Sin3A Cos3A Tan3A formulas

In this post, we’re going to discuss Sin3a,cos3a, and tan3a formulas. This is the concept of Trigonometric function at multiplies of angle such sin2x,sin\frac{x}{2}…etc.We’ll see sin3a ,co3a and tan3a derivation with example .

But, to derive given formulas you must know other basic trigonometric formulas.

                              sin(x+y)              sinxcosy+cosxsiny 
                              cos(x+y)               cosxcosy-sinxsiny
                              sin2x               2sinxcosx
                              cos2x               1-2sin²x
                              tan2x              \frac{2tanx}{1-tan²x}

Sin3x in term of x

sin3x=3sinx-4sin³x

Sin3x derivation 

Sin3x=3sinx-4sin³x

Derivation:

We know ,sin(x+y)=sinxcosy+cosxsiny 

Substitute y=2x 

Sin(x+2x)=sinxcos2x+cosxsin2x 

Sin(x+2x)=sinx(1-2sin²x)+cosx(2sinxcosx)   [Using : Cos2x= 1-2sin²x ,Sin2x=2sinxcosx]

Sin(3x)= sinx-2sin³x + 2sinxcos²x

Sin(3x)=sinx-2sin³x + 2sinx(1-sin²x)   [Using : sin²x+cos²x=1] 

Sin(3x)= sinx-2sin³x+2sinx-2sin³x

Sin(3x)= 3sinx-4sin³x

Cos3x in the term of x 

Cos3x=4cos³x-3cosx 

Cos3x derivation 

To Derive : Cos3x=4cos³x-3cosx 

Derivation:

We know ,cos(x+y)=cosxcosy-sinxsiny 

Substitute y=2x 

Cos(x+2x)=CosxCos2x-SinxSin2x

Cos(x+2x)=Cosx(2cos²x-1)-Sinx(2sinxcosx) [Using: Cos2x=2cos²x-1 ,Sin2x=2sinxcosx]

Cos(3x)=2Cos³x-Cosx -2Sin²xCosx 

Cos(3x)= 2Cos³x-Cosx-2Cosx(1-Cos²x)       [Using : sin²x+cos²x=1]

Cos(3x)=2Cos³x-Cosx-2Cosx+2Cos³x

Cos(3x)= 4Cos³x-3Cosx

Tan3x in the term of x 

Tan3x=\frac{3tanx-tan³x}{1-3tan²x}

Tan3x Derivation 

To Derive: Tan3x=\frac{3tanx-tan³x}{1-3tan²x}

Derivation: 

We know ,Tan(x+y)=\frac{tanx+tany}{1-tanxtany}

Substitute y=2x

Tan(x+2x)=\frac{tanx+tan2x}{1-tanxtan2x}

Replace tan2x= \frac{2tanx}{1-tan²x}

Simplifying it ,we get 

Tan(3x)= \frac{tanx(1-tan²x)+2tanx}{1-tan²x-2tan²x}

Tan(3x)=\frac{3tanx-tan³x}{1-3tan²x}

Problem:01 If SinA=\frac{8}{10} ,then find the value of Sin3A 

Solution:

We have SinA=\frac{8}{10}

And We know Sin3x=3sinx-4sin³x

3sinA-4sin³A=Sin3A

Substitute SinA=\frac{8}{10} in the equation 

3(\frac{8}{10})-4(\frac{8}{10})³=Sin3x 

\frac{24}{10}-4\frac{512}{1000}=Sin3x

\frac{24}{10}\frac{2048}{1000}=Sin3x 

\frac{2400-2048}{1000}=Sin3x 

\frac{448}{1000}=Sin3x

Hence , Sin3x=\frac{448}{1000}

Problem:02 \frac{Sin3x}{1+2Cos2x}=Sinx 

Solution:

\frac{Sin3x}{1+2Cos2x}

We know Sin3x=3sinx-4sin³x and cos2x=1-2sin²x

Substitute these values in \frac{Sin3x}{1+2Cos2x}

\frac{3sinx-4sin³x}{1+2(1-2sin²x)}

\frac{3sinx-4sin³x}{1+2-4 sin²x)}

\frac{3sinx-4sin³x}{3-4 sin²x)}

\frac{sinx(3-4sin²x)}{3-4 sin²x)}=Sinx 

Check other stuff as well 

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