x) = $\frac{sinx}{cosx}$
Therefore, tan2x = $\frac{\sin 2x}{\cos 2x}$
Now, tan2x = $\frac{2sinxcosx}{\cos 2x}$ / $\frac{\cos 2x}{\cos 2x}$ – $\frac{\sin 2x}{\sin 2x}$
= $\frac{2 \sin ?(x)}{\cos (x)}$ / 1 – $(\frac{\sin (x)}{\cos (x)})2$
= $\frac{2 tanx}{1-\tan 2x}$
Another Method:
tan2x = $\frac{\sin 2x}{\cos 2x}$
= $\frac{\sin (x+x)}{\cos (x+x)}$
= As we know Sin (a + b) = Sin (a). Sin (b) + Cos (a).Cos (b)
Therefore,
Sin(x + x) = Sin(x) Cos(x) + Cos (x) Sin(x)
Also, sin(x + x) = 2sin(x) cos(x)
& Likewise, Cos (a + b) = Cos (a). Cos (b) – Sin (a). Sin (b)
So,
Cos(x+x) = Cos(x) Cos(x) − Sin(x) Sin(x)
Also, cos (x + x) = cos ² (x)−sin ² (x)
Hence,
Tan (2x) = $\frac{2\sin (x) \cos (x)}{\cos 2 (x)- \sin 2 (x)}$
4. Sin3x
Proof: To prove Sin3x = 3sinx−4sin³x
Sin 3x = Sin (x + 2x) = Sinx. Cos2x + Cosx. Sin2x
When you substitute the values of Sin2x & Cos2x, we will get,
sin3x = (sinx).(1−2sinx)+(cosx).(2sinxcosx)
Now using, Sin²x + Cos²x = 1
We get, Sin3x = 3sinx−4sin³x
5. Cos3x
cos3x = cos(x+2x) It can also be written in this form
= cosxcos2x−sinxsin2x {as per the identity: Cos(x+x) = Cos(x) Cos(x) − Sin(x) Sin(x)}…Eq1
= Now as we know,
Cos2x = 2Cos ²x – 1;
Sin2x = 2SinxCosx.
Therefore,
Putting the values in Eq.1
= cosx(2cos ²x−1)−sinx(2sinxcosx)
= 2cos ³ x−cosx−2sin² xcosx
= 2cos³ x−cosx−2(1−cos² x)cosx [sin² x+cos² x=1]
= 2cos³x−cosx−2(cosx−cos³x) (opening the brackets)
= 2cos³x−cosx−2cosx+2cos³x
= 4cos³x−3cosx
Hence, Proved!
6. Tan3x
Proof: We are having,
Tan3x = tan(x+2x)
Tan(x+2x) = $\frac{tan(x)+tan(2x)}{1 – tan(x)tan(2x)}$
Since $\frac{2 tanx}{1-\tan 2x}$
Now, putting all the things together:
tan(3x) = tan(x) + $\frac{2\tan ?(x)}{1-\tan 2 x }$/1-tan2(x) / 1 – tan(x) . 2$(\frac{2tanx}{1-\tan 2x})$
Multiplying the numerator and denominator by 1−tan (x)
So, tan (3x) = $\frac{tan(x)-tan3(x)+2tan(x)}{1-\tan 2(x)-2tan2(x)}$
Hence,
Tan(3x) = $\frac{tan(x)\cdot (3-\tan 2(x))}{1-3tan2(x)}$