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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 = 2tanx1+tan2x

To Prove Sin2x in the form of tanx x which is equal to 2tanx1+tan2x

Now let us start the proof from the right-hand side and hence, prove it as LHS = RHS

RHS = 2tanx1+tan2x

⇒ 2.sinxcosx / Sec²x

⇒ 2.sinxcosx / 1cos2x

⇒ 2.sinxcosx . cos2x1

⇒ 2sinxcosx

⇒ sin2x

Hence Proved LHS = RHS

2. cos2x

Cos 2x = (1tan2x)(1+tan2x)

Proof: To prove LHS = RHS

We are solving RHS which is equal to (1tan2x)(1+tan2x)

1sin2xcos2x1+sin2xcos2x

cos2xsin2xcos2x / cos2x+sin2xcos2x

cos2xsin2xcos2xcos2xcos2x+sin2x (Since cos2x+sin2x=1)

cos2xsin2xcos2xcos2xcos2x+sin2x

⇒ Hence, cos2xsin2x=Cos2x

Another Method of proving – cos2x = cos²x – sin²x

Now, cos2xsin2xcos2xcos2x

cos2x(1sin2x)cos2x

⇒ cos²x (cos²x – sin²x /cos²x)

⇒ cos²x (1 – tan²x )

(1tan2x)sec2x

(1tan2x)(1+tan2x)

Hence, RHS is proved

3. Tan 2x

Proof:

As we know, tan(x) = sinxcosx

Therefore, tan2x = sin2xcos2x

Now, tan2x = 2sinxcosxcos2x / cos2xcos2xsin2xsin2x

= 2sin?(x)cos(x) / 1 – (sin(x)cos(x))2

= 2tanx1tan2x

Another Method:

tan2x = sin2xcos2x

= 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) = 2sin(x)cos(x)cos2(x)sin2(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) = tan(x)+tan(2x)1tan(x)tan(2x)

 

Since 2tanx1tan2x

Now, putting all the things together:

tan(3x) = tan(x) + 2tan?(x)1tan2x/1-tan2(x) / 1 – tan(x) . 2(2tanx1tan2x)

Multiplying the numerator and denominator by 1−tan (x)

So, tan (3x) = tan(x)tan3(x)+2tan(x)1tan2(x)2tan2(x)

Hence,

Tan(3x) = tan(x)(3tan2(x))13tan2(x)