How to Simplify the Expression sin60° + sin20°?

[rum2563]

[SOLVED] Simplifying Trig Products

Homework Statement

Express the following as a product and simplify.
sin60° + sin20°

Homework Equations

The Attempt at a Solution

I don't understand what the question is trying to say. For example, do I convert sin60° and then add. I don't understand. Someone please help.

 

[Moridin]

You are suppose to use a standard trigonometric identity that can be applied to sin a + sin b = some product. All you have to do is to find 'some product'.

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

 

[rum2563]

So, $$\sqrt{3}/2$$ + sin20°
= 1.208

Could this be right?

 

[Evalesco]

Hi rum2563,
I don't think you are required to solve it, you just ned to rewrite it in a different form.
You will need to use a trig identity which converts the sum to a product.
Moridin has pointed you in the right direction with the hint: "sin a + sin b = some product".
Look on the webpage Moridin has provided a link to and the equation you need is located under Sum-to-Product Formulas.

 

[rum2563]

Thanks Evalesco.

So, here it is:

sin60° + sin20° = 2sin(60+20/2)cos(60+20/2)

Is this right?

Also, because of the fact that you pointed this to me, there is another question I have.

The question states:
Using the method developed in Example 3 (my note: we don't have the book so we don't know what this method is) of this section, prove each of the following Transformation Formulas.

sinx - siny = 2cos(x+y/2)sin(x-y/2)

Could someone please help me with this? Also, thanks Evalesco and Moridin for helping me out.

 

[Moridin]

sin60° + sin20° = 2sin(60+20/2)cos(60+20/2)

Not quite; apply the correct formula under the headline "Sum-to-Product Formulas" and do not forget the signs or what should be divided with two.

 

[rum2563]

http://hh4.hollandhall.org/kluitwieler/pages/Advanced_Trig/Trig%20Frog%20Homepage/sumtoproduct.htm

About 10 seconds ago, I looked at the above webpage.

Here is what I came up with now:

sin60° + sin20° = 2sin40°cos20°

I hope finally it is correct.

Also, can you please help me with my second question which I posted in the last post? Thanks.

 

[Evalesco]

Yep it's correct.

sinx - siny = 2cos(x+y/2)sin(x-y/2)

Did you mean sinx - siny = 2cos((x+y)/2)sin((x-y)/2)?

I will start you out on one way of showing the above

Start out with sinx - siny

$$Then \ let \ x=\frac{x+y}{2}+\frac{x-y}{2} \ and \ let \ y=\frac{x+y}{2}-\frac{x-y}{2}$$

You will find the identities sin(a+b) = sin(a)cos(b)+cos(a)sin(b) and sin(a-b) = sin(a)cos(b)-cos(a)sin(b) will come in handy for the next step.

Let me know how you get on.

 

[rum2563]

Yes, I finally got it.

sin60° + sin20° = 2sin((x+y)/2)cos((x-y)/2)

A + B = 60°
A - B = 20°
---------- +
2A = 80°
A = 40°

A + B = 60°
A - B = 20°
---------- -
2B = 40°
B = 20°

sin60° + sin20° = 2sin(40°)cos(20°)

Yes, I finally got it. But could there have been an easier way? Or is this good enough? I tried my best anyways.

Thanks to Evalesco and Moridin for helping me out.