Question

cos2A - cos2B - cos2C = -1 + 4cosA sinB sinC​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{cos(2A)-cos(2B)-cos(2C)}$

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

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

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

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

$\sf{=-1+2\,sin\left(C\right)\,\big\{sin\left(C\right)-sin\left(A-B\right)\big\}}$

$\sf{=-1+2\,sin\left(C\right)\,\big\{sin\big(\pi-(A+B)\big)-sin\left(A-B\right)\big\}}$

$\sf{=-1+2\,sin\left(C\right)\,\big\{sin(A+B)-sin\left(A-B\right)\big\}}$

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

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

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

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