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|[cos(A-P),cos(A-Q),cos(A-R)],[cos(B-P),...

|[cos(A-P),cos(A-Q),cos(A-R)],[cos(B-P),cos(B-Q),cos(B-R)],[cos(C-P),cos(C-Q),cos(C-R)]|

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cos(-A+B+C)+cos(A-B+C)+cos(A+B-C)+cos(A+B+C)=4cos A cos B cos C

Prove that: cos(A+B+C)+cos(A-B+C)+cos(A+B-C)+cos(-A+B+C)=4cos A cos B cos C

(c-b cos A)/(b-c cos A)=(cos B)/(cos C)

Prove that: cos(A+B+C)+cos(A-B+C)+cos(A+B-C)+cos(-A+B+C)=4cos A\ cos B\ cos C

Prove that: cos(A+B+C)+cos(A-B+C)+cos(A+B-C)+cos(-A+B+C)=4cos A\ cos B\ cos C

If A+B+C = 2S , then P.T cos(S-A)+cos(S-B)+cos(S-C)+cosS=4cos.(A)/(2)cos.(B)/(2)cos.(C)/(2)

If A,B and C are the angles of a triangle, then |[-1+cos B, cos C+ cos B, cos B],[cos C+ cos A,-1+cos A, cos A],[-1+cos B,-1+cos A,-1]|

If : A+B+C=pi, "then" : 1-cos^(2) A-cos^(2)B-cos^(2)C= A) 2 cos A * cos B * cos C B) 2 sin A * cos B * cos C C) 2 cos A * sin B * cos C D) 2 cos A * cos B * sin C

4R cos ((A) / (2)) cos ((B) / (2)) cos ((C) / (2)) =