Question

What is the Value if Sin3thita=? please try it

Answer 1

There are several methods to show this. One of the basic method is by using the some of the simplest formulae of trigonometry. So let's start.

You must be aware of the formula

sin(A+B)=sinAcosB+cosAsinBsin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B

cos(A+B)=cosAcosB−sinAsinBcos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B

Now first find sin2xsin⁡2x using the above formula.

Replace both A and B with xx in 1st formula

We get sin(x+x)=sinxcosx+cosxsinx.sin⁡(x+x)=sin⁡xcos⁡x+cos⁡xsin⁡x.

Therefore sin2x=2sinxcosx…(1)sin⁡2x=2sin⁡xcos⁡x…(1)

Similarly

cos2x=cos2x−sin2xcos⁡2x=cos2⁡x−sin2⁡x

Using sin2x+cos2x=1…(2)sin2⁡x+cos2⁡x=1…(2)

cos2x=2cos2x−1cos⁡2x=2cos2⁡x−1

Again using eqn (2)(2)

cos2x=1−2sin2x…(3)cos⁡2x=1−2sin2⁡x…(3)

Now for sin3xsin⁡3x replace A and B with xx and 2x2x or vice versa in the first formula.

We get sin(x+2x)=sinxcos2x+cosxsin2xsin⁡(x+2x)=sin⁡xcos⁡2x+cos⁡xsin⁡2x

Subtituting the values of sin2xsin⁡2x and cos2xcos⁡2xwe get

sin3x=(sinx)(1−2sin2x)+(cosx)(2sinxcosx)sin⁡3x=(sin⁡x)(1−2sin2⁡x)+(cos⁡x)(2sin⁡xcos⁡x)

Using eqn (2)(2) and rearranging terms we get

sin3x=3sinx−4sin3xsin⁡3x=3sin⁡x−4sin3⁡x

Similarly

cos3x=4cos3x−3cosxcos⁡3x=4cos3⁡x−3cos⁡x

There's a second interesting Method and it involves complex number. For that you must know De Moivre's theorem which can be stated as

(cosθ+isinθ)n=(cosnθ+isinnθ)(cos⁡θ+isin⁡θ)n=(cos⁡nθ+isin⁡nθ)

So (cosθ+isinθ)3=cos3θ−isin3θ+3icos2θsinθ−3sin2θcosθ(cos⁡θ+isin⁡θ)3=cos3⁡θ−isin3⁡θ+3icos2⁡θsin⁡θ−3sin2⁡θcos⁡θ

Now using eqn (2)(2) we get

(cos3θ+isin3θ)=(4cos3θ−3cosθ)+i(3sinθ−4sin3θ)(cos⁡3θ+isin⁡3θ)=(4cos3⁡θ−3cos⁡θ)+i(3sin⁡θ−4sin3⁡θ)

When two complex number are equal then their real and imaginary part are also equal

Hence we get

cos3θ=4cos3θ−3cosθcos⁡3θ=4cos3⁡θ−3cos⁡θ

sin3θ=3sinθ−4sin3θsin⁡3θ=3sin⁡θ−4sin3⁡θ

I hope it helps

Answer 2

...may be we will use the formula of 'tripple one..'
but i don't know the formula......

so sryy

:)