Question

cos A + cos B - cos C = -1+4 cosA/2 cosB/2 sinC/2​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{cos(A)+cos(B)-cos(C)}$

$\tt{=2\,cos\left(\dfrac{A+B}{2}\right)\,cos\left(\dfrac{A-B}{2}\right)-1+2\,sin^2\left(\dfrac{C}{2}\right)}$

$\tt{=2\,cos\left(\dfrac{\pi-C}{2}\right)\,cos\left(\dfrac{A-B}{2}\right)-1+2\,sin^2\left(\dfrac{C}{2}\right)}$

$\tt{=-1+2\,cos\left(\dfrac{\pi}{2}-\dfrac{C}{2}\right)\,cos\left(\dfrac{A-B}{2}\right)+2\,sin^2\left(\dfrac{C}{2}\right)}$

$\tt{=-1+2\,sin\left(\dfrac{C}{2}\right)\,cos\left(\dfrac{A-B}{2}\right)+2\,sin^2\left(\dfrac{C}{2}\right)}$

$\tt{=-1+2\,sin\left(\dfrac{C}{2}\right)\left\{cos\left(\dfrac{A-B}{2}\right)+sin\left(\dfrac{C}{2}\right)\right\}}$

$\tt{=-1+2\,sin\left(\dfrac{C}{2}\right)\left\{cos\left(\dfrac{A-B}{2}\right)+sin\left(\dfrac{\pi-A-B}{2}\right)\right\}}$

$\tt{=-1+2\,sin\left(\dfrac{C}{2}\right)\left\{cos\left(\dfrac{A-B}{2}\right)+sin\left(\dfrac{\pi}{2}-\dfrac{A+B}{2}\right)\right\}}$

$\tt{=-1+2\,sin\left(\dfrac{C}{2}\right)\left\{cos\left(\dfrac{A-B}{2}\right)+cos\left(\dfrac{A+B}{2}\right)\right\}}$

$\tt{=-1+2\,sin\left(\dfrac{C}{2}\right)\cdot2\,cos\left(\dfrac{A}{2}\right)\,cos\left(\dfrac{B}{2}\right)}$

$\tt{=-1+4\,cos\left(\dfrac{A}{2}\right)\,cos\left(\dfrac{B}{2}\right)\,sin\left(\dfrac{C}{2}\right)}$