In $\triangle ABC$, if $\sin^2{A}+\cos^2{C}=\cos^2{B}$,then $C=?$

Question

We have $\sin^2{A}+\cos^2{C}=\cos^2{B} \implies 2\sin^2{A}+2\cos^2{C}=2\cos^2{B} \implies 1-\cos{2A}+\cos{2C}-1=\cos{2B}-1 \implies 1+\cos{2C}=\cos{2B}+\cos{2A} \implies 1+\cos{2C}=2\cos(A+B)\cos(A-B) \implies 1+\cos{2C}=2\cos(\pi-A)\cos(A-B)$, since $A+B+C=\pi$ $\implies 1+\cos{2C}=2\cos{A}\cos(A-B)$

How can I find $C$ ? Please help.

Whenever you see such form of identity, my friend, the best optimization is that, identify this as a right-angle triangle. See, C=pi/2 makes sense, as, it leaves a= pi/2 + b and so, sin A = cos B. There are many approaches you may like, but be sure to identify. — Sheikh Asif Imran Shouborno, Oct 07 '14 at 06:37
Pardon, there is a typo in my previous comment. I did not know that if five minutes elapse, I would be rendered unable to edit my own comment. (Just to confess my mistake.) — Sheikh Asif Imran Shouborno, Oct 07 '14 at 06:45

score 3 · Accepted Answer · edited Apr 13 '17 at 12:20

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$$\sin^2A=\cos^2B-\cos^2C=\sin^2C-\sin^2B=\sin(C+B)\sin(C-B)$$

$$\implies\sin A\sin(\pi-\overline{B+C})=\sin(\pi-A)\sin(C-B)$$

$$\implies\sin A[\sin(B+C)-\sin(C-B)]=0$$

$$\implies\sin A[2\sin B\cos C]=0$$

$$\implies C = \frac\pi 2$$(as $ A,B,C\ne 0 $)

Alternatively, $$\cos^2C=\cos^2B-\sin^2A=\cos(B+A)\cos(B-A)$$

Again, $\cos(A+B)=\cos(\pi-C)=-\cos C$

$$\implies\cos C[-\cos(A+B)]=-\cos C\cos(B-A)\iff\cos C[\cos(B-A)-\cos(B+A)]=0$$

$$\iff\cos C[2\sin B\sin A]=0$$

$$\implies C = \frac\pi 2$$(as $ A,B,C\ne 0 $)

In $\triangle ABC$, if $\sin^2{A}+\cos^2{C}=\cos^2{B}$,then $C=?$

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