Identities related to sin 2x, cos2x, tan 2x, sin3x, cos3x, and tan3x

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) = $\frac{tan(x)\cdot (3-\tan 2(x))}{1-3tan2(x)}$

 

 

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