Question

The value of cos² x + cos² y – 2cos x × cos y × cos (x + y) is

Answer 1

Answer:

ur answer mate

Step-by-step explanation:

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Answer 2

The value of $\cos ^{2} x+\cos ^{2} y-2 \cos x \cdot \cos y \cdot \cos (x+y)$ is $\sin ^{2}(x+y)$.

Given:

An equation - $\cos ^{2} x+\cos ^{2} y-2 \cos x \cdot \cos y \cdot \cos (x+y)$

To find:

Value of $\cos ^{2} x+\cos ^{2} y-2 \cos x \cdot \cos y \cdot \cos (x+y)$

Solution:

$\rightarrow\cos ^{2} x+\cos ^{2} y-2 \cos x \cdot \cos y \cdot \cos (x+y) \\\\ \cos ^{2} x+1-\sin ^{2} y-2 \cos x \cdot \cos y \cdot \cos (x+y)$

$1+\cos ^{2} x-\sin ^{2} y-2 \cos x \cdot \cos y \cdot \cos (x+y) \\\\ 1+\cos (x+y) \cdot \cos (x-y)-2 \cos x \cdot \cos y \cdot \cos (x+y) \\\\ 1+\cos (x+y)\{\cos (x-y)-2 \cos x \cdot \cos y\} \\\\ 1+\cos (x+y)\{\cos x \cdot \cos y-\sin x \cdot \sin y-2 \cos x \cdot \cos y\} \\\\ 1+\cos (x+y)\{-\cos x \cdot \cos y+\sin x \cdot \sin y\} \\\\ 1-\cos (x+y)\{\cos x \cdot \cos y-\sin x \sin y\}$

$1-\cos (x+y) \cdot \cos (x+y) \\\\1-\cos ^{2}(x+y) \\\\\sin ^{2}(x+y)$

Therefore, the value of $\cos ^{2} x+\cos ^{2} y-2 \cos x \cdot \cos y \cdot \cos (x+y) is \sin ^{2}(x+y).$