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Question: How do you prove \(\sin (x + y + z) = \sin x\cos y\cos z + \cos x\sin y\cos z + \cos x\cos y\sin z -...

How do you prove sin⁡(x+y+z)=sin⁡xcos⁡ycos⁡z+cos⁡xsin⁡ycos⁡z+cos⁡xcos⁡ysin⁡z−sin⁡xsin⁡ysin⁡z\sin (x + y + z) = \sin x\cos y\cos z + \cos x\sin y\cos z + \cos x\cos y\sin z - \sin x\sin y\sin z?

Explanation

Solution

Here we have to prove the trigonometric function sin⁡(x+y+z)\sin (x + y + z) is equal to sin⁡xcos⁡ycos⁡z+cos⁡xsin⁡ycos⁡z+cos⁡xcos⁡ysin⁡z−sin⁡xsin⁡ysin⁡z\sin x\cos y\cos z + \cos x\sin y\cos z + \cos x\cos y\sin z - \sin x\sin y\sin z. In order to solve this question we first assume that A=x+yA = x + y and B=zB = z in the function sin⁡(x+y+z)\sin (x + y + z) then we will use trigonometric identities such as sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B and cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A\cos B - \sin A\sin B to get the required results

Complete step by step answer:
Here we have to prove the trigonometric function sin⁡(x+y+z)\sin (x + y + z) is equal to
sin⁡xcos⁡ycos⁡z+cos⁡xsin⁡ycos⁡z+cos⁡xcos⁡ysin⁡z−sin⁡xsin⁡ysin⁡z\sin x\cos y\cos z + \cos x\sin y\cos z + \cos x\cos y\sin z - \sin x\sin y\sin z
Now, consider the left side of the equation.
We have, sin⁡(x+y+z)\sin (x + y + z)
Let A=x+yA = x + y and B=zB = z
So, we can write sin⁡(x+y+z)\sin (x + y + z)= sin⁡(A+B)\sin (A + B)

Using the identity sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B.
Therefore, sin⁡(x+y+z)=sin⁡(x+y)cos⁡z+cos⁡(x+y)sin⁡z\sin (x + y + z) = \sin (x + y)\cos z + \cos (x + y)\sin z
Now, we will solve sin⁡(x+y)\sin (x + y) and cos⁡(x+y)\cos (x + y)
Using the identities sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B and cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A\cos B - \sin A\sin B. We get,
⇒sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y\Rightarrow \sin (x + y) = \sin x\cos y + \cos x\sin y
⇒cos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y\Rightarrow \cos (x + y) = \cos x\cos y - \sin x\sin y

Substituting these values in the function. we get,
⇒sin⁡(x+y+z)=(sin⁡xcos⁡y+cos⁡xsin⁡y)cos⁡z+(cos⁡xcos⁡y−sin⁡xsin⁡y)sin⁡z\Rightarrow \sin (x + y + z) = (\sin x\cos y + \cos x\sin y)\cos z + (\cos x\cos y - \sin x\sin y)\sin z
Solving the above equation. We get,
⇒sin⁡(x+y+z)=sin⁡xcos⁡ycos⁡z+cos⁡xsin⁡ycos⁡z+cos⁡xcos⁡ysin⁡z−sin⁡xsin⁡ysin⁡z\Rightarrow \sin (x + y + z) = \sin x\cos y\cos z + \cos x\sin y\cos z + \cos x\cos y\sin z - \sin x\sin y\sin z
Hence, the left side of the equation is equal to the right side of the equation. i.e.,
∴sin⁡(x+y+z)=sin⁡xcos⁡ycos⁡z+cos⁡xsin⁡ycos⁡z+cos⁡xcos⁡ysin⁡z−sin⁡xsin⁡ysin⁡z\therefore \sin (x + y + z) = \sin x\cos y\cos z + \cos x\sin y\cos z + \cos x\cos y\sin z - \sin x\sin y\sin z
Hence proved

Note: To solve these types of questions we have to split the angles first before using the identity. Here we can also split as A=xA = x and B=y+zB = y + z and after that use the trigonometric identities such as sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin (A + B) = \sin A\cos B + \cos A\sin B and cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A\cos B - \sin A\sin B. Note that the addition and subtraction identities of cos⁡\cos have different signs in it. In addition, the subtraction sign is between the angles i.e., cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A + B) = \cos A\cos B - \sin A\sin B. Whereas in subtraction, the addition sign is between the angles i.e., cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos (A - B) = \cos A\cos B + \sin A\sin B. Some students are confused between trigonometric identities such as sin⁡(A+B)\sin (A + B) andsin⁡A+sin⁡B\sin A + \sin B. These both are different identities in one there is only a sum of angles and in second there is a sum of angles of sin⁡\sin and similarly for cos⁡A+cos⁡B\cos A + \cos B and cos⁡A−cos⁡B\cos A - \cos B.