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Question: What sign has \(\sin {\text{A + cosA}}\)for the following value of \({\text{A}}\)? \( - {1125^\cir...

What sign has sinA + cosA\sin {\text{A + cosA}}for the following value of A{\text{A}}?
1125- {1125^\circ }

Explanation

Solution

Since they have told to find the value of sinA + cosA\sin {\text{A + cosA}} where A = - 1125{\text{A = - 112}}{{\text{5}}^\circ } first try to solve the trigonometric equation by bringing in in some basic trigonometric identity and after solving this equation try write the angle as a multiple of 360 as it will help us get our solution more accurately and quickly.

Formula used:
Trigonometric identities used in this sum are
(1) sin(a+b)=sinacosb+cosasinb\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b
(2) sin(θ)=sin(θ)\sin ( - \theta ) = - \sin (\theta )

Complete step by step answer:
We have been asked to find the sign of sinA + cosA\sin {\text{A + cosA}} when A = - 1125{\text{A = - 112}}{{\text{5}}^\circ }
Since the given equation is in trigonometric form let us try to bring the equation in basic trigonometric identity if possible so that it becomes easier for us to further solve the equation.
Let us first start with multiplying and dividing the equation by 2\sqrt 2 .
We get
sinA + cosA = 22(sinA + cosA) sinA + cosA=2(12sinA + 12cosA)  \sin {\text{A + cosA = }}\dfrac{{\sqrt 2 }}{{\sqrt 2 }}\left( {\sin {\text{A + cosA}}} \right) \\\ \Rightarrow \sin {\text{A + cosA}} = \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\sin {\text{A + }}\dfrac{1}{{\sqrt 2 }}\cos {\text{A}}} \right) \\\
We know thatcos45=sin45=12\cos 45 = \sin 45 = \dfrac{1}{2}
Therefore substituting this value above we get,
sinA + cosA = 2[sinAcos45 + cosAsin45] sinA + cosA = 2[sin(A + 45)](1)  \sin {\text{A + cosA = }}\sqrt 2 \left[ {\sin {\text{Acos45 + cosAsin45}}} \right] \\\ \Rightarrow \sin {\text{A + cosA = }}\sqrt 2 \left[ {\sin \left( {{\text{A + 45}}} \right)} \right] - - - \left( 1 \right) \\\
Since sin(a+b)=sinacosb+cosasinb\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b
Now since we have been given the value of A = - 1125{\text{A = - 112}}{{\text{5}}^\circ } substituting it in equation (1) we get
sinA + cosA = 2sin(1125+45) sinA + cosA = 2sin(1080) sinA + cosA = 2sin(3×360)  \sin {\text{A + cosA = }}\sqrt 2 \sin ( - 1125 + 45) \\\ \Rightarrow \sin {\text{A + cosA = }}\sqrt 2 \sin ( - 1080) \\\ \Rightarrow \sin {\text{A + cosA = }}\sqrt 2 \sin ( - 3 \times 360) \\\
Since we know that sin(θ)=sin(θ)\sin ( - \theta ) = - \sin (\theta ) , we get
sinA + cosA = - 2sin(3×360)\sin {\text{A + cosA = - }}\sqrt 2 \sin (3 \times 360)
Here we know that value of sin(360)=0\sin (360) = 0 that implies that the value of sin(3×360)=0\sin (3 \times 360) = 0
sinA + cosA = - 2×0 sinA + cosA = 0  \sin {\text{A + cosA = - }}\sqrt 2 \times 0 \\\ \Rightarrow \sin {\text{A + cosA = }}0 \\\

Therefore we can conclude from the above calculations that the value of sinA + cosA\sin {\text{A + cosA}} is 0

Note: While trying to solve this kind of trigonometric equation and asking for signs of the equation try to use identities and bring the solution in the basic form. Also the angle given also determines the sign of the final equation so writing these huge angles as multiples of 360 gives us a better view to find the sign more accurately.