Here you will learn what is the formula of cos 2A in terms of sin and cos and also in terms of tan with proof and examples.<\/p>\n
Let’s begin –<\/p>\n
Given below are all the formulas for cos 2A.<\/p>\n
\n(i) cos 2A = \\(cos^2 A\\) – \\(sin^2 A\\)<\/p>\n
(ii) cos 2A = \\(2cos^2 A – 1\\) or, 1 + cos 2A = \\(2cos^2 A\\)<\/p>\n
(iii) cos 2A = \\(1 – 2sin^2 A\\) or, 1 – cos 2A = \\(2sin^2 A\\)<\/p>\n<\/blockquote>\n
Proof :<\/strong><\/p>\n
(i) We have,<\/p>\n
Cos (A + B) = cos A cos B – sin A sin B<\/p>\n
Replacing B by A,<\/p>\n
\\(\\implies\\) cos 2A = cos A cos A + sin A sin A<\/p>\n
\\(\\implies\\) cos 2A = \\(cos^2 A\\) – \\(sin^2 A\\)<\/p>\n
(ii) We have,<\/p>\n
cos 2A = \\(cos^2 A\\) – \\(sin^2 A\\)<\/p>\n
\\(\\implies\\) cos 2A = \\(cos^2 A\\) – \\(1 – cos^2 A\\)<\/p>\n
\\(\\implies\\) cos 2A = \\(2cos^2 A – 1\\)<\/p>\n
Again, cos 2A = \\(2cos^2 A – 1\\)<\/p>\n
\\(\\implies\\) 1 + cos 2A = \\(2cos^2 A\\)<\/p>\n
(iii) We have,<\/p>\n
cos 2A = \\(cos^2 A\\) – \\(sin^2 A\\)<\/p>\n
\\(\\implies\\) cos 2A = \\(1 – sin^2 A\\) – \\(sin^2 A\\)<\/p>\n
\\(\\implies\\) cos 2A = \\(1- 2sin^2 A\\)<\/p>\n
Again, cos 2A = \\(1- 2sin^2 A\\)<\/p>\n
\\(\\implies\\) 1 – cos 2A = \\(2ain^2 A\\)<\/p>\n
We can also write above relation in terms of angle A\/2, just replace A by A\/2, we get<\/p>\n
\n(i) cos A = \\(cos^2 ({A\\over 2})\\) – \\(sin^2 ({A\\over 2})\\)<\/p>\n
(ii) cos A = \\(2cos^2 ({A\\over 2}) – 1\\) or, 1 + cos A = \\(2cos^2 ({A\\over 2})\\)<\/p>\n
(iii) cos A = \\(1 – 2sin^2 ({A\\over 2})\\) or, 1 – cos A = \\(2sin^2 ({A\\over 2})\\)<\/p>\n<\/blockquote>\n
(ii) Cos 2A Formula in Terms of Tan :<\/h3>\n
\nCos 2A = \\(1 – tan^2 A\\over 1 + tan^2 A\\)<\/p>\n<\/blockquote>\n
Proof :<\/strong><\/p>\n
We have,<\/p>\n
cos 2A = \\(cos^2 A\\) – \\(sin^2 A\\)<\/p>\n
\\(\\implies\\) cos 2A = \\(cos^2 A – sin^2 A\\over sin^2 A + cos^2 A\\)<\/p>\n
[ \\(\\because\\) \\(sin^2 A + cos^2 A\\) = 1 ]<\/p>\n
Now, Dividing numerator and denominator by \\(cos^2 A\\),<\/p>\n
\\(\\implies\\) cos 2A = \\({cos^2 A – sin^2 A\\over cos^2 A}\\over {sin^2 A + cos^2 A\\over cos^2 A}\\)<\/p>\n
\\(\\implies\\) cos 2A = \\(1 – tan^2 A\\over 1 + tan^2 A\\)<\/p>\n
We can also write above relation in terms of angle A\/2, just replace A by A\/2, we get<\/p>\n
\ncos A = \\(1 – tan^2 ({A\\over 2})\\over 1 + tan^2 ({A\\over 2})\\)<\/p>\n<\/blockquote>\n