trigonometry

[r267747]

I havea difficulties to solve these questions, so please first see my work and then help me to solve these questions.
My work
1.PROVE THE FOLLOWING
a) sin alpha +sin(alpha+2pi/3)+ sin(alpha+4pi/3)=0
sol. using "C,D" FORMULA,
i.e,sinC+ sinD=2sin C/2+D/2 cosC/2-D/2
Actually i have a problem that in the above question which two term i will combine to form the formula so that i will get the answer zero. So please help me. simillarly i have a problem in two more questions which are:
2.sin10 degree+sin20degree+sin40degree+sin50degree=sin70degree=sin80degree
3.4sinA sin(60degree+A) sin(60degree-A)=sin3A

 

[Deleted member 4993]

I havea difficulties to solve these questions, so please first see my work and then help me to solve these questions.
My work
1.PROVE THE FOLLOWING
a) sin alpha +sin(alpha+2pi/3)+ sin(alpha+4pi/3)=0
sol. using "C,D" FORMULA,
i.e,sinC+ sinD=2sin C/2+D/2 cosC/2-D/2
Actually i have a problem that in the above question which two term i will combine to form the formula so that i will get the answer zero.

Do not understand what you are saying. Please post work - so that we know where you are having problem.

If you add, using the formula suggested, the second term and the third term of the Left-hand-side - the answer falls off.
So please help me. simillarly i have a problem in two more questions which are:
2.sin10 degree+sin20degree+sin40degree+sin50degree=sin70degree=sin80degree
3.4sinA sin(60degree+A) sin(60degree-A)=sin3A

 

[BigGlenntheHeavy]

1.a

$\displaystyle sin(\alpha)+sin(\alpha+\frac{2\pi}{3})+sin(\alpha+\frac{4\pi}{3}) \ = \ 0$

$\displaystyle sin(\alpha)+sin(\alpha)cos(\frac{2\pi}{3})+cos(\alpha)sin(\frac{2\pi}{3})+sin(\alpha)cos(\frac{4\pi}{3})+cos(\alpha)sin(\frac{4\pi}{3}) \ = \ 0, \ Sum \ Formula.$

$\displaystyle sin(\alpha)-\frac{sin(\alpha)}{2}+\frac{\sqrt3 cos(\alpha)}{2}-\frac{sin(\alpha)}{2}-\frac{\sqrt3 cos(\alpha)}{2} \ = \ 0$

$\displaystyle sin(\alpha)-sin(\alpha)+\frac{\sqrt3 cos(\alpha)}{2}-\frac{\sqrt3 cos(\alpha)}{2} \ = \ 0$

$\displaystyle 0 \ = \ 0$

 

[BigGlenntheHeavy]

3), don't understand 2

$\displaystyle 4sin(A)[sin(\pi/3+A)][sin(\pi/3-A)] \ = \ sin(3A)$

$\displaystyle 4sin(A)[sin(\pi/3)cos(A)+cos(\pi/3)sin(A)][sin(\pi/3)cos(A)-cos(\pi/3)sin(A)] \ = \ sin(3A), \ (Sum$

$\displaystyle and \ Differences \ Formulas).$

$\displaystyle 4sin(A)\bigg[\frac{\sqrt3 cos(A)}{2}+\frac{sin(A)}{2}\bigg]\bigg[\frac{\sqrt3 cos(A)}{2}-\frac{sin(A)}{2}\bigg] \ = \ sin(3A)$

$\displaystyle 4sin(A)\bigg[\frac{3cos^{2}(A)}{4}-\frac{\sqrt3 sin(A)cos(A)}{4}+\frac{\sqrt3 sin(A)cos(A)}{4}-\frac{sin^{2}(A)}{4}\bigg] \ = \ sin(3A)$

$\displaystyle 4sin(A)\bigg[\frac{3cos^{2}(A)}{4}-\frac{sin^{2}(A)}{4}\bigg] \ = \ sin(3A)$

$\displaystyle sin(A)[3cos^{2}(A)-sin^{2}(A)] \ = \ sin(3A)$

$\displaystyle sin(A)[3(1-sin^{2}(A))-sin^{2}(A)] \ = \ sin(3A), \ (sin^{2}(A)+cos^{2}(A) \ = \ 1).$

$\displaystyle sin(A)[3-4sin^{2}(A)] \ = \ sin(3A)$

$\displaystyle sin(A)\bigg[3-4\bigg(\frac{1-cos(2A)}{2}\bigg)\bigg] \ = \ sin(3A) \ (Power-Reducing \ Formulas).$

$\displaystyle sin(A)[3-2+2cos(2A)] \ = \ sin(3A)$

$\displaystyle sin(A)[1+2cos(2A)] \ = \ sin(3A)$

$\displaystyle sin(A)+2sin(A)cos(2A) \ = \ sin(3A)$

$\displaystyle sin(A)+2\frac{1}{2}[sin(3A)+sin(-A)] \ = \ sin(3A), \ (Product-to-Sum \ Formulas).$

$\displaystyle sin(A)+sin(3A)-sin(A) \ = \ sin(3A), \ (sin(-A) \ = \ -sin(A), \ odd \ function).$

$\displaystyle Hence, \ sin(3A) \ = \ sin(3A) \ QED$