{"id":9200,"date":"2022-01-03T01:59:37","date_gmt":"2022-01-02T20:29:37","guid":{"rendered":"https:\/\/mathemerize.com\/?p=9200"},"modified":"2022-01-03T02:13:01","modified_gmt":"2022-01-02T20:43:01","slug":"prove-that-sin2-theta-cos2-theta-1","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-sin2-theta-cos2-theta-1\/","title":{"rendered":"Prove that \\(sin^2 \\theta\\) + \\(cos^2 \\theta\\) = 1."},"content":{"rendered":"
In right angled triangle ABC,<\/p>\n
\\(sin \\theta\\) = \\(BC\\over AC\\) \\(\\implies\\) \\(sin^2 \\theta\\) = \\(BC^2\\over AC^2\\)<\/p>\n
\\(cos \\theta\\) = \\(AB\\over AC\\) \\(\\implies\\) \\(cos^2 \\theta\\) = \\(AB^2\\over AC^2\\)<\/p>\n
On adding,<\/p>\n
\\(sin^2 \\theta\\) + \\(cos^2 \\theta\\) = \\(BC^2\\over AC^2\\) + \\(AB^2\\over AC^2\\)<\/p>\n
\\(sin^2 \\theta\\) + \\(cos^2 \\theta\\) = \\(BC^2 + AB^2\\over AC^2\\) = \\(AC^2\\over AC^2\\) = 1<\/p>\n
[ By Pythagoras theorem, \\(AC^2\\) = \\(BC^2 + AB^2\\) ]<\/p>\n
\\(\\implies\\) \\(sin^2 \\theta\\) + \\(cos^2 \\theta\\) = 1<\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : In right angled triangle ABC, \\(sin \\theta\\) = \\(BC\\over AC\\) \\(\\implies\\) \\(sin^2 \\theta\\) = \\(BC^2\\over AC^2\\) \\(cos \\theta\\) = \\(AB\\over AC\\) \\(\\implies\\) \\(cos^2 \\theta\\) = \\(AB^2\\over AC^2\\) On adding, \\(sin^2 \\theta\\) + \\(cos^2 \\theta\\) = \\(BC^2\\over AC^2\\) + \\(AB^2\\over AC^2\\) \\(sin^2 \\theta\\) + \\(cos^2 \\theta\\) = \\(BC^2 + AB^2\\over AC^2\\) = \\(AC^2\\over AC^2\\) …<\/p>\n