高専の数学解説

第14章 14.12

(1)

4sin⁡A2sin⁡B2cos⁡C2=4sin⁡A2⋅12{sin⁡B+C2+sin⁡B−C2}=2{sin⁡A2sin⁡B+C2+sin⁡A2sin⁡B−C2}4\sin\frac{A}{2}\sin\frac{B}{2}\cos\frac{C}{2} = 4\sin\frac{A}{2}\cdot\frac{1}{2}\left\{\sin\frac{B+C}{2}+\sin\frac{B-C}{2}\right\} = 2\left\{\sin\frac{A}{2}\sin\frac{B+C}{2}+\sin\frac{A}{2}\sin\frac{B-C}{2}\right\}

=2{sin⁡A2sin⁡180−A2+sin⁡A2sin⁡B−C2}=2{sin⁡A2cos⁡A2−12cos⁡A+B−C2+12cos⁡A−B+C2}= 2\left\{\sin\frac{A}{2}\sin\frac{180-A}{2}+\sin\frac{A}{2}\sin\frac{B-C}{2}\right\} = 2\left\{\sin\frac{A}{2}\cos\frac{A}{2}-\frac{1}{2}\cos\frac{A+B-C}{2}+\frac{1}{2}\cos\frac{A-B+C}{2}\right\}

=2{sin⁡A2cos⁡A2−12cos⁡180−2C2+12cos⁡180−2B2}= 2\left\{\sin\frac{A}{2}\cos\frac{A}{2}-\frac{1}{2}\cos\frac{180-2C}{2}+\frac{1}{2}\cos\frac{180-2B}{2}\right\}

=2{12sin⁡A−12sin⁡C+12sin⁡B}=sin⁡A+sin⁡B−sin⁡C= 2\left\{\frac{1}{2}\sin A-\frac{1}{2}\sin C+\frac{1}{2}\sin B\right\} = \sin A+\sin B-\sin C

(2)

4sin⁡A2sin⁡B2sin⁡C2+1=−2{cos⁡A+B2−cos⁡A−B2}sin⁡C2+14\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}+1 = -2\left\{\cos\frac{A+B}{2}-\cos\frac{A-B}{2}\right\}\sin\frac{C}{2}+1

=−2{cos⁡180−C2−cos⁡A−B2}sin⁡C2+1=−2sin⁡C2sin⁡C2+sin⁡A−B+C2+sin⁡A−B−C2+1= -2\left\{\cos\frac{180-C}{2}-\cos\frac{A-B}{2}\right\}\sin\frac{C}{2}+1 = -2\sin\frac{C}{2}\sin\frac{C}{2}+\sin\frac{A-B+C}{2}+\sin\frac{A-B-C}{2}+1

=cos⁡C−1+cos⁡B+cos⁡A+1=cos⁡A+cos⁡B+cos⁡C= \cos C-1+\cos B+\cos A+1 = \cos A+\cos B+\cos C

(3)

a−c⋅cos⁡Bb−c⋅cos⁡A=a−c⋅a2+c2−b22acb−c⋅b2+c2−a22bc=b(2a2−a2−c2+b2)a(2b2−b2−c2+a2)=b(a2−c2+b2)a(b2−c2+a2)=ba=sin⁡Bsin⁡A\frac{a-c\cdot\cos B}{b-c\cdot\cos A} = \frac{a-c\cdot\dfrac{a^2+c^2-b^2}{2ac}}{b-c\cdot\dfrac{b^2+c^2-a^2}{2bc}} = \frac{b(2a^2-a^2-c^2+b^2)}{a(2b^2-b^2-c^2+a^2)} = \frac{b(a^2-c^2+b^2)}{a(b^2-c^2+a^2)} = \frac{b}{a} = \frac{\sin B}{\sin A}

解説: r31bn1z