Question

What is the formula for \[\left( {{{\sin }^2}A + {{\sin }^2}B} \right)\]?

Answer 1

Hint: Here in this question, we have to find the formula of given trigonometric function. this can be solve by, using a formula of double angle of cosine trigonometric function i.e., \[\cos 2\theta = 1 - 2{\sin ^2}\theta \], where \[\theta \]be the angle and by further simplification by the basic arithmetic operation we get the required solutions.

Complete step-by-step answer:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric function Also called circular function.
Consider the given question:
We have to find the formula of
\[ \Rightarrow \,\,{\sin ^2}A + {\sin ^2}B\]--------(1)
Let us consider a double angle formula of cosine function i.e., \[\cos 2\theta = 1 - 2{\sin ^2}\theta \], where \[\theta \] be the angle.
At an angle \[\theta = A\], then
\[ \Rightarrow \,\,\cos 2A = 1 - 2{\sin ^2}A\]
On rearranging, we have
\[ \Rightarrow \,\,2{\sin ^2}A = 1 - \cos 2A\]
Divide both side by 2, then we get
\[ \Rightarrow \,\,{\sin ^2}A = \dfrac{{1 - \cos 2A}}{2}\] ---------(2)
Similarly, at an angle
At an angle \[\theta = B\], then
\[ \Rightarrow \,\,\cos 2B = 1 - 2{\sin ^2}B\]
On rearranging, we have
\[ \Rightarrow \,\,2{\sin ^2}B = 1 - \cos 2B\]
Divide both side by 2, then we get
\[ \Rightarrow \,\,{\sin ^2}B = \dfrac{{1 - \cos 2B}}{2}\] ---------(3)
Substitute equation (2) and (3) in equation (1), we have
\[ \Rightarrow \,\,{\sin ^2}A + {\sin ^2}B = \dfrac{{1 - \cos 2A}}{2} + \dfrac{{1 - \cos 2B}}{2}\]
On simplification, we have
\[ \Rightarrow \,\,{\sin ^2}A + {\sin ^2}B = \dfrac{{1 - \cos 2A + 1 - \cos 2B}}{2}\]
\[ \Rightarrow \,\,{\sin ^2}A + {\sin ^2}B = \dfrac{{2 - \left( {\cos 2A + \cos 2B} \right)}}{2}\]
Separate the fraction in RHS, then
\[ \Rightarrow \,\,{\sin ^2}A + {\sin ^2}B = \dfrac{2}{2} - \dfrac{{\cos 2A + \cos 2B}}{2}\]
\[ \Rightarrow \,\,{\sin ^2}A + {\sin ^2}B = 1 - \dfrac{{\cos 2A + \cos 2B}}{2}\]
By the transformation trigonometric formula i.e., \[\cos a + \cos b = 2\cos \left( {\dfrac{{a + b}}{2}} \right) \cdot \cos \left( {\dfrac{{a - b}}{2}} \right)\], then RHS becomes
\[ \Rightarrow \,\,{\sin ^2}A + {\sin ^2}B = 1 - \dfrac{{2\cos \left( {\dfrac{{2A + 2B}}{2}} \right) \cdot \cos \left( {\dfrac{{2A - 2B}}{2}} \right)}}{2}\]
By simplification, we have
\[ \Rightarrow \,\,{\sin ^2}A + {\sin ^2}B = 1 - \cos \left( {\dfrac{{2A + 2B}}{2}} \right) \cdot \cos \left( {\dfrac{{2A - 2B}}{2}} \right)\]
\[ \Rightarrow \,\,{\sin ^2}A + {\sin ^2}B = 1 - \cos \left( {\dfrac{{2\left( {A + B} \right)}}{2}} \right) \cdot \cos \left( {\dfrac{{2\left( {A - B} \right)}}{2}} \right)\]
Again simplification, we get
\[ \Rightarrow \,\,{\sin ^2}A + {\sin ^2}B = 1 - \cos \left( {A + B} \right) \cdot \cos \left( {A - B} \right)\]
Hence, the required formula is \[{\sin ^2}A + {\sin ^2}B = 1 - \cos \left( {A + B} \right) \cdot \cos \left( {A - B} \right)\]

Note: When solving the trigonometry based questions, we have to know the definitions of ratios and always remember the standard angles and formulas are useful for solving certain integration problems where a double formula may make things much simpler to solve. Thus, in math as well as in physics, these formulae are useful to derive many important identities.