Question

If \[\sin\left( {\theta + \alpha } \right) = a\], and \[\sin\left( {\theta + \beta } \right) = b\]. Then prove that \[\cos 2\left( {\alpha - \beta } \right) - 4ab \cos\left( {\alpha - \beta } \right) = 1 - 2{a^2} - 2{b^2}\].

Answer 1

Hint: First using the trigonometric identity \[\sin^{2}A + \cos^{2}A = 1\] calculate the values of \[\cos\left( {\theta + \alpha } \right)\], and \[\cos\left( {\theta + \beta } \right)\]. Then rewrite the angle \[\cos\left( {\alpha - \beta } \right)\] as \[\cos\left[ {\left( {\theta + \alpha } \right) - \left( {\theta + \beta } \right)} \right]\] and simplify it using the trigonometric identity \[\cos\left( {A - B} \right) = \cos A \cos B + \sin A \sin B\]. After that, solve the left-hand side of the required equation by using the trigonometric formula \[\cos 2A = 2\cos^{2}A - 1\] and the value of \[\cos\left( {\alpha - \beta } \right)\] to prove the required answer.

Formula used:
\[\sin^{2}A + \cos^{2}A = 1\]
\[\cos\left( {A - B} \right) = \cos A \cos B + \sin A \sin B\]
\[\cos 2A = 2\cos^{2}A - 1\]

Complete step by step solution:
The given equations are \[\sin\left( {\theta + \alpha } \right) = a\], and \[\sin\left( {\theta + \beta } \right) = b\].
To Prove: \[\cos 2\left( {\alpha - \beta } \right) - 4ab \cos\left( {\alpha - \beta } \right) = 1 - 2{a^2} - 2{b^2}\]
Let’s find out the values of \[\cos\left( {\theta + \alpha } \right)\], and \[\cos\left( {\theta + \beta } \right)\].
Apply the trigonometric identity \[\sin^{2}A + \cos^{2}A = 1\].
We get,
\[\cos^{2}\left( {\theta + \alpha } \right) = 1 – \sin^{2}\left( {\theta + \alpha } \right)\]
\[ \Rightarrow \cos^{2}\left( {\theta + \alpha } \right) = 1 - {a^2}\]
Take square root on both sides.
\[\cos\left( {\theta + \alpha } \right) = \sqrt {1 - {a^2}} \]
Similarly, for \[\cos\left( {\theta + \beta } \right)\] we get
\[\cos\left( {\theta + \beta } \right) = \sqrt {1 - {b^2}} \]
Now calculate the value of \[\cos\left( {\alpha - \beta } \right)\].
Rewrite the angle \[\cos\left( {\alpha - \beta } \right)\] as \[\cos\left[ {\left( {\theta + \alpha } \right) - \left( {\theta + \beta } \right)} \right]\]
\[\cos\left( {\alpha - \beta } \right) = \cos\left[ {\left( {\theta + \alpha } \right) - \left( {\theta + \beta } \right)} \right]\]
Apply the trigonometric identity \[\cos\left( {A - B} \right) = \cos A \cos B + \sin A \sin B\].
\[\cos\left( {\alpha - \beta } \right) = \cos\left( {\theta + \alpha } \right)\cos\left( {\theta + \beta } \right) + \sin\left( {\theta + \alpha } \right)\sin\left( {\theta + \beta } \right)\]
Substitute the values in the above equation.
\[\cos\left( {\alpha - \beta } \right) = \left( {\sqrt {1 - {a^2}} } \right)\left( {\sqrt {1 - {b^2}} } \right) + ab\]
\[ \Rightarrow \cos\left( {\alpha - \beta } \right) = \sqrt {\left( {1 - {a^2}} \right)\left( {1 - {b^2}} \right)} + ab\]
\[ \Rightarrow \cos\left( {\alpha - \beta } \right) = \sqrt {1 - {a^2} - {b^2} + {a^2}{b^2}} + ab\]
Now solve the left-hand side of the given equation \[\cos 2\left( {\alpha - \beta } \right) - 4ab \cos\left( {\alpha - \beta } \right) = 1 - 2{a^2} - 2{b^2}\].
Let consider,
\[LHS = \cos 2\left( {\alpha - \beta } \right) - 4ab \cos\left( {\alpha - \beta } \right)\]
Apply the trigonometric formula \[\cos 2A = 2\cos^{2}A - 1\] in the above equation.
\[LHS = \left[ {2\cos^{2}\left( {\alpha - \beta } \right) - 1} \right] - 4ab \cos\left( {\alpha - \beta } \right)\]
Substitute the value of \[\cos\left( {\alpha - \beta } \right)\] in the above equation.
\[LHS = \left[ {2{{\left( {\sqrt {1 - {a^2} - {b^2} + {a^2}{b^2}} + ab} \right)}^2} - 1} \right] - 4ab\left( {\sqrt {1 - {a^2} - {b^2} + {a^2}{b^2}} + ab} \right)\]
Apply the formula of the square of the sum \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[ \Rightarrow LHS = \left[ {2\left( {\left( {1 - {a^2} - {b^2} + {a^2}{b^2}} \right) + {a^2}{b^2} + 2ab\sqrt {1 - {a^2} - {b^2} + {a^2}{b^2}} } \right) - 1} \right] - 4ab\sqrt {1 - {a^2} - {b^2} + {a^2}{b^2}} - 4{a^2}{b^2}\]
\[ \Rightarrow LHS = \left[ {2\left( {1 - {a^2} - {b^2} + {a^2}{b^2}} \right) + 2{a^2}{b^2} + 4ab\sqrt {1 - {a^2} - {b^2} + {a^2}{b^2}} - 1} \right] - 4ab\sqrt {1 - {a^2} - {b^2} + {a^2}{b^2}} - 4{a^2}{b^2}\]
Cancel out the common terms with opposite sides.
\[LHS = \left[ {2 - 2{a^2} - 2{b^2} + 2{a^2}{b^2} - 1} \right] - 2{a^2}{b^2}\]
\[ \Rightarrow LHS = 1 - 2{a^2} - 2{b^2}\]
\[ \Rightarrow LHS = RHS\]
Hence, proved.

Note: While solving any trigonometric equation with double angles and exponent convert the equation into the basic trigonometric ratios by using the trigonometric identities.
Students often get confused about the formulas of the sum and difference of angels for a cosine function.
The formulas are as follows:
\[\cos\left( {A - B} \right) = \cos A \cos B + \sin A \sin B\]
\[\cos\left( {A + B} \right) = \cos A \cos B - \sin A \sin B\]