Question

if \[\left[ {\cos (A + B) - \sin (A + B)\cos 2B} \right]\],\[\left[ {\sin A\cos A\sin B} \right]\],\[\left[ { - \cos A\sin A\cos B} \right]\]\[ = 0\],then \[B = \]
A. \[(2n + 1)\dfrac{\pi }{2}\]
B. \[n\pi \]
C. \[(2n + 1)\dfrac{\pi }{2}\]
D. \[2n\pi \]

Answer 1

Hint: To solve this question, we will use the trigonometry formulas sin\[\theta \],cos\[\theta \].When the ratio of sine, cosine, or tangent is given for each individual angle, we may apply the sum and difference formulas to determine the sum or difference of angles.

Formula Used: Cosine has the following formula for subtraction: \[\cos (AB) = \cos A\cos B + \sin A\sin B(AB) = \cos \] Sine's addition formula is \[\sin (A + B) = \sin A\cos B + \cos A\sin B\] where \[(A + B) = \sin \] Sine's subtraction formula is \[\sin (A,B) = \sin A,\cos B,\cos A\sin B\], where \[(A,B) = \sin \].

The cosine of sums and differences of angles can be calculated using these formulas.
\[\cos (A + B)(\cos A\cos B - \sin A\sin B) + S{\sin ^2}(A + B) + \cos 2B + 0\]
\[{\cos ^2}(A + B) + {\sin ^2}(A + B) + \cos 2B = 0\]
\[\cos (2B) = - 1\]

Complete step by step solution: \[[\cos (A + B) - \sin (A + B)\cos 2B]\]
\[[\sin A\cos A\sin B]\]
\[[ - \cos A\sin A\cos B]\]
Are \[ = 0\]
\[ \Rightarrow \]\[\cos (A + B)(\cos A.\cos B - \sin A.\sin B) + \]\[\sin (A + B)(\sin A\cos B + \cos A\sin B) + \]\[\cos 2B({\sin ^2}A + {\cos ^2}A) = 0\]
\[ \Rightarrow \]\[\cos (A + B)\cos (A + B) + \sin (A + B)\sin (A + B) + \cos 2B = 0\]
On expanding determinant,
\[ \Rightarrow \]\[{\cos ^2}(A + B) + {\sin ^2}(A + B) = - \cos 2B\]
\[ \Rightarrow \]\[1 = - \cos 2B\]
\[\cos 2B = - 1\]
\[2B = (2n + 1)\pi \]
\[B = (2n + 1)\dfrac{\pi }{2}\]
\[n \in Z\].
The numbers being added are known as addends, the addition process is denoted by the \[( + )\] symbol, and the outcome is denoted by the sum. Minuend \[ - \] Subtrahend \[ = \] Difference is the formula for subtraction. Therefore, we must first determine whether an expression of the form \[y = A\sin B\cos \] can be converted to an expression that only contains sine functions. Take \[\sin \] from \[A\sin \] and \[\cos \] from \[B\cos \], respectively. Sine and cosine are equal when we square and sum the two elements, giving us one.
The angles are shifted by \[2\pi \], \[\pi \], \[\dfrac{\pi }{2}\]etc. using periodic identity-based trigonometry calculations. Since all trigonometric identities have a cyclic nature, they eventually repeat themselves.

Option ‘A’ is correct

Note: The product between sines and or cosines is rewritten into a sum or difference using the product-to-sum identities. These identities are created by adding or subtracting the sine and cosine sum and difference formulas. With the exception of the inverted signs, addition and subtraction are nearly identical operations. By expressing trigonometric functions in terms of relatively equal functions, the sum and difference formulas calculate the values of trigonometric functions.