draw

Let's first prove that cos(A-B)=cosAcosB+sinAsinB.

In a standard circle, AB is the diameter and the length is 1. According to the nature of the circle, the angle ADB and the angle ACB are 90 degrees. Another vertical line CE was drawn on the advertisement.

Let angle a be angle BAC.

Angle b is angle DAC

Then angel (A-B) is angel bad.

Proved as follows:

cos(A-B)= AD/AB = AD①cosA = AC/AB = AC②sinA = BC/AB = BC③cosB = AE/AC？ ④sinB=CE/AC

Simultaneous ① ③ cosB=AE/cosA is known to mean cosAcosB=AE.

So to prove cos(A-B)=cosAcosB+sinAsinB is to prove AD=AE+sinAsinB.

AD=AE+ED, that is, it is only proved that sinAsinB=ED.

That is to prove BC*CE/AC=ED.

That is to say, it is necessary to prove that CE/AC=ED/BC.

Note that the triangle CEF is similar to the triangle BDF (the three angles are the same), so we can know that ED/BC=EF/CF (similar triangles's theorem).

So to prove the proposition, we only need to prove ce/AC = ef/cf.

Note that angle ECF+ angle ECA=90 degrees, angle ECA+ angle CAE=90 degrees. Known angle ECF= angle EAC. And angle CEF= angle AEC=90 degrees. It can be concluded that the triangle AEC is similar to the triangle CEF.

That is, it can be proved that ce/AC = ef/cf.

That is to say, it proves that cos(A-B)=cosAcosB+sinA+sinB.

By sinθ=cos(-θ)?

De: sin(α+β)=cos[-(α+β)]

=cos[(-α)-β]？

=cos(-α)cosβ+sin(-α)sinβ？

∫cos(-α)= sinα？

sin(-α)=cosα？

∴sin(α+β)=sinαcosβ+cosαsinβ Agree 10|? comment