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Question

Question: Simplify the given expression \(\cos \theta \left[ \begin{matrix} \cos \theta & \sin \theta ...

Simplify the given expression
cosθ[cosθsinθ sinθcosθ ]+sinθ[sinθcosθ cosθsinθ ]\cos \theta \left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix} \sin \theta & -\cos \theta \\\ \cos \theta & \sin \theta \\\ \end{matrix} \right]

Explanation

Solution

We start solving this question by dividing the given expression as two parts and simplify them by taking the constants, cosθ\cos \theta and sinθ\sin \theta which are outside of matrix, into the matrix and multiplying elements inside the matrix with them. Then we get two expressions for the two parts. Then we add them and obtain a single matrix with the trigonometric functions cosθ\cos \theta and sinθ\sin \theta . Then we use the trigonometric identity cos2θ+sin2θ=1{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1, and simplify the given expression to obtain a simplified version of the given expression.

Complete step by step answer:
Let us consider the given expression, cosθ[cosθsinθ sinθcosθ ]+sinθ[sinθcosθ cosθsinθ ]\cos \theta \left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix} \sin \theta & -\cos \theta \\\ \cos \theta & \sin \theta \\\ \end{matrix} \right].
Let us divide the given expression into two parts cosθ[cosθsinθ sinθcosθ ]\cos \theta \left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right] and sinθ[sinθcosθ cosθsinθ ]\sin \theta \left[ \begin{matrix} \sin \theta & -\cos \theta \\\ \cos \theta & \sin \theta \\\ \end{matrix} \right].
Let us consider cosθ[cosθsinθ sinθcosθ ]\cos \theta \left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right] as A.
Let us consider sinθ[sinθcosθ cosθsinθ ]\sin \theta \left[ \begin{matrix} \sin \theta & -\cos \theta \\\ \cos \theta & \sin \theta \\\ \end{matrix} \right] as B.
Now, let us consider A and simplify A.
cosθ[cosθsinθ sinθcosθ ] [cos2θsinθcosθ sinθcosθcos2θ ] \begin{aligned} & \Rightarrow \cos \theta \left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right] \\\ & \Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta & \sin \theta \cos \theta \\\ -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\\ \end{matrix} \right] \\\ \end{aligned}
Now, let us consider B and simplify it.
sinθ[sinθcosθ cosθsinθ ] [sin2θsinθcosθ sinθcosθsin2θ ] \begin{aligned} & \Rightarrow \sin \theta \left[ \begin{matrix} \sin \theta & -\cos \theta \\\ \cos \theta & \sin \theta \\\ \end{matrix} \right] \\\ & \Rightarrow \left[ \begin{matrix} {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\\ \sin \theta \cos \theta & {{\sin }^{2}}\theta \\\ \end{matrix} \right] \\\ \end{aligned}
As, we have simplified A and B, let us add them.
[cos2θsinθcosθ sinθcosθcos2θ ]+[sin2θsinθcosθ sinθcosθsin2θ ] [cos2θ+sin2θsinθcosθsinθcosθ sinθcosθ+sinθcosθcos2θ+sin2θ ] [cos2θ+sin2θ0 0cos2θ+sin2θ ]...............(1) \begin{aligned} & \Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta & \sin \theta \cos \theta \\\ -\sin \theta \cos \theta & {{\cos }^{2}}\theta \\\ \end{matrix} \right]+\left[ \begin{matrix} {{\sin }^{2}}\theta & -\sin \theta \cos \theta \\\ \sin \theta \cos \theta & {{\sin }^{2}}\theta \\\ \end{matrix} \right] \\\ & \Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta +{{\sin }^{2}}\theta & \sin \theta \cos \theta -\sin \theta \cos \theta \\\ -\sin \theta \cos \theta +\sin \theta \cos \theta & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \\\ \end{matrix} \right] \\\ & \Rightarrow \left[ \begin{matrix} {{\cos }^{2}}\theta +{{\sin }^{2}}\theta & 0 \\\ 0 & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \\\ \end{matrix} \right]...............\left( 1 \right) \\\ \end{aligned}
Now let us consider the trigonometric identity cos2θ+sin2θ=1{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1.
Using the above identity, we can write the expression in equation (1) as
[10 01 ]\Rightarrow \left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]
So, finally after simplification of the given expression we get that
cosθ[cosθsinθ sinθcosθ ]+sinθ[sinθcosθ cosθsinθ ]=[10 01 ]\cos \theta \left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right]+\sin \theta \left[ \begin{matrix} \sin \theta & -\cos \theta \\\ \cos \theta & \sin \theta \\\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right]
So, value of given expression is [10 01 ]\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right].
Hence answer is [10 01 ]\left[ \begin{matrix} 1 & 0 \\\ 0 & 1 \\\ \end{matrix} \right].

Note:
Here while solving this problem, one might make a mistake of multiplying the constant only to the first element of the matrix when the matrix is multiplied with a constant. For example, while simplifying cosθ[cosθsinθ sinθcosθ ]\cos \theta \left[ \begin{matrix} \cos \theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right], one might write it as [cos2θsinθ sinθcosθ ]\left[ \begin{matrix} {{\cos }^{2}}\theta & \sin \theta \\\ -\sin \theta & \cos \theta \\\ \end{matrix} \right]. But it is wrong. So, we need to remember that when a matrix is multiplied by a constant, every element in the matrix is multiplied by the constant.