Answer:
[tex]\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}[/tex]
Step-by-step explanation:
Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .
Given :-
• [tex]\sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}[/tex]
To Prove :-
•[tex]\sf\implies cos2A =\dfrac{\sqrt3}{2} [/tex]
Proof :-
We know that ,
[tex]\sf\implies cos2A = 1 - 2sin^2A [/tex]
Therefore , here substituting the value of sinA , we have ,
[tex]\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2 [/tex]
Simplify the whole square ,
[tex]\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8} [/tex]
Add the numbers in numerator ,
[tex]\sf\implies cos2A = 1-2\times \dfrac{4-2\sqrt3}{8} [/tex]
Multiply it by 2 ,
[tex]\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4} [/tex]
Take out 2 common from the numerator ,
[tex]\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4} [/tex]
Simplify ,
[tex]\sf\implies cos2A = 1 -\dfrac{ 2-\sqrt3}{2}[/tex]
Subtract the numbers ,
[tex]\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2} [/tex]
Simplify,
[tex]\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} } [/tex]
Hence Proved !