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Question: How do you multiply and simplify \[\left( \sin x-\cos x \right)\left( \sin x+\cos x \right)\]?...

How do you multiply and simplify (sin⁡x−cos⁡x)(sin⁡x+cos⁡x)\left( \sin x-\cos x \right)\left( \sin x+\cos x \right)?

Explanation

Solution

From the question given, we have been asked to multiply and simplify (sin⁡x−cos⁡x)(sin⁡x+cos⁡x)\left( \sin x-\cos x \right)\left( \sin x+\cos x \right).The given question can be multiplied by multiplying the second group with the first term of first group first and then multiplying the second group with the second term of the first group. And then add all the terms and then we can simplify it easily.

Complete answer:
From the question it have been given that (sin⁡x−cos⁡x)(sin⁡x+cos⁡x)\left( \sin x-\cos x \right)\left( \sin x+\cos x \right)
As we have already discussed above to multiply the above expression, first we have to multiply the every term of the second group with the first term of the first group.
By doing this process, we get
sin⁡x(sin⁡x)+sin⁡x(cos⁡x)\sin x\left( \sin x \right)+\sin x\left( \cos x \right)
⇒sin⁡2x+sin⁡xcos⁡x\Rightarrow {{\sin }^{2}}x+\sin x\cos x
Now, we have to multiply every term of the second group with the second term of the first group.
By doing this process, we get
−cos⁡x(sin⁡x)+−cos⁡x(cos⁡x)-\cos x\left( \sin x \right)+-\cos x\left( \cos x \right)
⇒−sin⁡xcos⁡x−cos⁡2x\Rightarrow -\sin x\cos x-{{\cos }^{2}}x
Now, we have to add all the terms we got above.
By adding all the terms, we get sin⁡2x+sin⁡xcos⁡x+(−sin⁡xcos⁡x−cos⁡2x){{\sin }^{2}}x+\sin x\cos x+\left( -\sin x\cos x-{{\cos }^{2}}x \right)
By the cancellation of terms, we get sin⁡2x−cos⁡2x{{\sin }^{2}}x-{{\cos }^{2}}x
Now, we have to simplify it.
In trigonometry, we have one identity that is sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1
From this identity, we get cos⁡2x=1−sin⁡2x{{\cos }^{2}}x=1-{{\sin }^{2}}x
Substitute this in the above simplified expression we got above.
sin⁡2x−(1−sin⁡2x){{\sin }^{2}}x-\left( 1-{{\sin }^{2}}x \right)
⇒2sin⁡2x−1\Rightarrow 2{{\sin }^{2}}x-1
Hence, (sin⁡x−cos⁡x)(sin⁡x+cos⁡x)=2sin⁡2x−1\left( \sin x-\cos x \right)\left( \sin x+\cos x \right)=2{{\sin }^{2}}x-1
Therefore, the given trigonometric expression is multiplied and simplified.

Note: We should be very careful while multiplying the terms. Also, we should be very careful while adding all the terms. Also, we should have the knowledge of the basic trigonometric identities which we have used in the above question to simplify. Also, we should be very careful while simplifying the expression. This question can also be answered as follows
(sin⁡x−cos⁡x)(sin⁡x+cos⁡x)=sin⁡2x−cos⁡2x ⇒(1−cos⁡2x)−cos⁡2x=1−2cos⁡2x \begin{aligned} & \left( \sin x-\cos x \right)\left( \sin x+\cos x \right)={{\sin }^{2}}x-{{\cos }^{2}}x \\\ & \Rightarrow \left( 1-{{\cos }^{2}}x \right)-{{\cos }^{2}}x=1-2{{\cos }^{2}}x \\\ \end{aligned}