(A) the purpose of the investigation

Sine theorem and cosine theorem are important trigonometric formulas in high school mathematics, which are widely used. However, the study of them in textbooks is relatively simple. In learning, in order to broaden our horizons and understand the mystery of mathematical flexibility, it is necessary for us to summarize, explore and expand it with the knowledge of trigonometric transformation. Therefore, we have explored some varieties and applications of it.

(2) Inquiry process, application and conclusion

(1) Sine and Cosine Theorem

1, sine theorem: a/sina = b a/sina = b/sinb = c/sinc = 2r.

2. Cosine theorem: A2 = B2+C2-2bcosacosa = (C2+B2-A2)/2bc.

b^2=a^2+c^2-2accosb cosb=(a^2+c^2-b^2)/2ac

c^2=a^2+b^2-2abcosc cosc=(a^2+b^2-c^2)/2ab

(2) Inference of sine and cosine theorem

Let the sides of three internal angles A, B and C of triangle ABC be A, B and C respectively, then

Inference 1, acosa+bcosb = CCOs (a-b) ≤ C. ...

bcosB+ccosC = acos(B-C) ≤ a......②

acosA+ccosC = bcos(A-C) ≤b......③

Proof: From sine theorem,

acosA+bcosB

=2RsinAcosA+2RsinBcosB

=R(2sinAcosA+2sinBcosB)

=R(sin2A+sin2B)

= R { sin[(A+B)+(A-B)]+sin[(A+B)-(A-B)]}

= R[sin(A+B)cos(A-B)+cos(A+B)sin(A-B)+sin(A+B)cos(A-B)-cos

(A+B) Crime (A-B)]

=2Rsin(A+B) cos(A-B)

=2Rsin(？ -C)A-B

=2RsinC cos(A-B)

=Ccos(A-B)

a，B∈(0，？ )，- 1≤cos(A-B) ≤ 1

∴ccos(A-B)≤C, take the equal sign if and only if A = B. 。

Similarly, formula ② ③ can be proved by the symmetry of three sides and three angles of a triangle.

Application: in ⊿ABC, verification: COSAcoSBOSC ≤ 1/8.

It is proved that: ① When ①⊿abc is an obtuse triangle or a right triangle, one of cosA, cosB and cosC must be less than or equal to 0, so the conclusion is valid.

② If ⊿ABC is an acute triangle, it is obtained by inference (1) and mean inequality.

A≥bcosB+ccosC≥2 times root BCOSBCOSC > 0 ...

B≥acosA+ccosC≥2 times root number Acosaccosc > 0...②

C≥acosA+bcosB≥2 root number acosabcosb > 0...③

①×× ②× ③ abC≥8abCcosAcosBcosC

∴cosAcosBcosC≤ 1/8

Conclusion: ① In a triangle, the sum of cosine values of any two sides and their diagonals is equal to the third side and two sides.

The product of cosine of diagonal difference of one side is less than or equal to the third side.

② The product of cosine values of three angles of a triangle is less than or equal to 1/8.

(3) Observing the formula, we can draw the following conclusions.

A, if the two angles and the corresponding two sides in a triangle are known, we can know the value range or minimum value of the third side.

B, if you know the two angles in a triangle, you can know the quantitative relationship between the three sides.

Inference 2, c/(a+b) = sin (c/2)/cos [(a-b)/2] ≥ sin (c/2). ...

B/(A+C)= sin(B/2)/cos[(A-C)/2]≥sin(B/2)......②

A/(B+C)= sin(A/2)/cos[(B-C)/2]≥sin(A/2)......③

Proof: Starting from sine theorem,

c/(a+b)=(2R sinc)/[2R(sinA+sinB)]

=sin(？ -c)/(sinA+sinB)

=sin(A+B)/ (sinA+sinB)

= sin[(A+B)/2+(A+B)/2]/{ sin[(A+B)/2+(A-B)/2]+

sin[(A+B)/2-(A-B)/2]}

= { 2 sin[(A+B)/2]cos[(A+B)/2]}/{ sin[(A+B)/2]cos[(A-B)/2]+sin[(A-B)/2]cos[(A+B)/2]+sin[(A+B)/2]cos

[(A-B)/2]—sin[(A-B)/2]cos[(A+B)/2]}

= { 2 sin[(A+B)/2]cos[(A+B)/2]}/{ 2 sin[(A+B)/2]cos[(A-B)/2]}

=cos[(A+B)/2]/ cos[(A-B)/2]

=sin[？ /2—(A+B)/2]/ cos[(A-B)/2]

=sin(C/2)/cos[(A-B)/2]

a，B∈(0，？ )∴ 00

∴3 cos2？ =3-2 cos2？ >0 ∴2? 、2? ∈(0,? /2)

3sin2 again? -2Sin2？ =0 ∴3/Sin2？ =2/sin2？

Construct ⊿ABC so that A=2? ，B=2？ ，BC=2，AC=3。

According to inference 6, AB=ACcos2? +BCcos2？

= 3cos2？ +2cos2？ =3

∴AB=AC ∴⊿ABC is an isosceles triangle.

∴C=B=2？

At ⊿ABC, A+B+C=2? +2? +2? = 180 degrees

∴? +2? =90 degrees

Conclusion: ① Inference 6 is a famous projective theorem.

② Application: It can handle the transformation of edges, angles and chords.