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Question: Prove that \(\cos A + \sin (270 - A) - \sin (270 - A) + \cos (180 + A) = 0\)...

Prove that cos⁡A+sin⁡(270−A)−sin⁡(270−A)+cos⁡(180+A)=0\cos A + \sin (270 - A) - \sin (270 - A) + \cos (180 + A) = 0

Explanation

Solution

As we know that if the angle say θ\theta is in the third quadrant then cos⁡θ\cos \theta will be negative and we know that sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin (A + B) = \sin A\cos B + \sin B\cos A. We are going to use this formula to get the required answer.

Complete step by step answer:
Here we are given the equation in terms of sin and cos and we need to prove that
cos⁡A+sin⁡(270−A)−sin⁡(270−A)+cos⁡(180+A)=0\cos A + \sin (270 - A) - \sin (270 - A) + \cos (180 + A) = 0
Now if we take LHS
So LHS=cos⁡A+sin⁡(270−A)−sin⁡(270−A)+cos⁡(180+A) = \cos A + \sin (270 - A) - \sin (270 - A) + \cos (180 + A) −−−−(1) - - - - (1)
So we need to take LHS
As we know that for the given angle A,BA,B
sin⁡(A+B)=sin⁡Acos⁡B+sin⁡Bcos⁡A\sin (A + B) = \sin A\cos B + \sin B\cos A
So we can find the value of sin⁡(270+A)\sin (270 + A)
⇒sin⁡(270+A)\Rightarrow \sin (270 + A) =sin⁡270cos⁡A+cos⁡270sin⁡A = \sin 270\cos A + \cos 270\sin A
We know that
sin⁡(2n+1)π2=(−1)n\sin (2n + 1)\dfrac{\pi }{2} = {( - 1)^n}
So for n=1n = 1
sin⁡3π2=(−1)1\sin \dfrac{{3\pi }}{2} = {( - 1)^1}
And we know that 3π2=270∘\dfrac{{3\pi }}{2} = 270^\circ
So sin⁡(270∘)=−1\sin (270^\circ ) = - 1
And also we know that cos⁡(2n+1)π2=0\cos (2n + 1)\dfrac{\pi }{2} = 0
Hence cos⁡(270∘)=0\cos (270^\circ ) = 0
Therefore we can say that
⇒sin⁡(270+A)\Rightarrow \sin (270 + A) =sin⁡270cos⁡A+cos⁡270sin⁡A = \sin 270\cos A + \cos 270\sin A
=(−1)cos⁡A+0.sin⁡A =−cos⁡A  = ( - 1)\cos A + 0.\sin A \\\ = - \cos A \\\
And we also know that sin⁡(A−B)=sin⁡Acos⁡B−sin⁡Bcos⁡A\sin (A - B) = \sin A\cos B - \sin B\cos A
Now we get that
⇒sin⁡(270−A)\Rightarrow \sin (270 - A) =sin⁡270cos⁡A−cos⁡270sin⁡A = \sin 270\cos A - \cos 270\sin A
=−cos⁡A−0.sin⁡A =−cos⁡A = - \cos A - 0.\sin A \\\ = - \cos A
Therefore we get that
⇒sin⁡(270+A)\Rightarrow \sin (270 + A) =−cos⁡A = - \cos A
⇒sin⁡(270−A)=−cos⁡A\Rightarrow \sin (270 - A) = - \cos A
Also we know that we need to findcos⁡(180+A)\cos (180 + A) and we know that
As 180+A180 + A is in the third quadrant, then we know that here cos is negative
So cos⁡(180+A)=−cos⁡A\cos (180 + A) = - \cos A
Now putting the value in equation (1) we get that
⇒LHS=cos⁡A+sin⁡(270−A)−sin⁡(270−A)+cos⁡(180+A)\Rightarrow LHS = \cos A + \sin (270 - A) - \sin (270 - A) + \cos (180 + A)
=cos⁡A−cos⁡A+cos⁡A−cos⁡A =0 =RHS = \cos A - \cos A + \cos A - \cos A \\\ = 0 \\\ = RHS

Hence proved.

Note:
We know that sin⁡(90+θ)=cos⁡θ\sin (90 + \theta ) = \cos \theta
Similarly for sin⁡(3π2+θ)=−cos⁡θ\sin \left( {\dfrac{{3\pi }}{2} + \theta } \right) = - \cos \theta as 3π2\dfrac{{3\pi }}{2} lies in the fourth quadrant. For the first quadrant all are positive, for the second, sin and cosec are positive, for the third, tan and cot are positive while in the fourth quadrant cos and sec are positive.