Question

If condition is given by \[A+B+C={{180}^{\circ }}\] then prove the following:
\[a.{{\sin }^{2}}A+{{\sin }^{2}}B-{{\sin }^{2}}C=2\sin A\sin B\cos C\]
\[b.{{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=2\left( 1+\cos A\cos B\cos C \right)\]

Answer 1

Hint: We have to start by rearranging the given condition as \[A+B={{180}^{\circ }}-C\]. Then we have to take sine on both sides, use identities \[\sin (180-\theta )=\sin \theta \] , \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\] and using suitable operations, convert it to the form of \[\left( {{\sin }^{2}}A{{\cos }^{2}}B+{{\cos }^{2}}A{{\sin }^{2}}B+2\sin A\cos A\sin B\cos B \right)={{\sin }^{2}}\left( C \right)\]. Then, we can take the LHS of the first expression to be proved and substitute this result and simplify further. Similarly, we will consider LHS of second expression and again substitute the result in it.

Complete step-by-step answer:
We have been given that \[A+B+C={{180}^{\circ }}\].
Let us subtract C on both sides,
\[A+B={{180}^{\circ }}-C\]
Applying sin on both sides, we get
\[\sin \left( A+B \right)=\sin \left( {{180}^{\circ }}-C \right)\]
We know that \[\sin (180-\theta )=\sin \theta \], so we can write
\[\Rightarrow \sin \left( A+B \right)=\sin \left( C \right)\]
We know that, \[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\] substitute in the above equation
\[\Rightarrow \sin A\cos B+\cos A\sin B=\sin \left( C \right)\]\[.........(1)\]
Squaring on both the sides for equation 1, we get
\[\Rightarrow {{\left( \sin A\cos B+\cos A\sin B \right)}^{2}}={{\sin }^{2}}\left( C \right)\]
\[\Rightarrow \left( {{\sin }^{2}}A{{\cos }^{2}}B+{{\cos }^{2}}A{{\sin }^{2}}B+2\sin A\cos A\sin B\cos B \right)={{\sin }^{2}}\left( C \right)\]\[.........(2)\]
From the statement (a)
\[{{\sin }^{2}}A+{{\sin }^{2}}B-{{\sin }^{2}}C=2\sin A\sin B\cos C\]
Consider \[LHS={{\sin }^{2}}A+{{\sin }^{2}}B-{{\sin }^{2}}C....\left( 3 \right)\]
Now substitute equation 2 in equation 3 and we get
\[\begin{align}
  & LHS={{\sin }^{2}}A+{{\sin }^{2}}B-\left( {{\sin }^{2}}A{{\cos }^{2}}B+{{\cos }^{2}}A{{\sin }^{2}}B+2\sin A\cos A\sin B\cos B \right) \\
 & LHS={{\sin }^{2}}A+{{\sin }^{2}}B-{{\sin }^{2}}A{{\cos }^{2}}B-{{\cos }^{2}}A{{\sin }^{2}}B-2\sin A\cos A\sin B\cos B \\
 & LHS={{\sin }^{2}}A-{{\sin }^{2}}A{{\cos }^{2}}B+{{\sin }^{2}}B-{{\cos }^{2}}A{{\sin }^{2}}B-2\sin A\cos A\sin B\cos B \\
\end{align}\]
\[{{\sin }^{2}}A\] and \[{{\sin }^{2}}B\] is taken common and we get,
\[={{\sin }^{2}}A(1-{{\cos }^{2}}B)+{{\sin }^{2}}B(1-{{\cos }^{2}}A)-2\sin A\cos A\sin B\cos B\]
Since we know that we can write $ {{\sin }^{2}}x=1-{{\cos }^{2}}x $ ,
\[={{\sin }^{2}}A{{\sin }^{2}}B+{{\sin }^{2}}A{{\sin }^{2}}B-2\sin A\cos A\sin B\cos B\]
\[=2{{\sin }^{2}}A{{\sin }^{2}}B-2\sin A\cos A\sin B\cos B\]
Now we will take \[2\sin A\sin B\] common as,
\[\begin{align}
  & =2\sin A\sin B(\sin A\sin B-\cos A\cos B) \\
 & =-2\sin A\sin B(\cos A\cos B-\sin A\sin B) \\
\end{align}\]
Since we know that \[\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)\] and also from question, \[A+B={{180}^{\circ }}-C\], we get that
\[=-2\sin A\sin B\cos (180{}^\circ -C)\]
We know that, \[\cos ({{180}^{\circ }}-C)=-\cos C\] . Substituting in the above equation, we get
\[=2\sin A\sin B\cos C\]
Hence, proved that LHS=RHS.
Hence, proved that \[{{\sin }^{2}}A+{{\sin }^{2}}B-{{\sin }^{2}}C=2\sin A\sin B\sin C\]
From the statement (b):
\[{{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=2\left( 1+\cos A\cos B\cos C \right)\]
Consider LHS as \[{{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C\]\[.........(3)\]
Now substitute equation 2 in equation 3
\[\begin{align}
  & LHS={{\sin }^{2}}A+{{\sin }^{2}}B+\left( {{\sin }^{2}}A{{\cos }^{2}}B+{{\cos }^{2}}A{{\sin }^{2}}B+2\sin A\cos A\sin B\cos B \right) \\
 & LHS={{\sin }^{2}}A+{{\sin }^{2}}A{{\cos }^{2}}B+{{\sin }^{2}}B+{{\cos }^{2}}A{{\sin }^{2}}B+2\sin A\cos A\sin B\cos B \\
\end{align}\]
Taking \[{{\sin }^{2}}A\] and \[{{\sin }^{2}}B\] as common and we get,
\[={{\sin }^{2}}A(1+{{\cos }^{2}}B)+{{\sin }^{2}}B(1+{{\cos }^{2}}A)+2\sin A\cos A\sin B\cos B\]
Now using identity $ {{\sin }^{2}}x=1-{{\cos }^{2}}x $ and substituting \[{{\sin }^{2}}A=1-{{\cos }^{2}}A\] and \[{{\sin }^{2}}B=1-{{\cos }^{2}}B\], we get
\[\begin{align}
  & =(1-{{\cos }^{2}}A)(1+{{\cos }^{2}}B)+(1-{{\cos }^{2}}B)(1+{{\cos }^{2}}A)+2\sin A\cos A\sin B\cos B \\
 & =1+{{\cos }^{2}}B-{{\cos }^{2}}A-{{\cos }^{2}}A{{\cos }^{2}}B+1+{{\cos }^{2}}A-{{\cos }^{2}}B-{{\cos }^{2}}B{{\cos }^{2}}A \\
 & \text{ }+2\sin A\cos A\sin B\cos B \\
\end{align}\]
Cancelling similar terms, we get
\[=2-2{{\cos }^{2}}A{{\cos }^{2}}B+2\sin A\cos A\sin B\cos B\]
Now taking \[\cos A\cos B\] common, we get
\[=2-2\cos A\cos B(\cos A\cos B-\sin A\sin B)\]
Again taking 2 common, we get
\[=2(1-\cos A\cos B(\cos A\cos B-\sin A\sin B))\]
Since we know that \[\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)\] and also from question, \[A+B={{180}^{\circ }}-C\], we get that
\[=2(1-\cos A\cos B\cos ({{180}^{\circ }}-C))\]
We know that, \[\cos ({{180}^{\circ }}-C)=-\cos C\] . Substituting in the above equation, we get
\[=2(1+\cos A\cos B\cos (C))\]
= RHS
Hence, proved that LHS=RHS.
Hence proved \[{{\sin }^{2}}A+{{\sin }^{2}}B+{{\sin }^{2}}C=2\left( 1+\cos A\cos B\cos C \right)\]

Note: This question requires a good understanding of trigonometric functions and where to apply them. A lot of care must be taken while simplifying the terms and making sure that no sign mistake or formula mistake is made. Most of the times, students get the formula wrong as \[\cos A\cos B-\sin A\sin B=\cos \left( A-B \right)\] instead of \[\cos A\cos B-\sin A\sin B=\cos \left( A+B \right)\]. If they do this, then they might never be able to simplify and prove the expressions.