Question

If \[\sin A+\cos A=\sqrt{2}\], then \[\sin A-\cos A\] is equal to?

Answer 1

Hint: In this problem, we have to find the value of \[\sin A-\cos A\], if \[\sin A+\cos A=\sqrt{2}\]. Here we have to take squaring on both sides to the given condition. We can then expand them using the algebraic whole square formula and simplify it step by step to find the value of the required data.

Complete step by step answer:
Here we have to find the value of \[\sin A-\cos A\].
We are given that,
\[\Rightarrow \sin A+\cos A=\sqrt{2}\]
We can now square on both sides of the given condition, we get
\[\Rightarrow {{\left( \sin A+\cos A \right)}^{2}}={{\left( \sqrt{2} \right)}^{2}}\]
 We can now use the algebraic whole square formula \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] to expand the expression in the left-hand side and cancel the square root and the whole square in the right-hand sides, we get
\[\Rightarrow {{\sin }^{2}}A+{{\cos }^{2}}A+2\sin A\cos A=2\]
We know that, \[{{\sin }^{2}}A=1-{{\cos }^{2}}A,{{\cos }^{2}}A=1-{{\sin }^{2}}A\], we can now substitute these identities in the above step, we get
 \[\Rightarrow 1-{{\cos }^{2}}A+1-{{\sin }^{2}}A+2\sin A\cos A=2\]
 We can now simplify the above step by adding the constant, we get
\[\Rightarrow 2-\left( {{\cos }^{2}}A+{{\sin }^{2}}A-2\sin A\cos A \right)=2\]
We can now write the above terms in the whole square format, we get
\[\Rightarrow 2-{{\left( \sin A-\cos A \right)}^{2}}=2\]
We can now subtract the number 2 on both sides, we get
\[\Rightarrow {{\left( \sin A-\cos A \right)}^{2}}=2-2\]
We can now simplify the above term, we get
\[\Rightarrow \sin A-\cos A=0\]
Therefore, if \[\sin A+\cos A=\sqrt{2}\], then the value of \[\sin A-\cos A\] is 0.

Note: We should always remember the trigonometric identities and formulas to substitute the equivalent form such as \[{{\sin }^{2}}A=1-{{\cos }^{2}}A,{{\cos }^{2}}A=1-{{\sin }^{2}}A\]. We should also remember the algebraic whole square formula such as the whole square of the addition \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\] to expand the terms for further simplification.