Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → A + B = π - C
→ B + C = π - A
→ C + A = π - B
→ C = π - (B + C)
Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]
Use the Sum/Difference Identity: cos (A - B) = cos A · cos B + sin A · sin B
Use the Double Angle Identity: sin 2A = 2 sin A · cos A
Use the Cofunction Identity: cos (π/2 - A) = sin A
<u>Proof LHS → Middle:</u>
![\text{Factor:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\sin \bigg(\dfrac{A+B}{4}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cquad%20%20%3D2%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5B%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B4%7D%5Cbigg%29%2B%5Csin%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5D)
![\text{Cofunction:}\quad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[ \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{\pi}{2}-\dfrac{A+B}{4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =2\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{A-B}{4}\bigg)+\cos \bigg(\dfrac{2\pi-(A+B)}{4}\bigg)\bigg]](https://tex.z-dn.net/?f=%5Ctext%7BCofunction%3A%7D%5Cquad%20%20%3D2%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5B%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B4%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi%7D%7B2%7D-%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D2%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7BA-B%7D%7B4%7D%5Cbigg%29%2B%5Ccos%20%5Cbigg%28%5Cdfrac%7B2%5Cpi-%28A%2BB%29%7D%7B4%7D%5Cbigg%29%5Cbigg%5D)
![\text{Sum to Product:}\ 2\cos \bigg(\dfrac{A+B}{4}\bigg)\bigg[2 \cos \bigg(\dfrac{2\pi-2B}{2\cdot 4}\bigg)\cdot \cos \bigg(\dfrac{2A-2\pi}{2\cdot 4}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =4\cos \bigg(\dfrac{A+B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi-B}{4}\bigg)\cdot \cos \bigg(\dfrac{\pi -A}{4}\bigg)](https://tex.z-dn.net/?f=%5Ctext%7BSum%20to%20Product%3A%7D%5C%202%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Cbigg%5B2%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B2%5Cpi-2B%7D%7B2%5Ccdot%204%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B2A-2%5Cpi%7D%7B2%5Ccdot%204%7D%5Cbigg%29%5Cbigg%5D%5C%5C%5C%5C%5C%5C.%5Cqquad%20%5Cqquad%20%5Cqquad%20%3D4%5Ccos%20%5Cbigg%28%5Cdfrac%7BA%2BB%7D%7B4%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi-B%7D%7B4%7D%5Cbigg%29%5Ccdot%20%5Ccos%20%5Cbigg%28%5Cdfrac%7B%5Cpi%20-A%7D%7B4%7D%5Cbigg%29)
LHS = Middle 
<u>Proof Middle → RHS:</u>
Middle = RHS
Answer:
yes
Step-by-step explanation:
I'm going to give you a slightly different answer, but it's going to make sense :-)
First, let's review what "sin" and "cos" really mean. They are functions that take as an input an angle, which we call theta. They output the base (cos) and height (sin) of a triangle which as a hypotenuse of length 1.
Now, let's pick some examples. If we happen to set theta to 45 degress, you will get a triangle that looks like this:
In this case, both sin(theta) and cos(theta) are the same number, the square root of 1/2. So cos(theta) + cos (theta) is 2 times the square tool of 1/2.
Now imagine that we now want to find cos (theta + theta). Remember that theta was 45 degrees, so this will be cos (45 + 45), or cos (90).
But remember that cos is the base of a triangle where theta is the angle with the base. Well, that's not a triangle at all, is it? It's just a vertical line. In fact, cos(90) will be zero.
Answer:
-4+2
Step-by-step explanation:
13,14,15,31,34,35,41,43,45,51,53,54 so 12
