cos4A-sin4A=cos squareA
sanaya865:
Pls give me solution
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cos^4 a - sin^4 a = cos2a
Let us rewrite:......
(cos^2 a)^2 - (sin^2 a)^2
We kno wthat:
a^2 - b^2 = (a-b)(a+b)
==> cos^4 a- sin^4 a=(cos^2 a - sin^2 a)(cos^2 a + sin^2 a)
Also:
We know that:
cos^2 a-sin^2 a = cos2a
cos^2 a + sin^2 a= 1
Now substitute:
cos^4 a - sin^4 a= cos2a * 1
= cos2a
Let us rewrite:......
(cos^2 a)^2 - (sin^2 a)^2
We kno wthat:
a^2 - b^2 = (a-b)(a+b)
==> cos^4 a- sin^4 a=(cos^2 a - sin^2 a)(cos^2 a + sin^2 a)
Also:
We know that:
cos^2 a-sin^2 a = cos2a
cos^2 a + sin^2 a= 1
Now substitute:
cos^4 a - sin^4 a= cos2a * 1
= cos2a
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0
✌✌hope it will help uh ..
nd plz mark as brainliest ..☺☺
nd plz mark as brainliest ..☺☺
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