![Hoe deze identiteit te bewijzen? sin ^ 2x + tan ^ 2x * sin ^ 2x = tan ^ 2x Hoe deze identiteit te bewijzen? sin ^ 2x + tan ^ 2x * sin ^ 2x = tan ^ 2x](https://img.go-homework.com/img/trigonometry/how-to-prove-this-identity-sin2xtan2x-sin2x-tan2x.png)
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Hoe te bewijzen (1 + sinx-cosx) / (1 + cosx + sinx) = tan (x / 2)?
![Hoe te bewijzen (1 + sinx-cosx) / (1 + cosx + sinx) = tan (x / 2)? Hoe te bewijzen (1 + sinx-cosx) / (1 + cosx + sinx) = tan (x / 2)?](https://img.go-homework.com/img/blank.jpg)
Zie onder. LHS = (1-cosx + sinx) / (1 + cosx + sinx) = (2sin ^ 2 (x / 2) + 2sin (x / 2) * cos (x / 2)) / (2cos ^ 2 (x / 2) + 2sin (x / 2) * cos (x / 2) = (2sin (x / 2) [sin (x / 2) + cos (x / 2)]) / (2cos (x / 2) * [ sin (x / 2) + cos (x / 2)]) = tan (x / 2) = RHS
Hoe verifieer je de identiteit sec ^ 4theta = 1 + 2tan ^ 2theta + tan ^ 4theta?
![Hoe verifieer je de identiteit sec ^ 4theta = 1 + 2tan ^ 2theta + tan ^ 4theta? Hoe verifieer je de identiteit sec ^ 4theta = 1 + 2tan ^ 2theta + tan ^ 4theta?](https://img.go-homework.com/calculus/how-do-you-verify-the-intermediate-value-theorem-over-the-interval-05-and-find-the-c-that-is-guaranteed-by-the-theorem-such-that-fc11-where-fxx2.jpg)
Bewijs hieronder Eerst zullen we bewijzen 1 + tan ^ 2theta = sec ^ 2theta: sin ^ 2theta + cos ^ 2theta = 1 sin ^ 2theta / cos ^ 2theta + cos ^ 2theta / cos ^ 2theta = 1 / cos ^ 2theta tan ^ 2theta + 1 = (1 / costheta) ^ 2 1 + tan ^ 2theta = sec ^ 2theta Nu kunnen we je vraag bewijzen: sec ^ 4theta = (sec ^ 2theta) ^ 2 = (1 + tan ^ 2theta) ^ 2 = 1 + 2tan ^ theta + tan ^ 4theta
Hoe verifieer je de identiteit 3sec ^ 2thetatan ^ 2theta + 1 = sec ^ 6theta-tan ^ 6theta?
![Hoe verifieer je de identiteit 3sec ^ 2thetatan ^ 2theta + 1 = sec ^ 6theta-tan ^ 6theta? Hoe verifieer je de identiteit 3sec ^ 2thetatan ^ 2theta + 1 = sec ^ 6theta-tan ^ 6theta?](https://img.go-homework.com/calculus/how-do-you-verify-the-intermediate-value-theorem-over-the-interval-05-and-find-the-c-that-is-guaranteed-by-the-theorem-such-that-fc11-where-fxx2.jpg)
Zie onder 3sec ^ 2thetatan ^ 2theta + 1 = sec ^ 6theta-tan ^ 6theta Rechterkant = sec ^ 6theta-tan ^ 6theta = (sec ^ 2theta) ^ 3- (tan ^ 2theta) ^ 3-> gebruik het verschil van twee kubussen formula = (sec ^ 2theta-tan ^ 2theta) (sec ^ 4theta + sec ^ 2thetatan ^ 2theta + tan ^ 4theta) = 1 * (sec ^ 4theta + sec ^ 2thetatan ^ 2theta + tan ^ 4theta) = sec ^ 4theta + sec ^ 2thetatan ^ 2theta + tan ^ 4theta = sec ^ 2theta sec ^ 2 theta + sec ^ 2thetatan ^ 2theta + tan ^ 2theta tan ^ 2 theta = sec ^ 2theta (tan ^ 2theta + 1) + sec ^ 2thetatan ^ 2theta + tan ^ 2theta (sec ^ 2theta-1) = sec ^ 2thetatan ^ 2theta + sec ^ 2theta +