Question

if √2cos(A-B)=1, then prove that, sin^2(A-B)= 1/2​

Answer 1

Answer:

$sin^2(A-B)= 1/2$

Step-by-step explanation:

$cos(A-B)=\frac{1}{\sqrt{2} }$

We know that,

$sin^{2}x+cos^{2} x=1$

Similarly here,

$sin^{2}x=1-cos^{2} x$

$sin^{2}(A-B)=1-(\frac{1}{\sqrt{2} } )^{2}\\sin^{2}(A-B)=1-\frac{1}{2} \\sin^{2} (A-B)=\frac{1}{2}$

Hence proved

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