Why doesn't COS40+COS30 = COS70?

Question

I have tried 3 or 4 times, but I cannot get them to be equal.

I even tried it on my Calculator [TI~83 Plus], which says

COS30 + COS40 = 1.632069847   

but COS70 = .3420201433

Answer 1

Cosine is the ratio of two sides of a right triangle. If you draw a right triangle with a base angle of 30°, you will find that the length of the hypotenuse is exactly twice the length of the side opposite the angle. This is the sine ratio for 30°. The length of the side adjacent to the 30° angle divided by the length of the hypotenuse is about 0.87. That's the cosine of 30°. Now look at the other angle of the right triangle. It's 60°, which has a cosine of 0.5 (=½, same as sin30°, because opposite and adjacent change places).

If we now rotate the hypotenuse through another 10°, making it 40° to the base, the base length will shrink. So the ratio of its length compared to the length of the hypotenuse (the cosine) gets smaller. If we reduce the size of the angle, this ratio gets bigger. So cos40° is smaller than cos30°. Therefore adding the two cosines has got to be less than cos70°. In fact cos70° is less than cos60°, that is, less than a half.

Sine, cosine and tangent are often known as circular measurements. If you draw a circle on graph paper with its centre at the origin, then draw a radius to any point on the circumference, and from that point you draw perpendiculars to the x and y axes, the ratio of the lengths of these perpendiculars to the fixed radius length changes as you move the point round the circumference. You should then be able to appreciate that these ratios (called trigonometric ratios) do not increase or decrease at a constant rate. Try this for yourself!

So, for example, cos60°+cos30° is not cos90°. In fact, cos90°=0, while cos60°=½ and cos30°=0.87 (approx) and 0.5+0.87=1.37, not zero. There is a formula for adding two cosines:

cos(A)+cos(B)=2cos(½(A+B))cos(½(A-B)).

If A=40° and B=30°, cos40°+cos30°=2cos35°cos5°.

Answer 2

1. First, think of a linear function such as y=f(x)=2x that hasn't any constants. The graph is a straight line that passes thru the origin(0,0) and the slope is   y/x=2 (x isn't 0).  The ratio of y to x is always constant(=2) means the function y is directly proportional to the variant x.                                                                CK : Substitute, e.g. 2, 4 and 2+4 for x.  f(2)=2*2=4, f(4)=2*4=8. So, f(2)+f(4)=4+8=12.                                              While f(2+4)=f(6)=2*6=12.  That is, f(2)+f(4)=f(2+4)  CKD.                Therefore, the equation f(a)+f(b)=f(a + b) is applicable to the function y=f(x)=2x.

2. Second, think of a quadratic function such as y=f(x)=2x² that hasn't any constants in it.  The graph is a parabola that is convex down and the bottom touches the origin tangentially.  The slope, rate of change or tangent, is f'(x)=4x.  So, the ratio of y to x changes as x varies.   That is, y is not proportional to x.        CK : Substitute, e.g. 2, 4 and 2+4 for x.  f(2)=2*2²=8 and f(4)=2*4²=32.            So, f(2)+f(4)=8+32=40.                                                                            While, f(2+4)=f(6)=2*6²=72.  That is, f(2)+f(4) is not equal to f(2+4).  CKD.      Therefor, the equation f(a)+f(b)=f(a + b) is not applicable to y=f(x)=2x².

3. In the same manner, we examine the cosine function y=f(x)=cosx.   The graph is a periodic repetition of a smooth curve that fluctuates in identical waves.            The amplitude is 1, the period is 2*(pi), y-intersects are {2*(pi)*n, 1) or {(2n+1)*(pi), -1} and x-intersects are {(n+½)*(pi) , 0}.                                    (Even a 1-period of rough sketch of the graph, y=f(x)=cosx on a x-y coordinate,    will help you understand this problem.)                                                              The slope, rate of change or tangent, is f'(x)=-sinx.  So, the ratio of y to x varies. That means y=cosx is not proportional to x.                                                     CK : f(40)=cos40=approx. 0.766, f(30)=cos30=approx. 0.866.  So, f(40)+f(30)=cos40+cos30=approx. 1.632.                                            While  f(40+30)=f(70)=cos70=approx.0.342  So that cos40+cos30 is not equal to cos70.  CKD.                                                                                               Therefore, the answer for this question is that the equation f(a)+f(b)=f(a + b) is not applicable to cos40+cos30.   

The formula for the sum of Cosine of 2 angles A and B to product form is :            cosA+cosB=2cos½(A+B)*cos½(A-B)  Put 40 and 30 into A and B respectively.      We have : cos40+cos30=2cos35*cos5=approx. 1.632