Question

If \[y = {\sin ^2}\alpha + {\cos ^2}\left( {\alpha + \beta } \right) + 2\sin \alpha \sin \beta \cos \left( {\alpha + \beta } \right)\], then \[\dfrac{{{d^3}y}}{{d{\alpha ^3}}}\] is, (keeping \[\beta \]as constant)
A. 1
B. 0
C. \[\cos \left( {\alpha + 3\beta } \right)\]
D. none of these

Answer 1

Hint: In the above question we need to find the value of \[\dfrac{{{d^3}y}}{{d{\alpha ^3}}}\]. Now we know the formula \[2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)\]. So, by applying this formula in the given equation and finding the derivative of the function with respect \[\alpha \] to get the desired result.

Formula used:
We have been using the following formulas:
1. \[2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)\]
2. \[\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\]

Complete step-by-step solution:
We are given that
\[y = {\sin ^2}\alpha + {\cos ^2}\left( {\alpha + \beta } \right) + 2\sin \alpha \sin \beta \cos \left( {\alpha + \beta } \right)...\left( 1 \right)\]
Now we know that \[2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)\]
Now we apply the above formula in equation (1), and we get
\[
  y = {\sin ^2}\alpha + {\cos ^2}\left( {\alpha + \beta } \right) + \left[ {\cos \left( {\alpha - \beta } \right) - \cos \left( {\alpha + \beta } \right)} \right]\cos \left( {\alpha + \beta } \right) \\
   = {\sin ^2}\alpha + {\cos ^2}\left( {\alpha + \beta } \right) + \cos \left( {\alpha - \beta } \right)\cos \left( {\alpha + \beta } \right) - {\left[ {\cos \left( {\alpha + \beta } \right)} \right]^2}...\left( 2 \right)
 \]
Now by differentiating the above function with respect to \[\alpha \], we get
\[
  \dfrac{{dy}}{{d\alpha }} = 2\sin \alpha \cos \alpha - 2\cos \left( {\alpha + \beta } \right)\sin \left( {\alpha + \beta } \right) - \sin \left( {\alpha - \beta } \right)\cos \left( {\alpha + \beta } \right) - \cos \left( {\alpha - \beta } \right)\sin \left( {\alpha + \beta } \right) \\
   + 2\cos \left( {\alpha + \beta } \right)\sin \left( {\alpha + \beta } \right) \\
   = 2\sin \alpha \cos \alpha - \sin \left( {\alpha - \beta } \right)\cos \left( {\alpha + \beta } \right) - \cos \left( {\alpha - \beta } \right)\sin \left( {\alpha + \beta } \right) \\
   = 2\sin \alpha \cos \alpha - \left[ {\sin \left( {\alpha - \beta } \right)\cos \left( {\alpha + \beta } \right) + \cos \left( {\alpha - \beta } \right)\sin \left( {\alpha + \beta } \right)} \right]
 \]
Now we know that \[\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\]
Now by applying the above formula in the above equation, we get
\[
  \dfrac{{dy}}{{d\alpha }} = 2\sin \alpha \cos \alpha - \left[ {\sin \left( {\alpha - \beta + \alpha + \beta } \right)} \right] \\
   = 2\sin \alpha \cos \alpha - \left[ {\sin \left( {2\alpha } \right)} \right] \\
   = 2\sin \alpha \cos \alpha - \sin \left( {2\alpha } \right)
 \]
Now we know that \[\sin 2\alpha = 2\sin \alpha \cos \alpha \]
Now by applying this formula in the above equation, we get
\[
  \dfrac{{dy}}{{d\alpha }} = \sin 2\alpha - \sin 2\alpha \\
   = 0
 \]
Therefore, the first derivative of the given function y is zero which means the second derivative and the third derivative are also zero.
That is \[\dfrac{{{d^3}y}}{{d{\alpha ^3}}} = 0\]
Hence, option (B) is correct option

Additional information: Trigonometric functions are also known as circular functions. These trigonometric functions define the relationship between the angles and sides of a triangle. Sine, cosine, tangent, cotangent, secant, and cosecant are the fundamental trigonometric functions.

Note: To answer the above question, the student must be familiar with all trigonometric formulas and identities. Students must exercise caution when calculating the derivative of a given function because errors can occur during the calculation process.