cos^2(x)-sin^2(x)=2cos^2(x)-1

Question

How do I prove that the identites are identified in this equation, with steps.

Answer 1

In right triangle ABC where ∠B=90°, AC²=AB²+BC² (Pythagoras), which can be written b²=c²+a², where a=length of BC, b=length of AC and c=length of AB.

sin(CÂB)=a/b; cos(CÂB)=c/b; sin²(CÂB)=a²/b²; cos²(CÂB)=c²/b²;

sin²(CÂB)+cos²(CÂB)=(a²+c²)/b²=b²/b²=1.

Let x=CÂB, then sin²(x)+cos²(x)=1, and sin²(x)=1-cos²(x).

cos²(x)-sin²(x)=cos²(x)-(1-cos²(x))=cos²(x)-1+cos²(x)=2cos²(x)-1.