(cos^4θ-sin^4θ)/cos^2θ = 1-tan^2θ
Can someone show me step by step how to solve this please?
Trigonometric identity question help?
2020-10-04 1:21 am
回答 (3)
2020-10-04 1:47 am
(cos(θ))^4 - (sin(θ))^4 / (cos(θ))^2
[((cos(θ))^2)^2 - (sin(θ))^4] / (cos(θ))^2
[((1 - (sin(θ))^2))^2 - (sin(θ))^4] / (cos(θ))^2
[1 - 2(sin(θ))^2 + (sin(θ))^4 - (sin(θ))^4] / (cos(θ))^2
[1 - 2(sin(θ))^2] / (cos(θ))^2
[1 - (sin(θ))^2 - (sin(θ))^2] / (cos(θ))^2
[(cos(θ))^2 - (sin(θ))^2] / (cos(θ))^2
1 - (tan(θ))^2
[((cos(θ))^2)^2 - (sin(θ))^4] / (cos(θ))^2
[((1 - (sin(θ))^2))^2 - (sin(θ))^4] / (cos(θ))^2
[1 - 2(sin(θ))^2 + (sin(θ))^4 - (sin(θ))^4] / (cos(θ))^2
[1 - 2(sin(θ))^2] / (cos(θ))^2
[1 - (sin(θ))^2 - (sin(θ))^2] / (cos(θ))^2
[(cos(θ))^2 - (sin(θ))^2] / (cos(θ))^2
1 - (tan(θ))^2
2020-10-04 2:22 am
(c^4 - s^4)/c^2 = 1 - t^2 multiply both sides by c^2
(c^2 - s^2)(c^2 + s^2) = (c^2 - s^2)
That is true because c^2 + s^2 = 1
(c^2 - s^2)(c^2 + s^2) = (c^2 - s^2)
That is true because c^2 + s^2 = 1
2020-10-04 1:39 am
(cos^4 𝚹 - sin^4 𝚹) / cos^2 𝚹 =
(cos^2 𝚹 + sin^2 𝚹)(cos^2 𝚹 - sin^2 𝚹) / cos^2 𝚹 =
(cos^2 𝚹 - sin^2 𝚹) / cos^2 𝚹 =
1 - tan^2 𝚹
(cos^2 𝚹 + sin^2 𝚹)(cos^2 𝚹 - sin^2 𝚹) / cos^2 𝚹 =
(cos^2 𝚹 - sin^2 𝚹) / cos^2 𝚹 =
1 - tan^2 𝚹
收錄日期: 2021-04-24 08:04:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20201003172157AAiPQYe