検索キーワード「pythagorean identities」に一致する投稿を日付順に表示しています。 関連性の高い順 すべての投稿を表示
検索キーワード「pythagorean identities」に一致する投稿を日付順に表示しています。 関連性の高い順 すべての投稿を表示

[コンプリート!] tan^2x = sin^2x/cos^2x 175629-Sin 2x cos 2x tan 2x

Sec2(2x) sec 2 ( 2 x) Because the two sides have been shown to be equivalent, the equation is an identity tan2(2x)sin2(2x) cos2(2x) = sec2 (2x) tan 2 ( 2 x) sin 2 ( 2 x) cos 2 ( 2 x) = sec 2 ( 2 xSin 2x = 2 sin x cos x • Cosine cos 2x = cos2 x – sin2 x = 1 – 2 sin2 x = 2 cos2 x – 1 • Tangent tan 2x = 2 tan x/1 tan2 x = 2 cot x/ cot2 x 1 = 2/cot x – tan x tangent doubleangle identity can be accomplished by applying the same methods, instead use the sum identity for tangent, first • Note sin 2x ≠ 2 sin x;Answer (1 of 6) Verify the following identity sin(x)^2 cos(x)^2 tan(x)^2 = 1/cos(x)^2 Hint Eliminate the denominator on the right hand side Multiply both sides by cos(x)^2 cos(x)^2 (cos(x)^2 sin(x)^2 tan(x)^2) = ^?1 Hint Express the left hand side in terms of sine and cosin

3

3

Sin 2x cos 2x tan 2x

close