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 = (1–tan2x)(1+tan2x)
Proof: To prove LHS = RHS
We are solving RHS which is equal to (1–tan2x)(1+tan2x)
⇒ 1–sin2xcos2x1+sin2xcos2x
⇒ cos2x−sin2xcos2x / cos2x+sin2xcos2x
⇒ cos2x−sin2xcos2xcos2xcos2x+sin2x (Since cos2x+sin2x=1)
⇒ cos2x−sin2xcos2xcos2xcos2x+sin2x
⇒ Hence, cos2x−sin2x=Cos2x
Another Method of proving – cos2x = cos²x – sin²x
Now, cos2x–sin2xcos2xcos2x
⇒ cos2x(1−sin2x)cos2x
⇒ cos²x (cos²x – sin²x /cos²x)
⇒ cos²x (1 – tan²x )
⇒ (1–tan2x)sec2x
⇒ (1–tan2x)(1+tan2x)
Hence, RHS is proved
3. Tan 2x
Proof:
As we know, tan(x) = sinxcosx
Therefore, tan2x = sin2xcos2x
Now, tan2x = 2sinxcosxcos2x / cos2xcos2x – sin2xsin2x
= 2sin?(x)cos(x) / 1 – (sin(x)cos(x))2
= 2tanx1−tan2x
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)1–tan(x)tan(2x)
Since 2tanx1−tan2x
Now, putting all the things together:
tan(3x) = tan(x) + 2tan?(x)1−tan2x/1-tan2(x) / 1 – tan(x) . 2(2tanx1−tan2x)
Multiplying the numerator and denominator by 1−tan (x)
So, tan (3x) = tan(x)−tan3(x)+2tan(x)1−tan2(x)−2tan2(x)
Hence,
Tan(3x) = tan(x)⋅(3−tan2(x))1−3tan2(x)